Limit Laws and Evaluating Limits Algebraically

The limit laws

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then

Law	Statement
Sum and difference	$\lim (f \pm g) = L \pm M$
Constant multiple	$\lim (c\,f) = cL$
Product	$\lim (f g) = LM$
Quotient	$\lim \tfrac{f}{g} = \tfrac{L}{M}$, provided $M \ne 0$
Power	$\lim f^{\,n} = L^n$
Root	$\lim \sqrt[n]{f} = \sqrt[n]{L}$, when the root is defined

Together with the two simplest limits, $\lim_{x \to a} c = c$ and $\lim_{x \to a} x = a$, these build every polynomial. So for any polynomial $p$,

$$\lim_{x \to a} p(x) = p(a)$$

and for a rational function $\tfrac{p}{q}$, the same holds wherever $q(a) \ne 0$. Evaluating a limit by plugging in is called direct substitution.

When substitution gives 0/0

Try direct substitution on $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 2x - 3}$: the top and bottom are both $0$. The quotient law does not apply, because the limit of the denominator is $0$.

$\tfrac{0}{0}$ is an indeterminate form. It does not mean the limit is $0$, or $1$, or that it fails to exist. It means this method gave no information, and the expression has to be rewritten before the limit shows.

Why canceling is allowed

Factor:

$$\frac{x^2 - 9}{x^2 - 2x - 3} = \frac{(x - 3)(x + 3)}{(x - 3)(x + 1)} = \frac{x + 3}{x + 1} \qquad (x \ne 3)$$

The two sides are not the same function: the left one is undefined at $x = 3$ and the right one is not. But they agree at every other $x$.

A limit as $x \to 3$ uses only values with $x \ne 3$. On those values the functions are identical, so their limits are identical. Canceling is legal precisely because a limit ignores the point it approaches.

$$\lim_{x \to 3} \frac{x + 3}{x + 1} = \frac{6}{4} = \frac{3}{2}$$

Conjugates and compound fractions

Radicals often hide the canceling factor. Multiplying by the conjugate brings it out:

$$\frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} = \frac{(x + 4) - 4}{x\left(\sqrt{x + 4} + 2\right)} = \frac{1}{\sqrt{x + 4} + 2}$$

Now $x \to 0$ gives $\tfrac{1}{4}$. Compound fractions work the same way: combine into one fraction, then look for the common factor.

The squeeze theorem

Some functions resist algebra entirely. If $f$ is trapped between two functions that approach the same limit, it has no choice:

Squeeze theorem

If $g(x) \le f(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$), and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$.

• y = x²
• y = −x²
• y = x² sin(1/x)

For $x^2 \sin\!\left(\tfrac{1}{x}\right)$: since $-1 \le \sin\!\left(\tfrac{1}{x}\right) \le 1$, multiplying by $x^2 \ge 0$ gives $-x^2 \le x^2 \sin\!\left(\tfrac{1}{x}\right) \le x^2$. Both bounds approach $0$, so the function does too.

The limit of sin x / x

For $0 < x < \tfrac{\pi}{2}$, compare three areas in the unit circle: the triangle inside the sector of angle $x$, the sector itself, and the larger triangle reaching up to the tangent line. They give

$$\frac{\sin x}{2} < \frac{x}{2} < \frac{\tan x}{2}$$

Dividing through by $\tfrac{\sin x}{2}$ and taking reciprocals, $\cos x < \tfrac{\sin x}{x} < 1$. The same holds for $-\tfrac{\pi}{2} < x < 0$, because every term is even. As $x \to 0$, $\cos x \to 1$, so the squeeze theorem gives

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

This is the limit the derivative of $\sin x$ rests on. It requires radians: in degrees the sector’s area is not $\tfrac{x}{2}$, and the limit is $\tfrac{\pi}{180}$ instead.

Worked examples

Common mistakes

Practice problems