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Calculus I Limits and Derivatives

Sunil Karna

Section 2.5 Example B

Example 2.5.1.

Is the function \(f(x) = 3x + 2\) continuous at \(x = 1\text{.}\)
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Solution.
Yes. It is a polynomial, so it is continuous everywhere (including at \(x = 1\)).
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Check: At \(x=1\text{,}\) \(f(1) = 3 \times 1 + 2 = 5 \) and
\begin{equation*} \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x + 2) = 5 \end{equation*}
\begin{equation*} \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3x + 2) = 5 \end{equation*}
\begin{equation*} \therefore \, \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1). \end{equation*}
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Example 2.5.2.

Prove that the function \(f(x)=5x-3\) is continuous at \(x=0,\) \(x=-3,\) and \(x=5.\)
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Solution.
At \(x=0\)
\begin{equation*} f(x)= f(0) = 5\times 0-3= -3. \end{equation*}
and
\begin{equation*} \lim_{x \to 0} f(x) = \lim_{x \to 0} (5x-3) = 5\times 0 -3 = -3. \end{equation*}
Hence, the function is continuous at \(x=0.\)
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Similarly check function at \(x=-3,\) and \(x=5.\)
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Example 2.5.3.

Examine the continuity of the function \(f(x)=2x^2-1\) at \(x=3.\)
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Solution.
Check: at \(x=3, \) \(f(x)=f(3) = 2\cdot 3^2-1 = 17.\)
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also,
\begin{equation*} \lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2-1) = 17. \end{equation*}
Hence, the function is continuous at \(x=3.\) Note: The polynomial functions are contnuous everywhere.
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Example 2.5.4.

Find all points of discontinuity of the given function, where \(f\) is defined by
  1. \begin{equation*} f(x) = \begin{cases} 2x+3 & x \leq 2 \\ 2x-3 & x \gt 2 \end{cases} \end{equation*}
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  2. \begin{equation*} f(x) = \begin{cases} \lvert x \rvert +3 & x \leq -3 \\ -2x & -3\lt x \lt 3 \\ 6x+2 & x\geq 3. \end{cases} \end{equation*}
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  3. \begin{equation*} f(x) = \begin{cases} \frac{\lvert x \rvert}{x} & x \neq 0 \\ 0 & x=0. \end{cases} \end{equation*}
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  4. \begin{equation*} f(x) = \begin{cases} \frac{x}{\lvert x \rvert} & x \lt 0 \\ -1 & x \geq 0. \end{cases} \end{equation*}
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  5. \begin{equation*} f(x) = \begin{cases} x+1 & x \geq 1 \\ x^2+1 & x \lt 1. \end{cases} \end{equation*}
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  6. \begin{equation*} f(x) = \begin{cases} x^3-3 & x \leq 2 \\ x^2+1 & x \gt 2. \end{cases} \end{equation*}
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Solution.
  1. do your self.
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  2. At \(x=-3,\)
    \begin{equation*} f(x)=\lvert x \rvert +3= \end{equation*}
    \begin{equation*} so, \, f(-3)=\lvert -3 \rvert +3 3+3 =6. \end{equation*}

    Note 2.5.5.

    \(\lvert x \rvert =-x \) for negative \(x\) and \(\lvert x \rvert =x \) for positive \(x\text{.}\)
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    now, check:
    \begin{equation*} \lim_{x\to -3^-} f(x)= \lim_{x\to -3^-} \lvert x \rvert +3 = \lim_{x\to -3^-} (-x +3) = +3+3 =6. \end{equation*}
    and
    \begin{equation*} \lim_{x\to -3^+} f(x)= \lim_{x\to -3^+} (-2x) = -2\times -3 =6. \end{equation*}
    Since, \(\lim_{x\to -3^-} f(x) = \lim_{x\to -3^+} f(x) = f(-3) \text{.}\) The function \(f(x)\) is continuous at \(x=-3.\)
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    Again, at \(x=3,\)
    \begin{equation*} f(x)=6x+2= \end{equation*}
    \begin{equation*} \therefore \, f(3)= 6\times 3 +2 =20. \end{equation*}
    and
    \begin{equation*} \lim_{x\to 3^-} f(x)= \lim_{x\to 3^-} (-2x) = -2\times 3 =-6. \end{equation*}
    Since \(f(3) \neq \lim_{x\to 3^-} f(x)\text{.}\) The function is not contunuous at \(x\to 3.\)
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  3. do your self.
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  4. at \(x=0,\)
    \begin{equation*} f(x)=-1 \end{equation*}
    \begin{equation*} \therefore \, f(0)= -1. \end{equation*}
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    and
    \begin{equation*} \lim_{x\to 0^-} f(x)= \lim_{x\to 0^-} \frac{x}{|x|} =\lim_{x\to 0^-} \frac{x}{-x} =-1. \end{equation*}
    \begin{equation*} \lim_{x\to 0^+} f(x)= \lim_{x\to 0^+} (-1) =-1. \end{equation*}
    Hence, the function is continuous at \(x=0.\)
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Example 2.5.6.

Determine if \(f(x) = |x|\) is continuous at \(x = 0\text{.}\)
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Solution.
Yes. \(\lim_{x \to 0^-} |x| = 0\text{,}\) \(\lim_{x \to 0^+} |x| = 0\text{,}\) and \(f(0) = 0\text{,}\) so the limit exists and equals the function value.
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Example 2.5.7.

Is \(f(x) = \frac{1}{x}\) continuous at \(x = 0\text{?}\)
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Solution.
No. The function is undefined at \(x = 0\text{,}\) so it cannot be continuous there.
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Example 2.5.8.

Check continuity of
\begin{equation*} f(x) = \begin{cases} x^2 & x \lt 1 \\ 2 & x = 1 \\ x + 1 & x \gt 1 \end{cases} \, \text{ at } x = 1. \end{equation*}
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Solution.
Yes. Left limit:
\begin{equation*} \lim_{x \to 1^-} x^2 = 1\text{.} \end{equation*}
Right limit:
\begin{equation*} \lim_{x \to 1^+} (x + 1) = 2\text{.} \end{equation*}
Since \(1 \neq 2\text{,}\) the limit does not exist \(\Rightarrow\) discontinuous.
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Example 2.5.9.

Is \(f(x) = \sqrt{x}\) continuous on \([0, \infty)\text{?}\)
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Solution.
Yes. It is defined for \(x \geq 0\text{,}\) and the limit equals the function value at every point (including \(x = 0\text{:}\) \(\lim_{x \to 0^+} \sqrt{x} = 0 = f(0)\)).
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Example 2.5.10.

For what value of \(c\) is
\begin{equation*} f(x) = \begin{cases} cx + 1 & x \leq 2 \\ 8 & x \gt 2 \end{cases} \, \text{ continuous at } x = 2? \end{equation*}
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Solution.
Set left limit = function value at 2:
\begin{equation*} \lim_{x \to 2^-} (cx + 1) = c(2) + 1 = 2c + 1. \end{equation*}
For continuity,
\begin{equation*} 2c + 1 = 8 \Rightarrow c = \frac{7}{2}. \end{equation*}
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Example 2.5.11.

Is \(f(x) = \sin(x)\) continuous at \(x = \frac{\pi}{2}\text{?}\)
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Solution.
Yes. Sine is continuous everywhere.
\begin{equation*} \lim_{x \to \pi/2} \sin(x) = 1 = f(\pi/2). \end{equation*}
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Example 2.5.12.

Determine if
\begin{equation*} f(x) = \frac{x^2 - 4}{x - 2} \end{equation*}
is continuous at \(x = 2\text{.}\)
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Solution.
No. The function is undefined at \(x = 2\) (division by zero), so it is discontinuous there. (Note: removable discontinuity if redefined as \(f(2) = 4\text{.}\))
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Example 2.5.13.

Is the function defined by \(f(x) = x^2-\sin x +5\) continuous at \(x=\pi?\)
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Solution.
At \(x=\pi,\)
\begin{equation*} f(x)= x^2-\sin x +5 \end{equation*}
\begin{equation*} or, \, f(\pi)= \pi^2-\sin \pi +5 = \pi^2+5\text{.} \end{equation*}
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\begin{equation*} \lim_{x \to \pi} f(x) = \lim_{x \to \pi} x^2-\sin x +5 =\pi^2+0+5 = \pi^2+5. \end{equation*}
Hence, the function is continuous.
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Example 2.5.14.

Examine the continuity of function \(f(x)\text{,}\) where \(f\) is defined by
\begin{equation*} f(x) = \begin{cases} \sin x -\cos x & x\neq 0 \\ -1 & x = 0. \end{cases} \end{equation*}
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Solution.
At \(x=0,\) \(f(x)=f(0)=-1.\)
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\begin{equation*} \lim_{x \to 0^-} f(x)= \lim_{x \to 0^-} (\sinx -\cos x) = -1 \end{equation*}
\begin{equation*} \lim_{x \to 0^+} f(x)= \lim_{x \to 0^+} (\sinx -\cos x) = -1 \end{equation*}
Hence, the function is continuous.
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Example 2.5.15.

Is the function \(f(x) = e^x \) continuous at \(x = 0\text{?}\)
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Solution.
Yes. The exponential function is continuous everywhere.
\begin{equation*} \lim_{x \to 0} e^x = 1 = e^0. \end{equation*}
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Example 2.5.16.

Check continuity of
\begin{equation*} f(x) = \begin{cases} 1 & x \text{ rational} \\ 0 & x \text{ irrational} \end{cases} \, \text{ at } x = 0. \end{equation*}
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Solution.
No. In any neighborhood of 0, \(f(x)\) takes both values 0 and 1 infinitely often, so \(\lim_{x \to 0} f(x)\) does not exist.
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Example 2.5.17.

Consider the function
\begin{equation*} f(x) = \begin{cases} x \sin\left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x = 0 \end{cases} \end{equation*}
Is \(f \) continuous at \(x = 0?\)
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Solution.
Yes. We need to check \(\lim_{x \to 0} f(x) = f(0) = 0 \text{.}\) Use the squeeze theorem:
\begin{equation*} -|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x| \quad \Rightarrow \quad -|x| \leq f(x) \leq |x| \quad (x \neq 0) \end{equation*}
As \(x \to 0\text{,}\) both \(-|x| \to 0 \) and \(|x| \to 0\text{,}\) so
\begin{equation*} \lim_{x \to 0} f(x) = 0 = f(0) \quad \Rightarrow \quad \text{ continuous at } x = 0. \end{equation*}
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Example 2.5.18.

Let
\begin{equation*} f(x) = \begin{cases} x^2+1 & x \lt 1 \\ a & x = 1 \\ 2x+b & x \gt 1\end{cases} \end{equation*}
Find \(a\) and \(b\) such that \(f\) is continuous at \(x = 1.\)
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Solution.
For continuity: \(lim_{\to 1^{-}}f(x)=lim_{\to 1^{+}}f(x) = f(1)=a\)
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\begin{equation*} Left: \, \lim_{x \to 1^-} (x^2 + 1) = 1 + 1 = 2. \end{equation*}
\begin{equation*} Right: \, \lim_{x \to 1^+}(2x+b)=2(1)+b=2+b. \end{equation*}
So: \(2 = 2 + b \quad \Rightarrow \quad b = 0\)
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Thus, \(a = 2\text{,}\) \(b = 0.\)
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Example 2.5.19.

Determine all values of \(x\) where
\begin{equation*} f(x) = \frac{x^2 - 3x + 2}{x^2 - 4} \end{equation*}
is discontinuous. Classify the discontinuities.
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Solution.
A rational function is discontinuous exactly where the denominator is zero and the numerator is not zero at the same point (non-removable discontinuity) or where the numerator and denominator are both zero (potentially removable).
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Find points where the denominator is zero \(x^2-4=0 \Rightarrow (x-2)(x+2)=0 \Rightarrow x=2 or x=-2.\)
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Check the numerator at these pointsNumerator: \(x^{2}-3x+2=(x-1)(x-2)\text{.}\) At \(x=2\text{:}\) \((2-1)(2-2)=1⋅0=0.\) Both numerator and denominator are zero.
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At \(x=-2\text{:}\) \((-2-1)(-2-2)=(-3)(-4)=12 \neq 0.\) Numerator is non-zero.
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So the simplified function after factorization:
\begin{equation*} f(x)=\frac{(x-1)(x-2)}{(x-2)(x+2)} = \begin{cases} \frac{x-1}{x+2} & x \neq 2 \\ \text{ undefined } & x = -2 \end{cases} \end{equation*}
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At \(x=2\)
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Both numerator and denominator = 0. Limit exists:
\begin{equation*} \lim_{x\to 2}\frac{x-1}{x+2}=\frac{2-1}{2+2}=\frac{1}{4} \end{equation*}
If we define \(f(2)=\frac{1}{4}\text{,}\) the function becomes continuous at \(x=2\text{.}\)
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Removable (hole).
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At \(x=-2\text{.}\)
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Denominator = 0, numerator \(\neq\) 0. Left limit: \(\lim_{x\to-2^{-}}\frac{x-1}{x+2} \to - \infty\) Right limit: \(\lim_{x\to-2^{+}}\frac{x-1}{x+2} \to + \infty\)
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Non-removable (vertical asymptote).
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Example 2.5.20.

Let
\begin{equation*} g(x) = \begin{cases} \frac{\sin(2x)}{x} & x \neq 0 \\ k & x = 0 \end{cases} \end{equation*}
Find \(k\) such that \(g\) is continuous at \(x = 0\text{.}\)
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Solution.
\begin{equation*} \lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{x \to 0} 2 \cdot \frac{\sin(2x)}{2x} = 2 \cdot 1 = 2 \end{equation*}
So \(k = 2\) \(\rightarrow\) continuous at 0.
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Example 2.5.21.

Let
\begin{equation*} f(x) = \frac{|x-1|}{x-1} \end{equation*}
for \(x \neq 1\text{.}\) Can \(f(1)\) be defined to make \(f\) continuous at \(x = 1.\)
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Solution.
\begin{equation*} \lim_{x \to 1^+} \frac{|x-1|}{x-1} = \frac{x-1}{x-1} = 1 \end{equation*}
\begin{equation*} \lim_{x \to 1^-} \frac{|x-1|}{x-1} = \frac{1-x}{x-1} = -1 \end{equation*}
Left \(\neq\) right \(\rightarrow\) limit does not exist \(\rightarrow\) cannot make continuous.
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Example 2.5.22.

Find all points of discontinuity of
\begin{equation*} f(x) = \begin{cases} \lfloor x \rfloor & x \lt 2 \\ x - 1 & x \geq 2 \end{cases} \end{equation*}
and classify them.
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Solution.
Floor function \(\lfloor x \rfloor\) is discontinuous at integers \(x = n\) for \(n \leq 1\) (but only \(x \lt 2\))
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At \(x = 2\text{:}\)
\begin{equation*} \lim_{x \to 2^-} \lfloor x \rfloor = 1 \end{equation*}
\begin{equation*} \lim_{x \to 2^+} (x-1) = 1 \end{equation*}
\begin{equation*} f(2) = 1 \Rightarrow \, \text{ continuous } \end{equation*}
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Discontinuities at integers \(x = 0, 1\) (jump discontinuities)
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\(x = 0\text{:}\)
\begin{equation*} \lim_{x \to 0^-} = -1\text{,} \end{equation*}
\begin{equation*} \lim_{x \to 0^+} = 0 \end{equation*}
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\(x = 1\text{:}\)
\begin{equation*} \lim_{x \to 1^-} = 0\text{,} \end{equation*}
\begin{equation*} \lim_{x \to 1^+} = 1 \end{equation*}
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Example 2.5.23.

Find the values of \(k\) so that the function \(f\) is continuous at the indicated point.
\begin{equation*} f(x)= \begin{cases} \frac{k \cos x}{\pi -2x} & x \neq \frac{\pi}{2} \\ 3 & x = \frac{\pi}{2} \end{cases} \, \text{ at } x=\pi/2. \end{equation*}
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Solution.
At \(x=\pi/2,\) \(f(x) = f(\frac{\pi}{2}) = =3\)
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\begin{equation*} \lim_{x \to \frac{\pi}{2}^-} f(x)= \lim_{h \to 0 } \frac{k \cos (\frac{\pi}{2}-h)}{\pi -2(\frac{\pi}{2}-h)} \end{equation*}
or,
\begin{equation*} \lim_{h \to 0 } \frac{k \sin h}{2h} =\frac{k}{2} \end{equation*}
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\begin{equation*} \lim_{x \to \frac{\pi}{2}^+} f(x)= \lim_{h \to 0 } \frac{k \cos (\frac{\pi}{2}+h)}{\pi -2(\frac{\pi}{2}+h)} \end{equation*}
or,
\begin{equation*} \lim_{h \to 0 } \frac{- k \sin h}{-2h} =\frac{k}{2} \end{equation*}
Hence, the function is continuous at \(x=\pi/2.\) Therefore,
\begin{equation*} \frac{k}{2} = 3 \, \Rightarrow \, k =6. \end{equation*}
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Example 2.5.24.

Find the value of \(k\) if
\begin{equation*} f(x)= \begin{cases} kx+1 & x \leq \pi \\ \cos x & x\gt \pi \end{cases} \, \text{ at } x=\pi. \end{equation*}
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Solution.
At \(x=\pi\text{,}\) \(f(x) = kx+1 \) or, \(f(\pi) =k\pi +1 \)
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Also,
\begin{equation*} \lim_{x \to \pi^-} f(x)= \lim_{h \to 0} f(\pi-h) =\lim_{h \to 0} [k(\pi-h)+1] = k\pi +1 \end{equation*}
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and,
\begin{equation*} \lim_{x \to \pi^+} f(x)= \lim_{h \to 0} f(\pi+h) =\lim_{h \to 0} \cos (\pi+h) = \lim_{h \to 0} [-\cos (h)] =-1 \end{equation*}
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Since the function is continuous at \(x=\pi\text{,}\) we have
\begin{equation*} k\pi +1 = -1 \Rightarrow \, k = \frac{-2}{\pi}. \end{equation*}
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Example 2.5.25.

Discuss the continuity of
\begin{equation*} f(x)= \begin{cases} \frac{\sin 3x}{\tan 2x} & x \lt 0 \\ \frac{3}{2} & x =0 \\ \frac{\log (1+3x)}{e^{2x}-1} & x\gt 0 \end{cases} \, \text{ check continuity at } x=0. \end{equation*}
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Solution.
\begin{equation*} LHL\rvert_0 = \lim_{x\to 0^-}f(x) = \lim_{x\to 0^-} \frac{\sin 3x}{\tan 2x} \end{equation*}
\begin{equation*} = \lim_{x\to 0^-} \frac{\left(\frac{\sin 3x}{3x}\right) 3x}{\left(\frac{\tan 2x}{2x}\right)2x } =\frac{3}{2}. \end{equation*}
[See some Standard Limits, Subsection 2.2.2]
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\begin{equation*} RHL\rvert_0 = \lim_{x\to 0^+}f(x) = \lim_{x\to 0^+} \frac{\log (1+3x)}{e^{2x}-1} \end{equation*}
\begin{equation*} = \lim_{x\to 0^+} \frac{\left(\frac{\log (1+3x)}{3x}\right) 3x}{\left(\frac{e^{2x}-1}{2x}\right)2x } =\frac{3}{2}. \end{equation*}
at \(x=0\)
\begin{equation*} f(0) = \frac{3}{2} \end{equation*}
Hence the function is continuous at \(x=0.\)
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Example 2.5.26.

Discuss the continuity of
\begin{equation*} f(x)= \begin{cases} \frac{\lfloor x^2 \rfloor -1}{x^2-1} & x^2 \neq 1 \\ 0 & x^2 =1 \end{cases} \, \text{ at } x=1. \end{equation*}
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Solution.
Step 1: Simplify the expression for \(x^2\neq 1\) and let \(t = x^2\text{.}\) The expression becomes
\begin{equation*} \frac{\lfloor t \rfloor -1}{t-1}; \, t\gt 0, t\neq 1 \end{equation*}
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Recall that for any real (y), \(y - 1 \lt \lfloor y \rfloor \leq y\text{,}\) so \(\lfloor y \rfloor - y\) is in \([0, 1)\text{.}\) Thus, \(\lfloor t \rfloor -1 =(\lfloor t \rfloor -t)+(t-1),\) and
\begin{equation*} \frac{\lfloor t \rfloor -1}{t-1}=\frac{\lfloor t \rfloor -t}{t-1}+1. \end{equation*}
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The solution is not as the standard of this class.
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Example 2.5.27.

Continuity of composite function.
\begin{equation*} f(x)= \begin{cases} 2x+1 & x \lt 0 \\ x^2+2 & 0\leq x\lt 2 \\ x+5 & x \geq 2 \end{cases} \, \quad \text{ and } g(x) = \begin{cases} x-1 & x \lt 2 \\ 2x-2 & x \geq 2 \end{cases} \end{equation*}
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Solution.
\begin{equation*} f \circ g (2) = f(g(2))=f(2) =7 \end{equation*}
as \(g(2)=2\times 2-2=2\) and \(f(2) = 2+5 =7.\)
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\begin{equation*} f \circ g (3) = f(g(3))=f(4) =9 \end{equation*}
\begin{equation*} f \circ g (1^-) = f(g(1^-))=f(0^-) =2x+1 =2\times 0+1=1 \end{equation*}
\begin{equation*} f \circ g (1^+) = f(g(1^+))=f(0^+) =x^2+2 = 0+2=2 \end{equation*}
\begin{equation*} f \circ g (2^-) = f(g(1))=f(0) = 0+2=2. \end{equation*}
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Example 2.5.28.

If \(f(x) \) is continuous and \(f\left(\frac{1}{2}\right) =3\text{,}\) then \(\lim_{x \to 0} f\left(\frac{\tan x-\sin x}{\sin^3 x}\right) =?\)
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Solution.
If function is continuous, then \(\lim_{x \to 0} f(x) = f\left(\lim_{x \to 0} x\right) \) Now,
\begin{align*} \lim_{x \to 0} f\left(\frac{\tan x-\sin x}{\sin^3 x}\right) \amp = \lim_{x \to 0}\left(\frac{\tan x-\sin x}{\sin^3 x}\right) \\ \amp = \lim_{x \to 0}\left(\frac{\frac{\sin x}{\cos x}-\sin x}{\sin^3 x}\right) = \lim_{x \to 0}\left(\frac{\sin x(1-\cos x)}{\cos x\sin^3 x}\right)\\ \amp = \lim_{x \to 0}\left(\frac{\cancel{\sin x}(1-\cos x)}{\cos x \sin^\cancelto{2}{3} x}\right) \\ \amp \lim_{x \to 0}\left(\frac{2\sin^2 \frac{x}{2}}{\sin^2 x\cos x }\right)\\ \amp \lim_{x \to 0}\left(\frac{2\sin^2 \frac{x}{2}}{4\times (\frac{x}{2})^2 \times \frac{\sin^2 x}{x^2} }\right)\frac{1}{\cos x} = \frac{1}{2}\\ \therefore \, \lim_{x \to 0} f\left(\frac{\tan x-\sin x}{\sin^3 x}\right) \amp = f(\frac{1}{2}) =3. \end{align*}
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Example 2.5.29.

Discuss the continuity of \(f(x) =\begin{cases} x^2+2; & x \lt 0 \\ 3x+2; & 0\leq x\lt 1 \\ 3e^x-2; & x\geq 1\end{cases} \,\quad \text{ at } x=0 \, \text{ and } x=1.\)
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Solution.
At \(x=0\)
\begin{equation*} f(x) = f(0) = 3\times 0+2 = 2 \end{equation*}
\begin{equation*} \lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} x^2+2 = 2 \end{equation*}
\begin{equation*} \lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} 3x+2 = 2 \end{equation*}
\begin{equation*} \therefore \, LHL\vert_0 = RHL\vert_0 =f(0) \end{equation*}
Hence, the function is continuous at \(x=0.\)
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Again \(x=1.\)
\begin{equation*} f(1) = 3e^1-2 = 3e-2 \end{equation*}
\begin{equation*} \lim_{x\to 1^-} f(x) = \lim_{x\to 1^-} 3x+2 = 5 \end{equation*}
\begin{equation*} \lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} 3e^x-2 = 3e-2 \end{equation*}
\begin{equation*} \because \,\lim_{x\to 1^-} f(x) \neq f(1) \end{equation*}
function is discontinuous at \(x=1.\)
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Example 2.5.30.

Discuss the continuity of \(f(x) =\begin{cases} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}; & x \neq 0 \\ 0; & x=0 \end{cases} \,\quad \text{ at } x=0. \)
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Solution.
\(f(0) = 0 \)
\begin{equation*} \lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1} = \frac{e^{-\infty}-1}{e^{-\infty}+1} = \frac{-1}{1} =-1. \end{equation*}
Since \(\lim_{x\to 0^-} f(x)\neq f(0)\) the function is discontinuous.
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Example 2.5.31.

Discuss contnuity of \(f(x) =\begin{cases} \frac{\sin 3x}{\tan 2x}; & x \lt 0 \\ \frac{3}{2}; & x=0 \\ \frac{\log (1+3x)}{e^{2x}-1} & x\gt 0 \end{cases} \, \quad \text{ at } x=0. \)
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Solution.
\(f(0) = \frac{3}{2}\)
\begin{equation*} \lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} \frac{\sin 3x}{\tan 2x} = \lim_{x\to 0^-} \frac{\frac{\sin 3x}{3x}\times 3x}{\frac{\tan 2x}{2x}\times 2x} =\frac{3}{2} \end{equation*}
\begin{equation*} \lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} \frac{\log (1+3x)}{e^{2x}-1} = \lim_{x\to 0^+} \frac{\frac{\log (1+3x)}{3x}\times 3x}{\frac{e^{2x}-1}{2x}\times 2x} =\frac{3}{2} \end{equation*}
[Logarithimic and Exponential] Therefore the function is continuous at \(x=0.\)
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Example 2.5.32.

Discuss continuity of
\begin{equation*} f(x) =\begin{cases} \frac{\lfloor x^2 \rfloor -1}{x^2-1}; & x^2 \neq 1 \\ 0; & x=1 \end{cases} \, \quad \text{ at } x=1. \end{equation*}
\(\lfloor \cdot \rfloor\) is a GIF.
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Solution.
\(f(1) = 0.\)
\begin{equation*} \lim_{x\to 1^-} f(x) = \lim_{x\to 1^-} \frac{\lfloor x^2 \rfloor -1}{x^2-1} = \frac{0-1}{-# \to 0} = -\infty \end{equation*}
\begin{equation*} \lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} \frac{\lfloor x^2 \rfloor -1}{x^2-1} = \frac{1-1}{+# \to 0} = 0 \end{equation*}
i.e., the function is discontinuous.
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