Limit Laws

Estimating limits from graphs and tables works, but it’s slow and only approximate. The limit laws let you find limits exactly, by breaking a complicated expression into simple pieces. They also explain why, for most functions you know, you can find a limit just by substituting.

Key ideas

Two basic limits

For any constants $a$ and $c$:

$$\lim_{x \to a} c = c \qquad\qquad \lim_{x \to a} x = a$$

A constant function stays at $c$, and the function $y = x$ approaches whatever $x$ approaches.

The limit laws

Suppose $\displaystyle\lim_{x \to a} f(x) = L$ and $\displaystyle\lim_{x \to a} g(x) = M$, where $L$ and $M$ are real numbers. Then:

Law	Rule
Sum	$\displaystyle\lim_{x \to a} [f(x) + g(x)] = L + M$
Difference	$\displaystyle\lim_{x \to a} [f(x) - g(x)] = L - M$
Constant multiple	$\displaystyle\lim_{x \to a} [k\,f(x)] = kL$
Product	$\displaystyle\lim_{x \to a} [f(x)\,g(x)] = LM$
Quotient	$\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$, as long as $M \ne 0$
Power	$\displaystyle\lim_{x \to a} [f(x)]^n = L^n$ for a positive integer $n$
Root	$\displaystyle\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$ (for even $n$, need $L \gt 0$)

In words: the limit of a sum is the sum of the limits, and so on. The big condition is that each separate limit must exist (and the bottom limit must not be $0$ for the quotient law).

Direct substitution

Using the laws over and over shows that for a polynomial $p$,

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

and for a rational function $\dfrac{p(x)}{q(x)}$, the limit is $\dfrac{p(a)}{q(a)}$ as long as $q(a) \ne 0$.

The same is true for roots, exponentials, logarithms, and trig functions at any point in their domains. So the first thing to try with any limit is to substitute. If you get a real number, that’s the answer. If you get $\tfrac{0}{0}$ or a nonzero number over $0$, you need more work (see algebraic techniques and infinite limits).

Composite functions

If $\displaystyle\lim_{x \to a} g(x) = L$ and $f$ is continuous at $L$, then

$$\lim_{x \to a} f(g(x)) = f\left(\lim_{x \to a} g(x)\right) = f(L)$$

You can “move the limit inside” a continuous outer function. If the outer function is not continuous at $L$ (for example, it has a jump there), you have to look at how $g(x)$ approaches $L$ from each side (see Practice 8).

Worked examples

Example 1: Using given limits

Suppose $\displaystyle\lim_{x \to 2} f(x) = 4$ and $\displaystyle\lim_{x \to 2} g(x) = -3$. Find:

(a) $\displaystyle\lim_{x \to 2} \left[3f(x) - g(x)^2\right]$

(b) $\displaystyle\lim_{x \to 2} \frac{f(x)}{g(x) + 1}$

(c) $\displaystyle\lim_{x \to 2} \sqrt{f(x)}\,g(x)$

Solution.

(a) Use the difference, constant multiple, and power laws:

$$\lim_{x \to 2} \left[3f(x) - g(x)^2\right] = 3(4) - (-3)^2 = 12 - 9 = 3$$

(b) The bottom approaches $-3 + 1 = -2$, which isn’t $0$, so the quotient law applies:

$$\lim_{x \to 2} \frac{f(x)}{g(x) + 1} = \frac{4}{-2} = -2$$

(c) Use the root and product laws:

$$\lim_{x \to 2} \sqrt{f(x)}\,g(x) = \sqrt{4}\,(-3) = -6$$

Example 2: A polynomial

Find $\displaystyle\lim_{x \to -1} (2x^3 - x + 5)$.

Solution. Here’s the full chain of laws once, so you can see why substitution works:

$$\begin{aligned} \lim_{x \to -1} (2x^3 - x + 5) &= 2\left(\lim_{x \to -1} x\right)^3 - \lim_{x \to -1} x + \lim_{x \to -1} 5 && \text{sum, difference, multiple, power} \\ &= 2(-1)^3 - (-1) + 5 \\ &= -2 + 1 + 5 = 4 \end{aligned}$$

From now on you can skip straight to substituting $x = -1$.

Example 3: A rational function

Find $\displaystyle\lim_{x \to 3} \frac{x^2 + 1}{2x - 1}$.

Solution. Check the denominator first: $2(3) - 1 = 5 \ne 0$. So substitute:

$$\lim_{x \to 3} \frac{x^2 + 1}{2x - 1} = \frac{9 + 1}{5} = 2$$

Example 4: Composite functions

Find $\displaystyle\lim_{x \to 0} \cos(x^2 + \pi)$ and $\displaystyle\lim_{x \to 0} e^{\cos x}$. (Radians.)

Solution. The inside of the first limit approaches $0 + \pi = \pi$, and cosine is continuous everywhere, so

$$\lim_{x \to 0} \cos(x^2 + \pi) = \cos \pi = -1$$

In the second, $\cos x \to \cos 0 = 1$, and $e^u$ is continuous everywhere, so

$$\lim_{x \to 0} e^{\cos x} = e^1 = e$$

Common mistakes

Using the quotient law when the bottom approaches 0. The quotient law needs $M \ne 0$. For $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$, substituting gives $\tfrac{0}{0}$, which is not an answer. It’s a signal to simplify first.

Writing 0/0 = 1 or 0/0 = 0. $\tfrac{0}{0}$ is called an indeterminate form: the limit could be any number, or not exist. You can’t tell until you do more work.

Splitting a limit when a piece doesn’t exist. The laws only work when each separate limit exists. $\displaystyle\lim_{x \to 0} \left(\frac{|x|}{x} - \frac{|x|}{x}\right) = 0$, even though $\displaystyle\lim_{x \to 0} \frac{|x|}{x}$ does not exist (see Practice 9).

Dropping the limit notation. Writing $\lim$ on one line and then a bare expression on the next (or $\lim$ in front of a number) loses marks on free-response questions. Keep $\displaystyle\lim_{x \to a}$ in front until you actually substitute.

Moving a limit inside a function that isn’t continuous. $\displaystyle\lim f(g(x)) = f(\lim g(x))$ needs $f$ to be continuous at the inside limit. Check this for piecewise outer functions.

Practice

1. (Warm-up) Find $\displaystyle\lim_{x \to 4} (x^2 - 3x + 2)$.

Solution

Substitute:

$$\lim_{x \to 4} (x^2 - 3x + 2) = 16 - 12 + 2 = 6$$

2. (Warm-up) If $\displaystyle\lim_{x \to 1} f(x) = 5$ and $\displaystyle\lim_{x \to 1} g(x) = 2$, find $\displaystyle\lim_{x \to 1} [2f(x) + g(x)]$.

Solution

$$\lim_{x \to 1} [2f(x) + g(x)] = 2(5) + 2 = 12$$

3. (Warm-up) Find $\displaystyle\lim_{x \to -2} \frac{x + 5}{x^2 + 1}$.

Solution

The denominator is $(-2)^2 + 1 = 5 \ne 0$, so substitute:

$$\lim_{x \to -2} \frac{x + 5}{x^2 + 1} = \frac{3}{5}$$

4. (Core) If $\displaystyle\lim_{x \to 1} f(x) = 5$ and $\displaystyle\lim_{x \to 1} g(x) = 2$, find $\displaystyle\lim_{x \to 1} \frac{f(x)\,g(x)}{f(x) - g(x)}$.

Solution

The bottom approaches $5 - 2 = 3 \ne 0$, so

$$\lim_{x \to 1} \frac{f(x)\,g(x)}{f(x) - g(x)} = \frac{5 \cdot 2}{3} = \frac{10}{3}$$

5. (Core) Find $\displaystyle\lim_{x \to \pi/3} 2\sin x \cos x$. (Radians: $\tfrac{\pi}{3}$ is $60^\circ$.)

Solution

Sine and cosine are continuous, so substitute the special-angle values:

$$2\sin\frac{\pi}{3}\cos\frac{\pi}{3} = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$$

6. (Core) Find $\displaystyle\lim_{x \to \ln 2} e^{2x}$.

Solution

The exponential function is continuous, so substitute:

$$e^{2\ln 2} = \left(e^{\ln 2}\right)^2 = 2^2 = 4$$

7. (Core) Suppose $\displaystyle\lim_{x \to 3} f(x) = 6$ and $\displaystyle\lim_{x \to 3} [f(x) + 2g(x)] = 10$. Find $\displaystyle\lim_{x \to 3} g(x)$.

Solution

Let $M = \displaystyle\lim_{x \to 3} g(x)$. (It exists, because $g(x) = \tfrac{1}{2}\left([f(x) + 2g(x)] - f(x)\right)$ is built from two limits that exist.) By the sum and constant multiple laws:

$$6 + 2M = 10 \quad\Rightarrow\quad M = 2$$

8. (Challenge) Let $g(x) = x^2 - 1$ and $f(u) = \begin{cases} u + 1, & u \lt 0 \\ u - 1, & u \ge 0 \end{cases}$.

• (a) Find $\displaystyle\lim_{x \to 0} f(g(x))$.
• (b) Find $\displaystyle\lim_{x \to 1} f(g(x))$, if it exists.

Solution

(a) As $x \to 0$, $g(x) \to -1$. Near $u = -1$, $f(u) = u + 1$, which is continuous there, so

$$\lim_{x \to 0} f(g(x)) = f(-1) = 0$$

(b) As $x \to 1$, $g(x) \to 0$, but $f$ has a jump at $u = 0$, so look at each side.

For $x$ slightly less than $1$, $x^2 \lt 1$, so $g(x) \lt 0$ and $f(g(x)) = g(x) + 1 \to 0 + 1 = 1$.

For $x$ slightly greater than $1$, $g(x) \gt 0$ and $f(g(x)) = g(x) - 1 \to 0 - 1 = -1$.

The one-sided limits are $1$ and $-1$, so the limit does not exist.

9. (Challenge) Give an example of functions $f$ and $g$ where $\displaystyle\lim_{x \to 0} f(x)$ and $\displaystyle\lim_{x \to 0} g(x)$ do not exist, but $\displaystyle\lim_{x \to 0} [f(x) + g(x)]$ does. Explain why this doesn’t break the sum law.

Solution

Take $f(x) = \dfrac{|x|}{x}$ and $g(x) = -\dfrac{|x|}{x}$. Each jumps between $-1$ and $1$ at $0$, so neither limit exists. But $f(x) + g(x) = 0$ for every $x \ne 0$, so

$$\lim_{x \to 0} [f(x) + g(x)] = 0$$

The sum law only says what happens when both limits exist. It says nothing when they don’t, so there’s no contradiction.