Limits

Section 2.1 Limits

A limit describes the behavior of a function as its input (or variable) approaches a specific value or infinity. It is a fundamental concept used to define continuity, derivatives, and integrals. The limit of a function \(f(x)\) as \(x\) approaches a value \(a\) is denoted as:

\begin{equation*} \lim_{x \to a} f(x) = L \end{equation*}

This means that as \(x\) gets arbitrarily close to \(a\text{,}\) \(f(x)\) gets arbitrarily close to \(L\text{,}\) provided the limit exists.

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Example 2.1.1.

Let’s say we have a function like \(f(x) =\frac{x^2}{x}\text{,}\) can I write this function as \(f(x) =x\text{,}\) the answer is no, I can not. Okay check if these simplification are correct:

\begin{equation*} y =\frac{5\times 2}{2} = 5\text{.} \end{equation*}

How about

\begin{equation*} y =\frac{5\times 0}{0} = 5\text{.} \end{equation*}

Good, so you know unless \(x\) is non-zero numbers we can not cancel them out. Meaning

\begin{equation*} f(x) =\frac{x^2}{x} \neq x \end{equation*}

because at \(x=0, \) our \(f(x)=f(0)\) is not defined. However, we can find the value of function just near to \(x=0.\) we call them \(x\to 0^+\) if approacing from right and \(x\to 0^-\) if approaching from left. That is \(f(x)\) can be approximated to some finite value as \(x \to 0^+\) or \(x \to 0^-\text{.}\)

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\begin{equation*} f(x) =\frac{x^2}{x} =\begin{cases} x; & \, \text { if } \, x \neq 0 \\ \text{ undefined }; & \, \text { if } \, x = 0 \end{cases} \end{equation*}

Here \(x=0\) makes a hole in the graph of \(f(x)\) which is a straight line passing through origin.

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Figure 2.1.2. Limits and Derivatives

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Example 2.1.3.

\begin{equation*} \lim_{x\to 1}{(x^2-4)} \end{equation*}

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Solution.

Direct substitution gives:

\begin{equation*} 1^2 - 4 = -3 \end{equation*}

Thus,

\begin{equation*} \lim_{x \to 1} (x^2 - 4) = -3\text{.} \end{equation*}

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Example 2.1.4.

\begin{equation*} \lim_{x\to -3}{\frac{2x+5}{x+3}} \end{equation*}

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Solution.

Direct substitution gives:

\begin{equation*} \frac{2(-3)+5}{-3+3} = \frac{-6+5}{0} = \frac{-1}{0} \end{equation*}

Since division by zero is undefined, the limit does not exist in the finite sense. However, we can analyze the behavior as \(x\) approaches \(-3\) from the left and right to determine if it approaches infinity or negative infinity.

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Example 2.1.5.

\begin{equation*} \lim_{x\to 5}{\frac{x^2-7x+10}{x-5}} \end{equation*}

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Solution.

Direct substitution gives:

\begin{equation*} \frac{5^2 - 7 \cdot 5 + 10}{5 - 5} = \frac{25 - 35 + 10}{0} = \frac{0}{0} \end{equation*}

This is an indeterminate form, so we simplify the function first:

\begin{equation*} \lim_{x \to 5} \frac{(x-5)(x-2)}{x-5} = \lim_{x \to 5} (x-2) \end{equation*}

Since \((x\neq 5)\text{,}\) we have:

\begin{equation*} \lim_{x \to 5} (x-2) = 5 - 2 = 3 \end{equation*}

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Example 2.1.6.

\begin{equation*} \lim_{x\to 9}{\frac{(x-9)}{(\sqrt{x}-3)}} \end{equation*}

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Solution.

Direct substitution gives:

\begin{equation*} \frac{9-9}{\sqrt{9}-3} = \frac{0}{3-3} = \frac{0}{0} \end{equation*}

This is an indeterminate form, so we simplify the function first by rationalizing the denominator:

\begin{equation*} \lim_{x \to 9} \frac{(x-9)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)} = \lim_{x \to 9} \frac{(x-9)(\sqrt{x}+3)}{x-9} \end{equation*}

Since \((x\neq 9)\text{,}\) we have:

\begin{equation*} \lim_{x \to 9} (\sqrt{x}+3) = \sqrt{9} + 3 = 3 + 3 = 6 \end{equation*}

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Example 2.1.7.

\begin{equation*} \lim_{x\to 0}{\frac{\sin x}{\sin 2x}} \end{equation*}

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Solution.

Direct substitution gives:

\begin{equation*} \frac{\sin 0}{\sin 0} = \frac{0}{0} \end{equation*}

This is an indeterminate form, so we simplify the function first using the small-angle approximation:

\begin{equation*} \lim_{x \to 0} \frac{\sin x}{\sin 2x} = \lim_{x \to 0} \frac{x}{2x} = \lim_{x \to 0} \frac{1}{2} = \frac{1}{2} \end{equation*}

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Example 2.1.8.

\begin{equation*} \lim_{x\to \infty}{\frac{5x^2+8x-3}{3x^2+2}} \end{equation*}

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Solution.

To evaluate this limit as \(x\) approaches infinity, we divide the numerator and denominator by \(x^2\text{,}\) the highest power of \(x\) in the denominator:

\begin{equation*} \lim_{x \to \infty} \frac{5 + \frac{8}{x} - \frac{3}{x^2}}{3 + \frac{2}{x^2}} = \frac{5 + 0 - 0}{3 + 0} = \frac{5}{3} \end{equation*}

Thus,

\begin{equation*} \lim_{x \to \infty} \frac{5x^2+8x-3}{3x^2+2} = \frac{5}{3}\text{.} \end{equation*}

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Subsection 2.1.1 One-Sided Limits

Left-hand limit: \(\lim_{x \to a^-} f(x)\) considers values of \(x\) approaching \(a\) from the left (smaller values).
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Right-hand limit: \(\lim_{x \to a^+} f(x)\) considers values of \(x\) approaching \(a\) from the right (larger values).
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For \(\lim_{x \to a} f(x)\) to exist, the left-hand and right-hand limits must equal each other.
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Limits at Infinity.
These describe the behavior of \(f(x)\) as \(x\) approaches \(\infty\) or \(-\infty\text{.}\) Denoted as \(\lim_{x \to \infty} f(x)\) or \(\lim_{x \to -\infty} f(x)\text{.}\)
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Non-Existent Limits.
A limit does not exist if the function’s values do not approach a single finite number (e.g., due to oscillation or undefined behavior).
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Indeterminate Forms.
Forms like [Subsection 2.1.6] are called indeterminate forms, such as \(\frac{0}{0}\text{,}\) \(\frac{\infty}{\infty}\) require special techniques to evaluate as shown below
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- Factorization Method:
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- Rationalization Method (involve square root):
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- L Hospital’s Rule: If the indeterminate form of functions are either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\text{,}\) then we can use this rule. Here we can find the derivative of the numerator, \(f^{\prime }(x),\) and the derivative of the denominator, \(g^{\prime }(x)\text{.}\)
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  Re-evaluate the Limit: Find the limit of the new fraction, \(f^{\prime }(x)/g^{\prime }(x)\text{.}\) Repeat if Necessary: If you still get an indeterminate form, repeat the process (differentiate again) until the limit is solvable.
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Subsection 2.1.2 Definition of Limit

The two small positive numbers \(\epsilon \) and \(\delta\) formalizes the idea of a limit. For

\begin{equation*} \lim_{x\to a} f(x) = L \end{equation*}

implies that for every \(\epsilon \gt 0,\) there exists a \(\delta \gt 0 \) such that if

\begin{equation*} 0\lt |x-a| \lt \delta, \, \text{ then } \, |f(x) -L| \lt \epsilon. \end{equation*}

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Here, we do not really determine the value of \(\epsilon\text{,}\) rather it is given arbitrarily to show us how close \(f(x)\) gets to \(L.\) There is no unique \(\delta \) rather it satisfies the given \(\epsilon \gt 0.\)

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Definition 2.1.9.

\begin{equation*} \lim_{x \to a} f(x) = L \end{equation*}

for every \(\epsilon \gt 0\text{,}\) there exists a \(\delta \gt 0 \) such that if \(0\lt |x-a| \lt \delta \) then \(|f(x)-L| \lt \epsilon \)

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To determine \(\delta \) in terms of \(epsilon\text{,}\) the following steps are taken:

Step I: Start with the target inequality \(|f(x) -L| \lt \epsilon.\)
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Step II: Express it in terms of \(|x-a| \lt \delta\) by assuming \(\delta \lt 1.\)
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Example 2.1.10.

Prove that

\begin{equation*} \lim_{x \to 3} (4x-1)=11 \end{equation*}

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Solution.

\begin{align*} |x-a| \lt \delta; \amp \text{ then } \, |f(x) -L| \lt \epsilon \\ \text{ here } |x-3| \lt \delta; \, \amp \, \text{ hence } \, |(4x-1)-11| \lt \epsilon \\ \text{ or } \, \amp \, |4x-12| \lt \epsilon \\ \text{ or } \, \amp \, 4|x-3| \lt \epsilon \\ but \, \amp |x-3| \lt \delta \\ \therefore \, \amp 4|x-3| \lt 4\delta = \epsilon \\ \text{ Hence, } \, \delta \amp = \frac{\epsilon}{4} \end{align*}

Therefore

\begin{equation*} \lim_{x \to 3} (4x-1)=11 \, \text{ is proved. } \end{equation*}

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Example 2.1.11.

Prove that

\begin{equation*} \lim_{x\to 2} \left(\frac{x^2-4}{x-2} \right) =4 \end{equation*}

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Solution.

\begin{align*} \text{ if } \, |x-a| \lt \delta; \amp \text{ then } \, |f(x) -L| \lt \epsilon \\ here, \, |x-2| \amp \lt \delta \\ Now, \, \bigg|\frac{x^2-4}{x-2} -4 \bigg| \amp \lt \epsilon \\ or, \, \bigg| x+2 -4 \bigg| \amp \lt \epsilon \\ or, \, \bigg| x-2 \bigg| \amp \lt \epsilon \\ but, \, \bigg| x-2 \bigg| \amp \lt \delta \\ \therefore \, \delta \amp = \epsilon \end{align*}

Hence, \(\lim_{x\to 2} \left(\frac{x^2-4}{x-2} \right) =4\) is proved.

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Example 2.1.12.

Prove

\begin{equation*} \lim_{x \to 2} x^2=4 \end{equation*}

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Solution.

\begin{align*} |x-a| \lt \delta; \amp \text{ then } \, |f(x) -L| \lt \epsilon \\ \text{ or } |x-2| \lt \delta; \, \amp |x^2-4| \lt \epsilon \\ \text{ or } \, \amp |(x-2)(x+2)| \lt \epsilon \\ \text{ or } \, \amp |(x-2)| |(x+2)| \lt \epsilon \\ \text{ but } \, \amp |x-2| \lt \delta \\ \text{ Hence, } \, \delta \amp = \frac{\epsilon}{|x+2|} \end{align*}

Remember

\begin{equation*} \delta \lt 1 \end{equation*}

and also in \(|(x-2)| |(x+2)| \lt \epsilon\) if \(|(x-2)|\) is decreasing then \(|(x+2)|\) starts increasing accordingly.

\begin{equation*} Now, \, |x-2| \lt 1 \end{equation*}

\begin{equation*} Hence, \, -1 \lt (x-2) \lt 1 \end{equation*}

Note: \(|x-2| = \begin{cases} (x-2); & (x-2) \geq 0 \\ -(x-2); & (x-2) \lt 0 \end{cases}\)

\begin{equation*} or, \, 1 \lt x \lt 3 \end{equation*}

\begin{equation*} i.e., \, 3 \lt |x+2| \lt 5 \end{equation*}

\begin{equation*} or, \, \frac{1}{5} \lt \frac{1}{|x+2|} \lt \frac{1}{3} \end{equation*}

\begin{equation*} or, \, \frac{\epsilon}{5} \lt \frac{\epsilon}{|x+2|} \lt \frac{\epsilon}{3} \end{equation*}

we can take any value \(\frac{1}{|x+2|}\) but since we are interested in smaller value of \(\delta\) we can choose \(\frac{1}{5}.\)

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Meaning choose \(\delta\) between \(1\) and \(\frac{\epsilon}{5}.\) Therefore, \(\delta = min\{1,\frac{\epsilon}{5}\}. \)

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Hence,

\begin{equation*} \lim_{x \to 2} x^2=4 \, \text{ is proved. } \end{equation*}

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Example 2.1.13.

How close to \(2\) do we have to take \(x\) so that \(4x-5\) is within a distance of (a) \(0.1\) and (b) \(0.01\) from \(3\text{?}\)

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Solution.

\begin{equation*} |x-2| \lt \delta \end{equation*}

then

\begin{equation*} |f(x)-L|\lt \epsilon \end{equation*}

\begin{equation*} or, \, |(4x-5)-3|\lt \epsilon \end{equation*}

\begin{equation*} or, \, |4x-8|\lt \epsilon \end{equation*}

\begin{equation*} or, \, |x-2|\lt \frac{\epsilon}{4} \end{equation*}

\begin{equation*} \therefore \, \delta = \frac{\epsilon}{4} \end{equation*}

Now (a) \(\epsilon = 0.1.\)

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Therefore \(\delta = \frac{0.1}{4} =\frac{1}{40} = 0.025.\)

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Checkpoint 2.1.14.

Prove that

\begin{equation*} \lim_{x \to 2} \frac{x}{7}=\frac{2}{7} \end{equation*}

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Checkpoint 2.1.15.

Prove that

\begin{equation*} \lim_{x \to 2} (x^2+2x-7) =1 \end{equation*}

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Checkpoint 2.1.16.

Prove

\begin{equation*} \lim_{x \to 3} x^2=9 \end{equation*}

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Checkpoint 2.1.17.

Prove that

\begin{equation*} \lim_{x \to -5^+} \sqrt[5]{5+x}=0 \end{equation*}

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Definition 2.1.18.

For

\begin{equation*} \lim_{x \to a} f(x) =\infty \end{equation*}

If \(0\lt |x-a| \lt \delta \) then \(f(x)\gt M\) where \(M\gt 0.\)

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Definition 2.1.19.

For

\begin{equation*} \lim_{x \to a} f(x) =-\infty \end{equation*}

If \(0\lt |x-a| \lt \delta \) then \(f(x)\lt N\) where \(N\lt 0.\)

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Subsection 2.1.3 Evaluate Limit

To evaluate limit of a function always try with direct substitution method first. If you get indeterminate forms then only look for other method to simply the function before apply the limit.

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Example 2.1.20.

Basic Polynomial Limit: Consider the function

\begin{equation*} f(x) = x^2 + 2x + 1\text{.} \end{equation*}

Find

\begin{equation*} \lim_{x \to 2} f(x)\text{.} \end{equation*}

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Solution.

Substitute \(x = 2\) directly since the function is a polynomial (continuous everywhere):

\begin{equation*} f(2) = 2^2 + 2 \cdot 2 + 1 = 4 + 4 + 1 = 9 \end{equation*}

Thus,

\begin{equation*} \lim_{x \to 2} (x^2 + 2x + 1) = 9\text{.} \end{equation*}

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Polynomials are continuous, so the limit at any point is simply the function’s value at that point.

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Example 2.1.21.

Rational Function with a Hole: Find

\begin{equation*} \lim_{x \to 1} \frac{x^2 - 1}{x - 1}\text{.} \end{equation*}

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Solution.

Direct substitution gives:

\begin{equation*} \frac{1^2-1 }{1 - 1} = \frac{0 }{0} \end{equation*}

This is an indeterminate form, so simplify the function.

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Simplify the function using all possible mathematical means:

\begin{equation*} \lim_{x \to 1}\frac{x^2 - 1}{x - 1} = \lim_{x \to 1}\frac{(x - 1)(x+1)}{x - 1} = \lim_{x \to 1}(x + 1). \end{equation*}

Since, \((x\neq 1)\) we have,

\begin{equation*} \lim_{x \to 1}(x + 1) = 1+1 =2. \end{equation*}

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Subsection 2.1.4 One-Sided Limits

The limit of a function \(f(x)\) as \(x\) approaches a value \(a\) from the left side is called the left-hand limit (LHL) and is denoted as

\begin{equation*} \lim_{x \to a^-} f(x). \end{equation*}

Similarly, the limit of \(f(x)\) as \(x\) approaches \(a\) from the right side is called the right-hand limit (RHL) and is denoted as

\begin{equation*} \lim_{x \to a^+} f(x). \end{equation*}

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The limiting values (LHL, RHL) and functional value could be different or could be same for a given function as can be described below:

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Left hand limit:

\begin{equation*} \lim_{x \to a^-}f(x) = n. \end{equation*}

Right hand limit:

\begin{equation*} \lim_{x \to a^+}f(x) = m. \end{equation*}

and the function at \(a\)

\begin{equation*} f(a)=n \end{equation*}

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The filled circle represents the functional value of the given function and open circle represents its limiting value.

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Left hand limit:

\begin{equation*} \lim_{x \to a^-}f(x) = n. \end{equation*}

Right hand limit:

\begin{equation*} \lim_{x \to a^+}f(x) = m. \end{equation*}

and the function at \(a\)

\begin{equation*} f(a)= p \end{equation*}

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Left hand limit:

\begin{equation*} \lim_{x \to a^-}f(x) = \lim_{h \to 0}f(a-h) = m. \end{equation*}

Right hand limit:

\begin{equation*} \lim_{x \to a^+}f(x) = \lim_{h \to 0}f(a+h) = n. \end{equation*}

and the function at \(a\)

\begin{equation*} \lim_{x \to a}f(x)= m \end{equation*}

Note: the small increment, \(h\) is always positive number.

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\begin{equation*} \text{ If } \, \lim_{x \to a^+}f(x) = \lim_{x \to a^-}f(x) = L \text{ (any number)}, \end{equation*}

then, we say the limit exist for the function at \(x\to a\text{.}\) Here, the limiting value of function is

\begin{equation*} LHL\vert_{a}= RHL\vert_{a} = m \end{equation*}

and the functional value of \(f(x) \) at \(a\) is

\begin{equation*} f(a)=n \end{equation*}

For the limit to exist,

\begin{equation*} \lim_{x \to a^+}f(x) = \lim_{x \to a^-}f(x) = finite. \end{equation*}

The functional value \(f(a)\) may or may not be the same as the limiting value.

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\begin{equation*} \text{ If } \, \lim_{x \to a^+}f(x) = \lim_{x \to a^-}f(x) = m, \end{equation*}

and

\begin{equation*} \lim_{x \to a}f(x) = f(a)= m \end{equation*}

i.e., the limiting value is same as the functional value, then also the limit of the function exist and function is called a continuous function.

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The filled circle represents the functional value of the given function and open circle represents its limiting value.

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one sided limit

\begin{equation*} \text{ For } \, \lim_{x \to 0^+}f(x) =\sqrt{x} \end{equation*}

is an example of one sided limit where the function is defined only for \(x \geq 0.\)

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The left-hand limit \(\lim_{x \to 0^-}f(x)\) does not exist as \(f(x)\) is not defined for \(x \lt 0.\)

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Hence, the limit \(\lim_{x \to 0^+}f(x)\) does exist only for \(x \geq 0\) and the function is continuous for \(x \geq 0.\text{.}\) Left side limit is ignored here.

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Point functions are continuous.

\begin{equation*} f(x) = \sqrt{1-x} +\sqrt{x-1} \end{equation*}

Here, the function is defined only at \(x=1\) and undefined elsewhere. Hence, both LHL and RHL do not exist still the limit exist at \(x=1\) and function is continuous at that point.

\begin{equation*} \lim_{x \to 1}f(x) =0 \end{equation*}

\begin{equation*} \lim_{x \to 1^-}f(x) = \text{ not defined. } \end{equation*}

\begin{equation*} \lim_{x \to 1^+}f(x) = \text{ not defined. } \end{equation*}

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Example 2.1.23.

Evaluate LHL and RHL of the given function.

\begin{equation*} f(x) = \begin{cases} \frac{|x-4|}{x-4}; & x\neq 4 \\ 0; & x=4 \end{cases} \,\quad \text {at } \, x=4. \end{equation*}

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Solution.

LHL,

\begin{align*} \lim_{x \to a^-}f(x) \amp = \lim_{x \to 4^-}\frac{|x-4|}{x-4} \\ = \lim_{h \to 0}f(a-h) \amp = \lim_{h \to 0} \frac{|(4-h)-4|}{(4-h)-4} =\lim_{h \to 0} \frac{|-h|}{-h} = \lim_{h \to 0} \frac{h}{-h} =-1 \end{align*}

RHL,

\begin{align*} \lim_{x \to a^+}f(x) \amp = \lim_{x \to 4^+}\frac{|x-4|}{x-4} \\ = \lim_{h \to 0}f(a+h) \amp = \lim_{h \to 0} \frac{|(4+h)-4|}{(4+h)-4} =\lim_{h \to 0} \frac{|h|}{h} = \lim_{h \to 0} \frac{h}{h} =1 \end{align*}

Since LHL \(\neq\) RHL, limit does not exist.

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Example 2.1.24.

Evaluate LHL and RHL of the given function.

\begin{equation*} f(x) = \begin{cases} 1+x^2; & \text{ if } \, 0 \leq x \leq 1 \\ 2-x; & \text{ if }\, x \gt 1 \end{cases} \,\quad \text {at } \, x=1. \end{equation*}

Also, show that \(\lim_{x\to1}f(x) \) does not exist.

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Solution.

LHL,

\begin{align*} \lim_{x \to a^-}f(x) \amp = \lim_{x \to 1^-} (1+x^2) \\ = \lim_{h \to 0}f(a-h) \amp = \lim_{h \to 0} [1+(1-h)^2] =\lim_{h \to 0} (2-2h+h^2) = 2. \end{align*}

RHL,

\begin{align*} \lim_{x \to a^+}f(x) \amp = \lim_{x \to 1^+}(2-x) \\ = \lim_{h \to 0}f(a+h) \amp = \lim_{h \to 0} [2-(1+h)] =\lim_{h \to 0} (1-h) = 1 \end{align*}

Again,

\begin{equation*} \lim_{x\to1}f(x) = \lim_{x\to1}(1+x^2) = 1+1=2 \end{equation*}

Since, \(\lim_{x \to 1^-}f(x) \neq \lim_{x \to 1^+} f(x) \neq \lim_{x \to 1} f(x)\text{,}\) limit does not exist.

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Checkpoint 2.1.25.

\begin{equation*} If\, \quad f(x) = \begin{cases} 5x-4; & 0 \lt x \leq 1 \\ 4x^3-3x; & 1 \lt x \lt 2 \end{cases} \end{equation*}

show that \(\lim_{x \to 1} f(x)\) exist.

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Checkpoint 2.1.26.

\begin{equation*} If\, \quad f(x) = \begin{cases} \frac{x-|x|}{x}; & x\neq 0 \\ 2; & x=0 \end{cases} \end{equation*}

show that \(\lim_{x \to 0} f(x)\) does not exist.

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Example 2.1.27.

Suppose

\begin{equation*} f(x) = \begin{cases} a+bx; & x\lt 1 \\ 4; & x=1 \\ b-ax; & x \gt 1 \end{cases} \, \text{and if } \, \lim_{x \to 1}f(x) =f(1) \end{equation*}

What are possible values of \(a\) and \(b?\)

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Solution.

Do your self.

Answer.

\(a=0,\, b=4.\)

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Example 2.1.28.

Evaluate:

\begin{equation*} \lim_{x \to 1^+} \lfloor x-3 \rfloor \end{equation*}

where \(\lfloor \,. \rfloor\) is GIF [Subsection 1.4.3].

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Answer.

\begin{equation*} \lim_{x \to 1^+} \lfloor x \rfloor -3 = (x-3) \end{equation*}

or,

\begin{equation*} \lim_{x \to 1^+} (x-3) = 1-3 =-2. \end{equation*}

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\(\lfloor x \rfloor\) is from right of \(1\text{,}\) e.g., \(\lfloor 1.1 \rfloor\) =\(\lfloor 1.99 \rfloor\) =\(\lfloor 1 \rfloor\text{.}\) Remeber, GIF (\(\lfloor \,. \rfloor\)) always gives lower integer value.

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Example 2.1.29.

Evaluate:

\begin{equation*} \lim_{x \to 5^+} \left(\frac{x-5}{x+5}\right) \end{equation*}

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Answer.

\begin{equation*} \lim_{x \to 5^+} \left(\frac{x-5}{x+5}\right) \end{equation*}

or,

\begin{equation*} \lim_{h \to 0} \left( \frac{(5+h)-5}{(5+h)+5} \right) \end{equation*}

or,

\begin{equation*} \lim_{h \to 0} \left(\frac{h}{10+h}\right) = 0 \end{equation*}

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Example 2.1.30.

Evaluate:

\begin{equation*} \lim_{x \to 3^+} \frac{\lfloor x \rfloor -3}{x-3} \end{equation*}

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Answer.

\begin{equation*} \lim_{x \to 3^+} \frac{\lfloor x \rfloor -3}{x-3} \end{equation*}

or,

\begin{equation*} \lim_{h \to 0} \frac{\lfloor (3+h) \rfloor -3}{(3+h)-3} \end{equation*}

\(\because \lfloor (3+h) \rfloor =3\text{.}\) or,

\begin{equation*} \lim_{h \to 0} \left(\frac{3-3}{h}\right) = 0. \end{equation*}

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Example 2.1.31.

Evaluate:

\begin{equation*} \lim_{x \to 4^+} \{ x \} \end{equation*}

where, \(\{ x \} \) is FPF [Subsection 1.4.3].

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Answer.

\begin{equation*} \lim_{x \to 4^+} \{ x \} \end{equation*}

or,

\begin{equation*} \lim_{x \to 4^+} x- \lfloor x \rfloor = 4-4 =0. \end{equation*}

\(\because \{ x \} = x - \lfloor x \rfloor \text{.}\)

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Example 2.1.32.

Evaluate:

\begin{equation*} \lim_{x \to 4^-} \{ x \} \end{equation*}

where, \(\{ x \} \) is FPF [Subsection 1.4.3].

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Answer.

\begin{equation*} \lim_{x \to 4^-} \{ x \} \end{equation*}

or,

\begin{equation*} \lim_{x \to 4^-} x- \lfloor x \rfloor = 4-3 =1. \end{equation*}

\(\because \{ x \} = x - \lfloor x \rfloor \) and \(\lfloor 3.99 \rfloor\) = \(\lfloor 3.1 \rfloor\) = \(3\text{.}\)

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Example 2.1.33.

Evaluate \(f(x)\)

\begin{equation*} f(x) = \begin{cases} x^2-4 & \text{ if } x \lt 2 \\ 3x-2 & \text{ if } x \geq 2 \\ \end{cases} \end{equation*}

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Answer.

\begin{equation*} \lim_{x \to 1^-} f(x)=1+x^2 = 1+1=2. \end{equation*}

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\begin{equation*} \lim_{x \to 1^+} f(x)=2-x = 2-1 =1. \end{equation*}

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Checkpoint 2.1.34.

Evaluate \(f(x)\)

\begin{equation*} f(x) = \begin{cases} 2x+1 & \text{ if } x \lt 1 \\ x^2-2x+3 & \text{ if } x \geq 1 \\ \end{cases} \end{equation*}

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Checkpoint 2.1.35.

Evaluate \(f(x)\)

\begin{equation*} f(x) = \begin{cases} x^2+3x & \text{ if } x \lt 0 \\ 3x+4 & \text{ if } x \geq 0 \\ \end{cases} \end{equation*}

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Checkpoint 2.1.36.

Evaluate \(f(x)\)

\begin{equation*} f(x) = \begin{cases} x^2+3x & \text{ if } x \lt 0 \\ 3x & \text{ if } x \geq 0 \\ \end{cases} \end{equation*}

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Example 2.1.37.

For what value of \(m\) does the \(\lim_{x \to 2} f(x) \) exist if

\begin{equation*} f(x) = \begin{cases} mx-3; & x \lt 2\\ \frac{x}{m}; & x \geq 2 \\ \end{cases} \end{equation*}

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Answer.

\begin{equation} \lim_{x \to 2^-} f(x) = mx -3 = 2m-3\tag{2.1.1} \end{equation}

and

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\begin{equation} \lim_{x \to 2^+} f(x) = \frac{2}{m}\tag{2.1.2} \end{equation}

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equating these two equations, we get -

\begin{align*} 2m-3 \amp =\frac{2}{m} \\ or, \, 2m^2 -3m -2 \amp =0 \\ or, \, m \amp = \frac{1}{2}, 2. \end{align*}

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Checkpoint 2.1.38.

Evaluate \(f(x)\)

\begin{equation*} f(x) = \begin{cases} 2x+5 & \text{ if } x \lt -1 \\ -x+3 & \text{ if } -1 \leq x \lt 2 \\ x^2-4x+7 & \text{ if } x \geq 2 \\ \end{cases} \end{equation*}

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Checkpoint 2.1.39.

Evaluate \(f(x)\)

\begin{equation*} f(x) = \begin{cases} x+6 & \text{ if } x \leq -3 \\ -2x-3 & \text{ if } -3 \lt x \leq 1 \\ (x-1)^2+2 & \text{ if } x \gt 1 \\ \end{cases} \end{equation*}

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Subsection 2.1.5 Limit at Infinity

The limit of a function \(f(x)\) as \(x\) approaches infinity (denoted \(\lim_{x \to \infty} f(x)\)) describes the behavior of \(f(x)\) as \(x\) grows without bound. It equals \(L\) (a real number,\(\infty, \) or \(-\infty\)) if \(f(x)\) gets arbitrarily close to \(L\) for sufficiently large \(x\text{.}\)

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Remark 2.1.40.

On solving a limit at \(x \to \infty\text{:}\)

Take common and try to eliminate the biggest terms from numerator and denominator.
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Evaluating the limit, when \(x \to -\infty\text{,}\) try substituting \(x=-t\) and take \(x \to \infty\text{.}\)
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If \(p(x)\) and \(q(x)\) are two polynomials of degree \(m\) and \(n\) respectively, then try:

\begin{equation*} \lim_{x \to \infty} \frac{p(x)}{q(x)} = \begin{cases} \frac{a_o}{b_o}; & \text{ if } m=n \\ 0; & \text{ if } m \lt n \\ \infty; & \text{ if } m \gt n & \frac{a_o}{b_o} \gt 0 \\ -\infty; & \text{ if } m \gt n & \frac{a_o}{b_o} \lt 0 \\ \end{cases} \end{equation*}

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If numerator and denominator or both contain square roots then rationalize them.
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Example 2.1.41.

Evaluate the following.

\begin{equation*} \lim_{x \to \infty} \frac{2x^3-3x^2+1}{4x^3+5x+3} \end{equation*}

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\begin{equation*} \lim_{x \to \infty} \frac{3x^3+2x-5}{5x^3+4x-2} \end{equation*}

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\begin{equation*} \lim_{x \to \infty} \frac{3x^5+5x^3+2}{2x^3+x-1} \end{equation*}

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\begin{equation*} \lim_{x \to \infty} \frac{2x^3-3x^2+1}{4x^3+5x+3} \end{equation*}

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\begin{equation*} \lim_{x \to \infty} \frac{-2x^4+5x^2+x+1}{5x^3+x-2} \end{equation*}

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Solution.

\begin{align*} \lim_{x \to \infty} \frac{2x^3-3x^2+1}{4x^3+5x+3} \amp = \lim_{x \to \infty} \frac{\cancel{x^3}\left(2-3/x+1/x^2\right)}{\cancel{x^3}\left(4+5/x^2+3/x^3\right)}\\ \amp = \lim_{x \to \infty} \frac{2-0+0}{4+0+0} =\frac{2}{4} = \frac{1}{2}. \end{align*}

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Example 2.1.42.

\begin{equation*} \lim_{n \to \infty} \frac{4^n-5^n}{4^n+5^n} \end{equation*}

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\begin{equation*} \lim_{n \to \infty} \left(4^n+6^n\right)^{(1/n)} \end{equation*}

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Solution.

\begin{equation*} \lim_{n \to \infty} \frac{4^n-5^n}{4^n+5^n} = \lim_{n \to \infty} \frac{\cancel{5^n}\left((4/5)^n-1\right)}{\cancel{5^n}\left((4/5)^n+1\right)} = \frac{-1}{1} \end{equation*}

\(\because \,(4/5)^\infty \to 0 \) as \(4/5 \lt 1.\)

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\begin{equation*} \lim_{n \to \infty} \left(4^n+6^n\right)^{(1/n)} \end{equation*}

\begin{equation*} = \lim_{n \to \infty} \{6^n[(\frac{4}{6})^n+1]\}^{(1/n)} \end{equation*}

\begin{equation*} = \lim_{n \to \infty} 6\{(\frac{4}{6})^n+1\}^{(1/n)} = 6(0+1)^{1/\infty} = 6\times 1^0 =6. \end{equation*}

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Example 2.1.43.

Evaluate:

\begin{equation*} \lim_{n \to \infty}\{\frac{n^2}{1+2+3+ \cdots +n}\} \end{equation*}

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Solution.

\begin{equation*} \lim_{n \to \infty}\{\frac{n^2}{\frac{n(n+1)}{2}}\} = \lim_{n \to \infty}\frac{2n^2}{n^2(1+1/n)} = 2/1 =2. \end{equation*}

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Subsection 2.1.6 Indeterminate Form

When evaluating limits of the form \(\lim_{x \to a} f(x)\text{,}\) direct substitution of \(x = a\) sometimes yields an indeterminate form. It is an expression that doesn’t immediately reveal the limit’s value (or existence). These forms require algebraic manipulation, L’Hôpital’s rule, or other techniques to resolve.

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Common Indeterminate Forms

\(\frac{0}{0}\text{:}\) Numerator and denominator both approach 0.
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\(\frac{\infty}{\infty}\) (or \(-\infty/\infty\text{,}\) etc.): Both approach infinity in absolute value.
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\(0 \cdot \infty\text{:}\) One factor approaches 0, the other \(\infty\text{.}\)
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\(\infty - \infty\text{:}\) Difference of two terms approaching infinity.
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\(1^\infty\text{:}\) Base approaches 1, exponent approaches \(\infty\text{.}\)
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\(0^0\text{:}\) Base approaches 0, exponent approaches 0.
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\(\infty^0\text{:}\) Base approaches \(\infty\text{,}\) exponent approaches 0.
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Note: \(\frac{0}{\infty} = 0\) and \(\frac{\infty}{0}\) is undefined (or \(\pm \infty\) if considering one-sided) these are not ndeterminate.

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Techniques to Resolve Indeterminate Forms

Algebraic Simplification Factor, rationalize, or use trig identities.
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L’Hôpital’s Rule (for \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\))
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Series Expansions or Known Limits
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Rewrite Other Forms
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Example 2.1.44.

\begin{equation*} \lim_{x \to 2} \, \frac{x^3-2x-4}{x^2-3x+2} \end{equation*}

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\begin{equation*} \lim_{t \to 1} \, \frac{t^3-1}{t^6-1} \end{equation*}

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\begin{equation*} \lim_{x \to 1} \, \frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{(x-1)^2} \end{equation*}

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Solution.

\begin{equation*} \lim_{x \to 2} \, \frac{x^3-2x-4}{x^2-3x+2} \end{equation*}

Numerator: \(2^3 - 2(2) - 4 = 8 - 4 - 4 = 0\text{,}\)

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Denominator: \(2^2 - 3(2) + 2 = 4 - 6 + 2 = 0\text{.}\)
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This is the indeterminate form \(\frac{0}{0}\text{.}\) Factor both polynomials.
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Denominator: \(x^2 - 3x + 2 = (x - 1)(x - 2)\text{.}\)
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Numerator: \(x^3 - 2x - 4\text{.}\)
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Use the factor theorem: test \(x = 2\text{:}\) \(2^3 - 2(2) - 4 = 8 - 4 - 4 = 0\text{,}\) so \(x - 2\) is a factor.
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Now, Numerator: \(x^3 - 2x - 4 = (x-2)(x^2+2x+2)\text{.}\)

\begin{equation*} \therefore \, \lim_{x \to 2} \, \frac{x^3-2x-4}{x^2-3x+2} = \lim_{x \to 2} \, \frac{\cancel{(x-2)}(x^2+2x+2)}{(x - 1)\cancel{(x - 2)}} \end{equation*}

\begin{equation*} or, \, \lim_{x \to 2} \, \frac{(x^2+2x+2)}{(x - 1)} = \frac{(2^2+2*2+2)}{(2 - 1)} = \frac{10)}{1} =10. \end{equation*}

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\begin{equation*} \lim_{t \to 1} \, \frac{t^3-1}{t^6-1} = \lim_{t \to 1} \, \frac{\cancel{(t-1)}(t^2 +t+1)}{\cancel{(t-1)}(t^5+t^4+t^3+t^2+t+1)} = 3/6 =1/2. \end{equation*}

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\begin{equation*} \lim_{x \to 1} \, \frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{(x-1)^2} \end{equation*}

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Let, \(\sqrt[3]{x} = x^{1/3} = t\) or, \(t^3 = x\) Hence, as \(x \to 1,\) we have \(t \to 1.\) Therefore,

\begin{equation*} \lim_{t \to 1} \, \frac{t^2-2t+1}{(t^3-1)^2} = \lim_{t \to 1} \, \frac{(t-1)^2}{(t-1)^2(t^2+t+1)^2} \end{equation*}

\begin{equation*} \lim_{t \to 1} \, \frac{\cancel{(t-1)^2}}{\cancel{(t-1)^2}(t^2+t+1)^2} = 1/9. \end{equation*}

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Subsection 2.1.7 Oscillating Function

Some functions oscillate (repeatedly go up and down) as (x) approaches a point or infinity. This can cause the limit to not exist, even if the function is bounded.

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Example 2.1.45.

Find:

\begin{equation*} \lim_{x \to 0} \sin\frac{\pi}{x} \end{equation*}

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\begin{equation*} \lim_{x \to 0} \sin\frac{1}{x} \end{equation*}

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\(\lim_{x \to 0} f(x) = \sin (1/x), \sin (\pi/x), x\sin (1/x), \sin (1/x^2), \sin(\ln|x|) \) etc. are some oscillatory functions whose limit do not exist at \(x \to 0.\)

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Subsection 2.1.8 Difference Quotient:

For a curve of given function

\begin{equation*} y=f(x) \end{equation*}

a Secant line is a line stretches between the two points on the given curve as shown in figure.

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The average rate of change for \(f(x)\) on the interval \([a,b]\) is the slope of the secant line between the two points \((a,f(a))\) and \((b,f(b)).\)

\begin{equation*} slope, \, m =\frac{\text{ rise } }{\text{ run }} =\frac{\Delta y}{\Delta x} \end{equation*}

\begin{equation*} or, \, m = \frac{f(b)-f(a)}{(b-a)} \end{equation*}

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For example: the average rate of change for the \(f(x) = \sqrt{x}\) on the interval \([1,4]\) is

\begin{equation*} m= \frac{f(4)-f(1)}{(4-1)} = \frac{\sqrt{4}-\sqrt{1}}{3} = \frac{2-1}{3} =\frac{1}{3}. \end{equation*}

A difference quotient represents the average rate of change of a function \(f(x)\) on the interval \([x, x+h].\) i.e., slope of secant line between the points \((x,f(x))\) and \(((x+h),f(x+h)).\)

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