Example 2.4.1.
\begin{equation*}
f(x) = \frac{x^2 - 1}{x - 1}, x \neq 1; \quad\, \text{ define } f(1) = 2 \text { to fix.}
\end{equation*}
Section 2.4 Continuity
A function is continuous at a point if there are no breaks, jumps, or holes on its graph. Intuitively, you can draw the graph without lifting your pencil.
A function is said to be continuous at \(x=a,\) if [Figure 2.1.22]
\begin{equation*}
\lim_{x\to a} f(x) = f(a).
\end{equation*}
i.e.,
\begin{equation*}
\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x)= f(a).
\end{equation*}
To see if the given function is continuous or not needs the following three conditions:
-
\(f(a)\) is defined.
-
\(\displaystyle \lim_{x \to a} f(x) \,\text{ exists. }\)
-
The limit equals the function value, i.e.,\begin{equation*} \lim_{x \to a} f(x) = f(a)\text{.} \end{equation*}
If any one fails, \(f\) is discontinuous at \(a\text{.}\)
Alternatively, using the difference quotient [Subsection 2.1.8], continuity at (c) means:
\begin{equation*}
\lim_{h \to 0} f(c+h) = f(c)
\end{equation*}
Subsection 2.4.1 Types of Continuous Functions
-
Polynomial (linear, square, cubic, etc. ) functions are continuous everywhere (in \(\mathbb R\)).
-
Rational functions are continuous everywhere (in \(\mathbb R\)) except where \(denominator =0\text{.}\)
-
The mode function \(\lvert x \rvert\) is continuous everywhere.
-
Logarithmic (\(\ln x, \log_b x\)) functions are continuous where they are defined (i.e., \(\forall x \gt 0\)).
-
Trigonometric functions (\(\sin x, \cos x\)) are continuous everywhere but \(\tan x\text{,}\) \(\sec x\) are continuous except at \(x=(2n+1)\frac{\pi}{2}; n\in \mathbb Z. \)
-
The GIF \(\lfloor x \rfloor \) and FPF \(\{x\}\) functions are continuous \(\forall x \in \mathbb R-\mathbb z.\)
From geometry or [Subsection 2.2.1] Squeeze theorem, we have some standard limits [Subsection 2.2.2]:
\begin{equation*}
\lim_{x \to 0}\frac{\sin x}{x} =1
\end{equation*}
Subsection 2.4.2 Types of Discontinuities
-
Removable Discontinuity Limit exists, but \(f(a)\) is missing or wrong.
-
Jump Discontinuity Left and right limits exist but differ.
Example 2.4.2.
\begin{equation*} f(x) = \begin{cases} 1 & x \lt 0 \\ 2 & x \geq 0 \end{cases} \quad \, \text{ at } x = 0. \end{equation*} -
Infinite Discontinuity Limit is \(\pm \infty\text{.}\)\begin{equation*} f(x) = \frac{1}{x} \quad \, \text{ at } x=0. \end{equation*}
-
Oscillatory Discontinuity Limit fails due to wild oscillation.\begin{equation*} f(x) = \sin\left(\frac{1}{x}\right) \quad \, \text{ at } x=0. \end{equation*}
Subsection 2.4.3 Holes
Removal Discontinuity While solving some rational functions, we need to restrict some variables from getting function undefined. This resitriction may introduce holes in the plot.
Example 2.4.3.
\begin{equation*}
f(x) = \frac{x-3}{x^2+3x-18}
\end{equation*}
Solution.
\begin{equation*}
f(x) = \frac{x-3}{x^2+3x-18} = \frac{\cancel{(x-3)}}{(x+6)\cancel{(x-3)}}; \, \text{ Restriction } \, x\neq -6,3
\end{equation*}
Here \(x=3 \) produces hole in the plot, which can be removed by simplyfying the function as
\begin{equation*}
f(x) = \frac{1}{x+6}; \, x\neq -6,3.
\end{equation*}
Here \(x=-6\) gives vertical asymptote.
Example 2.4.4.
\begin{equation*}
f(x) = \frac{x^2-9}{x^2+7x+12}
\end{equation*}
Subsection 2.4.4 Continuity on an Interval
\(f\) is continuous on \([a, b]\) if continuous at every point in \((a, b)\text{,}\) and right-continuous at \(a\text{,}\) left-continuous at \(b\text{.}\) That is:
\begin{equation*}
\lim_{x\to a^+} f(x)=f(a), \lim_{x\to b^-} f(x)=f(b)
\end{equation*}
Subsection 2.4.5 Properties of Continuous Functions
If \(f\) and \(g\) are two continuous functions at real number \(c\text{,}\) then
Subsection 2.4.6 Intermediate Value Theorem
If \(f\) is continuous on the closed interval \([a,b]\) and \(N\) is any number between \(f(a)\) and \(f(b),\) where \(f(a) \neq f(b),\) then there exists at least one number \(c \in (a,b)\) such that \(f(c) = N.\)
We use it to prove equations have roots (solutions) by showing f(a) and f(b) have opposite signs (so N = 0 is between them).
Example 2.4.5.
Basic root existence (polynomial) Consider \(f(x) = x^3 - 3x + 1\) on the interval \([0, 2]\text{.}\)
Does the equation \(f(x) = x^3 - 3x + 1\) have a solution in \([0, 2]\text{?}\) Evaluate the endpoints:
\begin{equation*}
f(x) = x^3 - 3x + 1 = 0^3 - 3(0) + 1 = 1
\end{equation*}
\begin{equation*}
f(2) = 2^3 - 3(2) + 1 = 8 - 6 + 1 = 3
\end{equation*}
Both positive → doesn’t help directly for root. But let’s check a better interval or adjust.
\begin{equation*}
f(0) = -2 \lt 0
\end{equation*}
\begin{equation*}
f(2) = 32 - 16 - 2 = 14 \gt 0
\end{equation*}
Since \(f\) is continuous (polynomial) and \(f(0) \lt 0 \lt f(2)\text{,}\) by IVT there is at least one \(c \in (0, 2)\) such that \(f(c) = 0.\)
\begin{equation*}
f(2) = 8 - 6 + 1 = 3
\end{equation*}
Both positive → doesn’t help directly for root.
But let’s check a better interval or adjust. Better example (classic one): \(f(x) = x^5 - 2x^3 - 2\) on \([0, 2]\text{.}\)
\begin{equation*}
f(0) = -2 \lt 0
\end{equation*}
\begin{equation*}
f(2) = 32 - 16 - 2 = 14 \gt 0
\end{equation*}
Since \(f\) is continuous (polynomial) and \(f(0) \lt 0 \lt f(2)\text{,}\) by IVT there is at least one \(c \in (0, 2)\) such that \(f(c) = 0.\)
Example 2.4.6.
Show there exists \(c \in (1, 4)\) such that \(f(c) = 10. \)
\begin{equation*}
f(1) = 1 - 6 + 2 = -3
\end{equation*}
\begin{equation*}
f(4) = 64 - 24 + 2 = 42
\end{equation*}
Since \(-3 \lt 10 \lt 42\) and \(f\) is continuous, by IVT there exists \(c \in (1, 4)\) such that \(f(c) = 10.\)
Example 2.4.7.
Show there is a solution to \(\sin(x) = 0.7\text{.}\) \(f(0) = 0\) \(f(\pi) \approx 0\) (actually exactly 0) — bad choice. Better interval: \([1, 2]\) (radians) \(\sin(1) \approx 0.8415\) \(\sin(2) \approx 0.9093\) Both \(\gt\) 0.7 → not helpful. Try \([0.5, 1.2]\) \(\sin(0.5) \approx 0.479\) \(\sin(1.2) \approx 0.932\) Since \(0.479 \lt 0.7 \lt 0.932\text{,}\) by IVT there exists \(c \in (0.5, 1.2)\) such that \(\sin(c) = 0.7.\)
Example 2.4.8.
Showing a root in a narrower interval Let \(f(x) = 2x^3 - 9x^2 + 12x - 5\) on \([0, 3]\text{.}\)
\begin{equation*}
f(0) = -5 \lt 0
\end{equation*}
\begin{equation*}
f(3) = 54 - 81 + 36 - 5 = 4 \gt 0
\end{equation*}
→ root exists in \((0,3)\text{.}\) Now narrow it: \(f(1) = 2-9+12-5 = 0\) → actually a root at \(x=1\text{!}\) (But if we didn’t notice, \(f(2) = 16 - 36 + 24 - 5 = -1 \lt 0\text{,}\) so another sign change in \((2,3)\text{.}\))
Example 2.4.9.
Transcendental equation (mix of functions) Show that \(x = cos(x)\) has a solution. Let \(f(x) = x - cos(x)\text{.}\)
\begin{equation*}
\end{equation*}
\begin{equation*}
f(0) = 0 - 1 = -1 \lt 0
\end{equation*}
\begin{equation*}
f(\pi/2) = \pi/2 - 0 \gt 0
\end{equation*}
Since \(f(0) \lt 0 \lt f(\pi/2)\) and \(f\) is continuous (composition of continuous functions), by IVT there is at least one \(c \in (0, \pi/2)\) such that \(f(c) = 0.\) Thus, \(c = \cos(c).\)
\begin{equation*}
f(\pi/2) \approx 1.57 - 0 = 1.57 \gt 0
\end{equation*}
f is continuous (composition of continuous functions), sign change → there is \(c \in (0, \pi/2)\) with \(f(c) = 0\) → \(c = \cos(c).\)
