Computing Limits

Main idea

Getting $\frac00$ when you plug in doesn't mean there's no limit. It means the top and bottom share something that's making both vanish. Simplify it away, by factoring, rationalizing or combining fractions, and then plugging in works.

Squeeze theorem

{\htmlData{sym=0}{g(x)}} \le f(x) \le {\htmlData{sym=1}{h(x)}} \text{ and } \lim g = \lim h = L \implies \lim f = L

What the symbols mean

$g(x)$
Lower bound
A function that stays below f near a.
$h(x)$
Upper bound
A function that stays above f near a. If both bounds approach L, f is squeezed to L.

Explanation

Computing a limit is simply finding where a function is heading, and most of the work comes when plugging in gives the useless answer $\frac00$ . All we're really doing is cleaning up the expression and trying again. For functions too wild to clean up, there's the squeeze theorem, which looks fancier than it is: $g(x)$ (a floor that stays under $f$ ), $h(x)$ (a ceiling that stays over $f$ ) and $L$ (the spot where floor and ceiling meet). If the floor and ceiling both close in on $L$ , the function stuck between them has to go there too. That's all it says, so don't let the inequalities scare you off.

The last lesson ended with plugging in for nice functions, and a warning about what happens when that gives $\frac00$ . Here's the good news: $\frac00$ is a signal, not an answer. If the top and bottom both go to $0$ at $x = a$ , they usually share a factor like $(x - a)$ . Remember $\dfrac{x^2 - 1}{x - 1}$ from earlier: both pieces vanish at $x = 1$ because both contain $x - 1$ .

Since a limit never looks at $x = a$ itself, you're allowed to cancel that shared piece. What's left is usually a nice function that's defined at $a$ , and then plugging in works again. So the whole game is finding a way to expose the shared piece.

Your toolkit:

Factor and cancel. $\dfrac{x^2 - 9}{x - 3} = x + 3$ for $x \ne 3$ .
Rationalize. A square root hides the factor, so multiply by the conjugate to bring it out: $\dfrac{\sqrt x - 2}{x - 4}\cdot\dfrac{\sqrt x + 2}{\sqrt x + 2} = \dfrac{1}{\sqrt x + 2}$ .
Combine fractions. When there are fractions stacked inside fractions, put them over one denominator first, and the shared factor shows up.
Squeeze. Some functions are too wild to simplify. Then you can trap them between two simple functions that head to the same place. Since $-1 \le \cos\frac1x \le 1$ , multiplying through by $x^2$ gives $-x^2 \le x^2\cos\frac1x \le x^2$ . Both bounds go to $0$ , and the function stuck between them has no choice but to go to $0$ too. That's the squeeze theorem.

See it

This is x² sin(1/x), shown between the parabolas x² and −x². Near 0 it wiggles faster and faster, so there's no way to simplify it, but notice it never escapes the two parabolas. They both pinch down to 0, so it has nowhere to go but 0. That's the squeeze theorem at work.

Worked examples

Example 1

Find $\displaystyle\lim_{x\to4}\frac{\sqrt x - 2}{x - 4}$ .

1
Plugging in gives $\frac00$ , and there's a square root on top, so multiply by $\frac{\sqrt x + 2}{\sqrt x + 2}$ . The top becomes $(\sqrt x)^2 - 4 = x - 4$ .

Example 2

Find $\displaystyle\lim_{x\to0} x^2\cos\frac1x$ .

1
$\cos\frac1x$ has no limit at $0$ , so we can't use the product rule. But cosine always stays between $-1$ and $1$ , so $-x^2 \le x^2\cos\frac1x \le x^2$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find $\displaystyle\lim_{x\to-3}\frac{x^2 + x - 6}{x + 3}$ .

Problem 2+5 XP

Find $\displaystyle\lim_{h\to0}\frac{(2 + h)^2 - 4}{h}$ .

Problem 3+5 XP

Find $\displaystyle\lim_{x\to9}\frac{\sqrt x - 3}{x - 9}$ .

Problem 4+5 XP

Find $\displaystyle\lim_{x\to0}\frac{\frac{1}{x+2} - \frac12}{x}$ .

Problem 5+5 XP

Find $\displaystyle\lim_{x\to0} x\sin\frac1x$ .

Problem 6+5 XP

A skateboard ramp follows $y = \sqrt x$ (in meters). Find its steepness at $x = 4$ by computing $\displaystyle\lim_{h\to0}\frac{\sqrt{4 + h} - 2}{h}$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find $\displaystyle\lim_{x\to8}\frac{\sqrt[3]x - 2}{x - 8}$ .

Where it's used

Physics formulas often look undefined at special points, like average velocity over a zero-length time interval, until you simplify and take the limit.

Why Small Swings Keep TimeCalc I · Physics

Limit Properties Infinite Limits