MATH 111 Calculus I \(\quad\;\;\) Instructor: Difeng Cai

2.2 The Limit of a Function

Concept of Limit

Suppose \(f(x)\) is defined when \(x\) is near the number \(a\), then \[\lim_{x\to a} f(x) = L\] if we can make values of \(f(x)\) arbitrarily close to \(L\) by restricting \(x\) to be sufficiently close \(a\) (on either side of \(a\)) but not equal to \(a\).

• \(f(a)\) may not be equal to \(L\). The function value has nothing to do with limit at that point.
• If \(f\) is defined on both sides of \(a\), the definition imposes condition on both sides.

One-Sided Limits

Consider the Heaviside function \[H(t)=\begin{cases} 0 \quad\text{if}\quad t<0, \\ 1\quad\text{if} \quad t\geq 0. \end{cases}\] \[\lim_{t\to 0^-}H(t)=0\quad\text{and}\quad \lim_{t\to 0^+}H(t)=1\] Here \(t\to 0^-\) means \(t\) approaches \(0\) from the left (\(t<0\)), and \(t\to 0^+\) from the right of \(0\) (\(t>0\)).

\((a). \lim_{x\to 2^-} f(x) \quad (b). \lim_{x\to 2^+} f(x)\quad (c). f(2)\quad (d). \lim_{x\to 2} f(x)\quad (e). \lim_{x\to 4} f(x) \)

Limit and one-sided limits: \[\lim_{x\to a} f(x)=L\quad\text{if and only if}\quad \lim_{x\to a^-}f(x)=L\text{ and } \lim_{x\to a^+}f(x)=L\]

Infinite Limits

\(\lim_{x\to a}f(x)=\infty\) means that the values of \(f(x)\) can be made arbitrarily large by taking \(x\) sufficiently close to \(a\), but not equal to \(a\). ( \(f(x)\) "blows up" when \(x\) approaches \(a\) )

Since \(\infty\) is not a number, \(\lim_{x\to a}f(x)=\infty\) implies that the limit doest not exist when \(x\) tends to \(a\).

Similarly, \(\lim_{x\to a}f(x)=\infty\) means that \(f(x)\) can be arbitrarily large negative as \(x\) approaches \(a\).

Example. \(\lim_{x\to 0}\frac{1}{x^2}=\infty\) and \(\lim_{x\to 0}\frac{-1}{x^2}=-\infty\)

One-sided infinite limits can be defined similarly (see graphs below).

If \(f(x)\) tends to \(\infty\) or \(-\infty\) as \(x\) approaches \(a\) (either one-sided or two-sided), the vertical line \(x=a\) is called a vertical asymptote of \(y=f(x)\).

Ex. Find the limits.
(a). \(\lim_{x\to 4^-} \frac{5-x}{x-4}\) (b). \(\lim_{x\to 1^+}\ln(x^2-1)\) (c). \(\lim_{x\to (\frac{\pi}{2})^-} \tan x\) (d). \(\lim_{x\to (\frac{\pi}{2})^+} \tan x\)
Answer: (a).\(-\infty\) (b). \(-\infty\) (c).\(\infty\) (d).\(-\infty\)

2.3 Limit Laws

Assume \(\lim_{x\to a} f(x)\) exists. Then

• \(\lim_{x\to a} [f(x)]^n = \left[\lim_{x\to a} f(x)\right]^n\) (\(n\) is a positive integer)
• \(\lim_{x\to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\to a} f(x)}\) (\(n\) is a positive integer; if \(n\) is even, we further assume \(\lim_{x\to a} f(x)>0\) because otherwise the limit does not exist)

Ex. (a). \(\lim_{x\to 1} \sqrt[3]{x^2-9}\) (b). \(\lim_{x\to -1} 3x-5x^2+1\)
Answer: (a). -2 (b). -7

Assume that the limits \(\lim_{x\to a} f(x)\) and \(\lim_{x\to a} g(x)\) exist. Then

• \[\lim_{x\to a} [ f(x)\pm g(x) ] = \lim_{x\to a} f(x) \pm \lim_{x\to a} g(x)\]
• \[\lim_{x\to a} [ f(x)g(x) ] = \lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x)\]
• \[\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{\lim_{x\to a} f(x)}{\lim_{x\to a} g(x)} \; \text{ if } \lim_{x\to a} g(x)\neq 0\]

What if \(\lim_{x\to a} g(x) =0\) ? See below.

Compute \(\lim_{x\to a} \frac{f(x)}{g(x)}\) when \(\lim_{x\to a} g(x)=0\)

In this case, it is very likely that the expression can be simlified by canceling out common factors from the top and the bottom.

Example 1. \(\lim_{x\to 1} \frac{1-x^2}{x-1}\)
Solution: \(\lim_{x\to 1} x-1 = 0\) so we cannot apply the law above. Instead, we first simplify the expression: \[\frac{1-x^2}{x-1} = \frac{(1+x)(1-x)}{x-1} = -(1+x)\] Now it follows that \(\lim_{x\to 1} \frac{1-x^2}{x-1} = \lim_{x\to 1} -(1+x)=-2\).

Example 2. \(\lim_{x\to 0} \frac{(1-h)^2-1}{h}\)
Solution: Since \((1-h)^2-1=-2h+h^2\), we have \[\lim_{x\to 0} \frac{(1-h)^2-1}{h} = \lim_{x\to 0} \frac{-2h+h^2}{h}=\lim_{x\to 0} -2+h = -2\]

Example 3. \(\lim_{t\to 9} \frac{9-t}{3-\sqrt{t}}\)
Solution: \[\lim_{t\to 9} \frac{9-t}{3-\sqrt{t}}=\lim_{t\to 9} \frac{9-t}{3-\sqrt{t}} \frac{3+\sqrt{t}}{3+\sqrt{t}}=\lim_{t\to 9} \frac{(9-t)(3+\sqrt{t})}{9-t}=\lim_{t\to 9}3+\sqrt{t}=6\]