3.3: Limits

The notion of a limit is closely related to that of a derivative, but it is more general. In this chapter $f$ will always be a real function of one variable. Let us recall the definition of the slope off at a point $a$:

$S$ is the slope of $f$ at $a$ if whenever $dx$ is infinitely close to but not equal to zero, the quotient $$\frac{f(a + \Delta x) - f(a)}{\Delta x} \nonumber$$is infinitely close to S.

We now define the limit. $c$ and $L$ are real numbers.

Definition:

$L$ is the limit of $f(x)$ as $x$ approaches $c$ if whenever $x$ is infinitely close to but not equal to $c$, $f(x)$ is infinitely close to $L$.

In symbols, $$\lim_{x \rightarrow c} f(x) = L \nonumber$$if whenever $x \approx c$ but $x \neq c$, $f(x) \approx L$. When there is no number $L$ satisfying the above definition, we say that the limit of $f(x)$ as $x$ approaches $c$ does not exist.

Notice that the limit $$\lim_{x \rightarrow c} f(x) = L \nonumber$$depends only on the values of $f(x)$ for $x$ infinitely close but not equal to $c$. The value $f(c)$ itself has no influence at all on the limit. In fact, it very often happens that $$\lim_{x \rightarrow c} f(x) = L \nonumber$$exists but $f(c)$ is undefined.

Figure $\PageIndex{1(a)}$ shows a typical limit. Looking at the point $(c, L)$ through an infinitesimal microscope, we can see the entire portion of the curve with $x \approx c$ because $f(x)$ will be infinitely close to $L$ and hence within the field of vision of the microscope.

In Figure $\PageIndex{1(b)}$, part of the curve with $x \approx c$ is outside the field of vision of the microscope, and the limit does not exist.

Our first example of a limit is the slope of a function.

Figure $\PageIndex{1}$: Points on a curve where limits exist and do not exist.

Theorem $\PageIndex{1}$

The slope of $f$ at $a$ is given by the limit $$f'(a) = \lim_{\Delta x \rightarrow 0} \frac{f(a + \Delta x) - f(a)}{\Delta x}. \nonumber$$

Verbally, the slope of $f$ at $a$ is the limit of the ratio of the change in $f(x)$ to the change in $x$ as the change in $x$ approaches zero. The theorem is seen by simply
comparing the definitions of limit and slope. The slope exists exactly when the limit exists; and when they do exist they are equal. Notice that the ratio $$\frac{f(a + \Delta x) - f(a)}{\Delta x} \nonumber$$ is undefined when $\Delta x = 0$.

The slope of $f$ at $a$ is also equal to the limit $$f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}. \nonumber$$

This is seen by setting $$\begin{align*} \Delta x &= x - a, \\ x &= a + \Delta x. \end{align*}$$

Then when $x \approx a$ but $x \neq a$, we have $\Delta x \approx 0$ but  $\Delta x \neq 0$; and $$\frac{f(x) - f(a)}{x - a} = \frac{f(a + \Delta x) - f(a)}{\Delta x} \approx f'(a). \nonumber$$

Sometimes a limit can be evaluated by recognizing it as a derivative and using Theorem $\PageIndex{1}$ above.

Example $\PageIndex{1}$

Evaluate $$\lim_{\Delta x \rightarrow 0} \frac{(3 + \Delta x)^{2} - 9}{\Delta x}. \nonumber$$

Solution

Let $F(x) = x^{2}$. The given limit is just $F'(3)$, $$\begin{align*} F'(3) &= \lim_{\Delta x \rightarrow 0} \frac{F(3 + \Delta x)^{2} - 9}{\Delta x} = 6. \\ F'(3) = 2 \cdot 3 = 6. \end{align*}$$

Therefore $$\lim_{\Delta x \rightarrow 0} \frac{(3 + \Delta x)^{2} - 9}{\Delta x} = 6. \nonumber$$

The symbol $x$ in $$\lim_{x \rightarrow c} f(x) \nonumber$$is an example of a "dummy variable." The value of the limit does not depend on $x$ at all. However, it does depend on $c$. If we replace $c$ by a variable $u$, we obtain a new function $$L(u) = \lim_{x \rightarrow u} f(x). \nonumber$$

A limit $\lim_{x \rightarrow c} f(x)$ is usually computed as follows.

Step 1: Let $x$ be infinitely close but not equal to $c$, and simplify $f(x)$.
Step 2: Compute the standard part $st(f(x))$.

Conclusion: If the limit $\lim_{x \rightarrow c} f(x)$ exists, it must equal $st(f(x))$.

Example $\PageIndex{1}$ (continued)

Instead of using the derivative, we can directly compute $$\lim_{\Delta x \rightarrow 0} \frac{(3 + \Delta x)^{2} - 9}{\Delta x}. \nonumber$$

Solution

Step 1
Let $\Delta x \approx 0$, but $\Delta x \neq 0$. Then $$\frac{(3 + \Delta x)^{2} - 9}{\Delta x} = \frac{(9 + 6 \Delta x + \Delta x^{2} - 9}{\Delta x} = \frac{6 \Delta x + \Delta x^{2}}{\Delta x} = 6 + \Delta x. \nonumber$$

Step 2
Taking standard parts, $$st \frac{(3 + \Delta x)^{2} - 9}{\Delta x} = st(6 + \Delta x) = 6. \nonumber$$

Therefore the limit is equal to 6. (See Figure \(\PageIndex{2}.)

Figure $\PageIndex{2}$: Graph of $f(\Delta x) = \dfrac{(3 + \Delta x)^{2} - 9}{\Delta x}$, showing the limit is 6 as $\Delta x$ approaches 0.

Example $\PageIndex{2}$

Find $\displaystyle \lim_{t \rightarrow 4} \left(t^{2} + 3t - 5\right)$.

Solution

Step 1
Let $t$ be infinitely close to but not equal to $4$.

Step 2
We take the standard part $$st \left(t^{2} + 3t - 5\right) = 4^{2} + 3 \cdot 4 - 5 = 23, \nonumber$$ so the limit is 23. 

Example $\PageIndex{3}$

Find $\displaystyle \lim_{x \rightarrow 2} = \dfrac{x^{2} + 3x - 10}{x^{2} - 4}$.

Solution

Step 1 
This time the term inside the limit is undefined at $x = 2$. Taking $x \approx 2$ but $x \neq 2$, we have $$\frac{x^{2} + 3x - 10}{x^{2} - 4} = \frac{(x - 2)(x+5)}{(x - 2)(x + 2)} = \frac{x + 5}{x + 2}. \nonumber$$

Step 2
$$st \left(\frac{x^{2} + 3x - 10}{x^{2} - 4}\right) = st \left(\frac{x+5}{x+2}\right) = \frac{2 + 5}{2 + 2} = \frac{7}{4}. \nonumber$$

Thus $$\lim_{x \rightarrow 2} \left(\frac{x^{2} + 3x - 10}{x^{2} - 4}\right) = \frac{7}{4}. \nonumber$$

Example $\PageIndex{4}$

Find $\displaystyle \lim_{x \rightarrow 0} \left( \dfrac{(2/x) + 3}{(3/x) - 1} \right)$.

Solution

Step 1
Taking $x \approx 0$ but $x \neq 0$, $$\frac{(2/x) + 3}{(3/x) - 1} = \frac{2 + 3x}{3 - x}. \nonumber$$

Step 2
$$st \left( \frac{(2/x) + 3}{(3/x) - 1} \right) = st \left(\frac{2 + 3x}{3 - x}\right) = \frac{2}{3}. \nonumber$$

Thus the limit exists and equals $\frac{2}{3}$.

Example $\PageIndex{5}$

Find $\displaystyle \lim_{x \rightarrow 9} \left( \dfrac{\sqrt{x} - 3}{x - 9} \right)$.

Solution

Step 1
Taking $x \approx 9$ but $x \neq 9$, $$\frac{\sqrt{x} - 3}{x - 9} = \frac{\left(\sqrt{x} - 3\right) \left(\sqrt{x} + 3\right)}{(x - 9) \left(\sqrt{x} + 3\right)} = \frac{x - 9}{(x - 9) \left(\sqrt{x} + 3\right)} = \frac{1}{\sqrt{x} + 3}. \nonumber$$

Step 2
$$st \left( \dfrac{\sqrt{x} - 3}{x - 9} \right) = st \left(\frac{1}{\sqrt{x} + 3}\right) = \frac{1}{\sqrt{9} + 3} = \frac{1}{6}, \nonumber$$so the limit exists and is $\frac{1}{6}$.

Our rules for standard parts in Chapter 1 lead at once to rules for limits. We list these rules in Table $\PageIndex{1}$. The limit rules can be applied whenever the two limits $\displaystyle \lim_{x \rightarrow c} f(x)$ and $\displaystyle \lim_{x \rightarrow c} g(x)$ exist.

Table $\PageIndex{1}$. Rules for standard parts as applied to limits.

Standard Part Rule	Limit Rule
$st(kb) = k \ st(b), \ k \text{ real}$	$\displaystyle \lim_{x \rightarrow c} kf(x) = k \lim_{x \rightarrow c} f(x)$
$st(a + b) = st(a) + st(b)$	$\displaystyle \lim_{x \rightarrow c} \left(f(x) + g(x)\right) = \lim_{x \rightarrow c} f(x) + \lim_{x \rightarrow c} g(x)$
$st(ab) = st(a) \cdot st(b)$	$\displaystyle \lim_{x \rightarrow c} \left(f(x) g(x)\right) = \lim_{x \rightarrow c} f(x) \cdot \lim_{x \rightarrow c} g(x)$
$st(a/b) = st(a) / st(b)$	$\displaystyle \lim_{x \rightarrow c} \left(f(x) / g(x)\right) = \lim_{x \rightarrow c} f(x) / \lim_{x \rightarrow c} g(x), \text{ if } \lim_{x \rightarrow c} g(x) \neq 0$
$st \left(\sqrt[n]{a}\right) = \sqrt[n]{st(a)}, \text{ if } a > 0$	$\displaystyle \lim_{x \rightarrow c} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \rightarrow c} f(x)}, \text{ if } \lim_{x \rightarrow c} f(x) > 0$

Example $\PageIndex{6}$

Find $\displaystyle \lim_{x \rightarrow 1} \left(x^{2} - 2x\right) \sqrt{\left(x^{2} - 1\right) / (x - 1)}.$

Solution

All the limits involved exist, so we can use the limit rules to compute the limit as follows. First we find the limit of the expression inside the radical. 

$$\lim_{x \rightarrow 1} \frac{x^{2} - 1}{x - 1} = \lim_{x \rightarrow 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \rightarrow 1} (x + 1) = 2. \nonumber$$

Now we find the answer to the original problem.

$$\begin{align*} \lim_{x \rightarrow 1} \left(x^{2} - 2x\right) \sqrt{\left(x^{2} - 1\right) / (x - 1)} &= \lim_{x \rightarrow 1} \left(x^{2} - 2x\right) \sqrt{\lim_{x \rightarrow 1} \left(x^{2} - 1\right) / (x - 1)} \\ &= (1 - 2) \sqrt{2} = - \sqrt{2}. \end{align*}$$

There are three ways in which a limit $\displaystyle \lim_{x \rightarrow c} f(x)$ can fail to exist:

1. $f(x)$ is undefined for some $x$ which is infinitely close but not equal to $c$.
2. $f(x)$ is infinite for some $x$ which is infinitely close but not equal to $c$.
3. The standard part of $f(x)$ is different for different numbers $x$ which are infinitely close but not equal to $c$.

Example $\PageIndex{7}$

$\displaystyle \lim_{x \rightarrow 0} \sqrt{x}$ does not exist because $\sqrt{x}$ is undefined for negative infinitesimal $x \neq 0$. (See Figure $\PageIndex{3(a)}$.)

Example $\PageIndex{8}$

$\displaystyle \lim_{x \rightarrow 0} 1/x^{2}$ does not exist because $1/x^{2}$ is infinite for infinitesimal $x \neq 0$. (See Figure $\PageIndex{3(b)}$.)

Figure $\PageIndex{3}$: Graphs of (a) $y = \sqrt{x}$, (b) $y = \dfrac{1}{x^{2}}$, and (c) $y = \dfrac{x}{|x|}$.

Example $\PageIndex{9}$

$\displaystyle \lim_{x \rightarrow 0} x/|x|$ does not exist because $$st \left(\frac{x}{|x|}\right) = \begin{cases} 1 \quad &\text{if } x > 0, \\ -1 &\text{if } x < 0. \end{cases} \nonumber$$

(See Figure $\PageIndex{3(c)}$).

In the above examples the function behaves differently on one side of the point 0 than it does on the other side. For such functions, one-sided limits are useful.

We say that $$\lim_{x \rightarrow c^{+}} f(x) = L \nonumber$$if whenever $x > c$ and $x \approx c$, $f(x) \approx L$.

$$\lim_{x \rightarrow c^{-}} f(x) = L \nonumber$$means that whenever $x < c$ and $x \approx c$, $f(x) \approx L$. These two kinds of limits, shown in Figure $\PageIndex{4}$, are called the limit from the right and the limit from the left.

Figure $\PageIndex{4}$: One-sided limits.

Theorem $\PageIndex{2}$

A limit has value $L$, $$\lim_{x \rightarrow c} f(c) = L, \nonumber$$

if and only if both one-sided limits exist and are equal to $L$, $$\lim_{x \rightarrow c^{-}} f(c) = \lim_{x \rightarrow c^{+}} f(c) = L. \nonumber$$

Proof

If \(\displaystyle \lim_{x \rightarrow c\) f(x) = L\), it follows at once from the definition that both one-sided limits are $L$.

Assume that both one-sided limits are equal to $L$. Let $x \approx c$, but $x \neq c$. Then either $x < c$ or $x > c$. If $x < c$, then because $\lim_{x \rightarrow c^{-}} f(x) = L$, we have $f(x) \approx L$. On the other hand if $x > c$, then $\lim_{x \rightarrow c^{+}} f(x) = L$ gives $f(x) \approx L$. Thus in either case $f(x) \approx L$. This shows that $\lim_{x \rightarrow c} f(x) = L$.

When a limit does not exist, it is possible that neither one-sided limit exists, that just one of them exists, or that both one-sided limits exist but have different values.

Example $\PageIndex{7}$ (continued)

$\displaystyle \lim_{x \rightarrow 0^{+}} \sqrt{x} = 0$, and $\displaystyle \lim_{x \rightarrow 0^{-}} \sqrt{x}$ does not exist.

Example $\PageIndex{8}$ (continued)

Neither $\displaystyle \lim_{x \rightarrow 0^{+}} 1/x^{2}$ nor $\displaystyle \lim_{x \rightarrow 0^{-}} 1/x^{2}$ exists.

Example $\PageIndex{9}$ (continued)

$\displaystyle \lim_{x \rightarrow 0^{+}} x/|x| = 1$, and  $\displaystyle \lim_{x \rightarrow 0^{-}} x/|x| = -1$. 

Problems for Section 3.3

In each problem below, determine whether or not the limit exists. When the limit exists, find its value. With a calculator, compute some values as $x$ approaches its limit, and see what happens.

1.	$\displaystyle \lim_{t \rightarrow 4} 3t^{2} + t + 1$	2.	$\displaystyle \lim_{\Delta x \rightarrow -1} \dfrac{(\Delta x)^{2} + 2 \Delta x + 1}{\Delta x + 1}$
3.	$\displaystyle \lim_{x \rightarrow c} \sqrt{c - x}$	4.	$\displaystyle \lim_{y \rightarrow 0} \dfrac{1}{y^{5}}$
5.	$\displaystyle \lim_{x \rightarrow 2} \dfrac{x}{x^{2} - 4}$	6.	$\displaystyle \lim_{x \rightarrow 2} \dfrac{x^{2} - 4}{x - 2}$
7.	$\displaystyle \lim_{v \rightarrow 8} \dfrac{\sqrt{8} - \sqrt{v}}{v - 8}$	8.	$\displaystyle \lim_{x \rightarrow 0} \dfrac{\sqrt{x + 1} - 1}{x}$
9.	$\displaystyle \lim_{u \rightarrow 1} \dfrac{\sqrt[3]{u} - 1}{u - 1}$	10.	$\displaystyle \lim_{t \rightarrow 0} \dfrac{t^{3} - 2t^{2} + 4}{3t^{2} - 5t + 7}$
11.	$\displaystyle \lim_{y \rightarrow 0} \left(\sqrt{1 + 1/y} - \sqrt{1/y}\right)$	12.	$\displaystyle \lim_{x \rightarrow 0} \dfrac{(a + x)^{2} - a^{2}}{x}$
13.	$\displaystyle \lim_{y \rightarrow -1} \dfrac{y^{2} + 1}{y + 1}$	14.	$\displaystyle \lim_{x \rightarrow 1} \dfrac{|x - 1|}{x - 1}$
15.	$\displaystyle \lim_{x \rightarrow 1^{+}} \dfrac{|x - 1|}{x - 1}$	16.	$\displaystyle \lim_{x \rightarrow c^{-}} \sqrt{c - x}$
17.	$\displaystyle \lim_{z \rightarrow 1} \sqrt{z + \sqrt{z + \sqrt{z}}}$	18.	$\displaystyle \lim_{x \rightarrow a} \sqrt{|a - x|}$
19.	$\displaystyle \lim_{x \rightarrow 0^{+}} x \sqrt{1 + x^{-2}}$	20.	$\displaystyle \lim_{x \rightarrow 0^{-}} x \sqrt{1 + x^{-2}}$
21.	$\displaystyle \lim_{t \rightarrow 0} \dfrac{1 + 2t^{-1}}{3 - 4t^{-1}}$	22.	$\displaystyle \lim_{x \rightarrow 0} \dfrac{3 + 4x^{-1} - 5x^{-2}}{6 - x^{-1} + 3x^{-2}}$
23.	$\displaystyle \lim_{\Delta x \rightarrow 0} \dfrac{(x + \Delta x)^{2} - x^{2}}{\Delta x}$	24.	$\displaystyle \lim_{\Delta x \rightarrow 0} \dfrac{\frac{1}{x + \Delta x} - \frac{1}{x}}{\Delta x} \ (x \neq 0)$
25.	$\displaystyle \lim_{\Delta t \rightarrow 0} \dfrac{\sqrt{t + \Delta t} - \sqrt{t}}{\Delta t} \ (t > 0)$	26.	$\displaystyle \lim_{\Delta t \rightarrow 0} \dfrac{(t + \Delta t)^{1/5} - t^{1/5}}{\Delta t} \ (t > 0)$
27.	$\displaystyle \lim_{\Delta x \rightarrow 0} \dfrac{(x - \Delta x)^{3} - x^{3}}{\Delta x}$	28.	$\displaystyle \lim_{\Delta x \rightarrow 0} \dfrac{\frac{x + \Delta x}{x + \Delta x + 1} - \frac{x}{x + 1}}{\Delta x}$
29.	$\displaystyle \lim_{\Delta x \rightarrow 0^{-}} \dfrac{|(1 + \Delta x)^{3} - (1 + \Delta x)|}{\Delta x}$	30.	$\displaystyle \lim_{\Delta x \rightarrow 0^{+}} \dfrac{|(1 + \Delta x)^{3} - (1 + \Delta x)|}{\Delta x}$
31.	$\displaystyle \lim_{\Delta x \rightarrow 0^{-}} \dfrac{\sqrt{1 - (1 + \Delta x)^{2}}}{\Delta x}$