Introduction to Limits

How fast is a falling stone moving at exactly $t = 2$ seconds? Speed is distance divided by time, but “at an instant” no time passes, so the formula gives $\tfrac{0}{0}$. Calculus gets around this with limits: instead of asking what happens at a point, you ask what happens as you get closer and closer to it. Every big idea in calculus (derivatives, integrals, continuity) is built on limits.

Key ideas

Can change happen at an instant?

A stone dropped from a bridge falls $d(t) = 4.9t^2$ metres in $t$ seconds. Its average speed from $t = 2$ to $t = 2 + h$ is

$$\frac{d(2 + h) - d(2)}{h}$$

You can’t put $h = 0$ (that divides by zero), but you can make $h$ tiny:

$h$ (s)	$0.1$	$0.01$	$0.001$
average speed (m/s)	$20.09$	$19.649$	$19.6049$

The averages close in on $19.6$ m/s. That number, the value the averages approach, is the stone’s speed at the instant $t = 2$. It’s a limit.

What a limit means

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

is read “the limit of $f(x)$ as $x$ approaches $a$ is $L$.” It means you can make $f(x)$ as close to $L$ as you like by taking $x$ close enough to $a$, from both sides, but not equal to $a$.

The key word is approaches. A limit describes what $f$ does near $a$, not at $a$. The value $f(a)$ might equal $L$, might be something else, or might not exist at all. None of that changes the limit.

One-sided limits

Sometimes a function does different things on each side of $a$, so we look at one side at a time.

Notation	Meaning
$\displaystyle\lim_{x \to a^-} f(x)$	left-hand limit: $x$ approaches $a$ from values less than $a$
$\displaystyle\lim_{x \to a^+} f(x)$	right-hand limit: $x$ approaches $a$ from values greater than $a$

The two-sided limit exists only when both one-sided limits exist and are equal:

$$\lim_{x \to a} f(x) = L \quad\text{exactly when}\quad \lim_{x \to a^-} f(x) = L \ \text{ and } \ \lim_{x \to a^+} f(x) = L$$

An open dot means “not included”; a filled dot shows the actual value of $f$.

When a limit does not exist

A two-sided limit fails to exist (DNE) in three typical ways:

1. Jump: the left and right limits are different numbers (like $x = 3$ in the graph).
2. Unbounded: $f(x)$ grows without bound near $a$, as near a vertical asymptote. We write $\displaystyle\lim_{x \to a} f(x) = \infty$ to describe how the limit fails, but $\infty$ is not a number, so the limit still does not exist. (See infinite limits.)
3. Oscillation: $f(x)$ keeps bouncing between values and never settles down, like $\sin\left(\dfrac{1}{x}\right)$ near $0$.

Worked examples

Example 1: A hole in the graph

Use the graph above to find $\displaystyle\lim_{x \to 1} f(x)$ and $f(1)$.

Solution. Trace the curve toward $x = 1$ from the left and from the right. From both sides, the $y$-values head toward the hole at height $2$:

$$\lim_{x \to 1^-} f(x) = 2, \qquad \lim_{x \to 1^+} f(x) = 2 \quad\Rightarrow\quad \lim_{x \to 1} f(x) = 2$$

The filled dot shows the actual value: $f(1) = 4$. The limit and the function value are different, and that’s allowed.

Example 2: A jump

Use the same graph to find the one-sided limits at $x = 3$, the two-sided limit, and $f(3)$.

Solution. Coming from the left, the curve rises toward the open dot at height $3$. Coming from the right, the line comes down toward the filled dot at height $1$:

$$\lim_{x \to 3^-} f(x) = 3, \qquad \lim_{x \to 3^+} f(x) = 1$$

The one-sided limits are different, so $\displaystyle\lim_{x \to 3} f(x)$ does not exist. The filled dot gives $f(3) = 1$.

Example 3: A piecewise function

Let

$$g(x) = \begin{cases} x^2 + 1, & x \lt 2 \\ 1, & x = 2 \\ 7 - x, & x \gt 2 \end{cases}$$

Find $\displaystyle\lim_{x \to 2} g(x)$.

Solution. For the left-hand limit, use the piece for $x \lt 2$. For the right-hand limit, use the piece for $x \gt 2$:

$$\lim_{x \to 2^-} (x^2 + 1) = 4 + 1 = 5, \qquad \lim_{x \to 2^+} (7 - x) = 7 - 2 = 5$$

Both sides agree, so $\displaystyle\lim_{x \to 2} g(x) = 5$. The value $g(2) = 1$ plays no part in the limit.

Example 4: Absolute value

Find $\displaystyle\lim_{x \to 0} \frac{|x|}{x}$.

Solution. The expression isn’t defined at $0$, so look at each side.

For $x \gt 0$, $|x| = x$, so $\dfrac{|x|}{x} = 1$. For $x \lt 0$, $|x| = -x$, so $\dfrac{|x|}{x} = -1$.

$$\lim_{x \to 0^-} \frac{|x|}{x} = -1, \qquad \lim_{x \to 0^+} \frac{|x|}{x} = 1$$

The one-sided limits differ, so the limit does not exist.

Common mistakes

Using f(a) as the limit. The limit is about values near $a$. In Example 1, the limit is $2$ even though $f(1) = 4$. Always trace the graph toward the point instead of reading the dot at the point.

Saying the limit doesn’t exist just because f(a) is undefined. A hole in the graph doesn’t stop a limit from existing. If both sides approach the same height, that height is the limit.

Checking only one side. For a piecewise function or an absolute value, you must find both one-sided limits. If they disagree, the two-sided limit does not exist.

Using the wrong piece for a one-sided limit. For $x \to 2^-$, use the piece whose condition includes values just less than $2$. The piece written for $x = 2$ itself is never used for a limit.

Treating infinity as a number. Writing $\displaystyle\lim_{x \to a} f(x) = \infty$ is a useful description, but the limit does not exist. On a test, if a question asks “does the limit exist?”, the answer is no.

Practice

1. (Warm-up) Use the graph of $f$ above to find $\displaystyle\lim_{x \to 1^-} f(x)$ and $\displaystyle\lim_{x \to 1^+} f(x)$.

Solution

From both sides, the curve heads toward the hole at $(1, 2)$:

$$\lim_{x \to 1^-} f(x) = 2, \qquad \lim_{x \to 1^+} f(x) = 2$$

2. (Warm-up) Write this sentence in limit notation: “As $x$ approaches $4$ from the left, $f(x)$ approaches $-2$.”

Solution

$$\lim_{x \to 4^-} f(x) = -2$$

3. (Warm-up) Suppose $\displaystyle\lim_{x \to 5^-} f(x) = 7$, $\displaystyle\lim_{x \to 5^+} f(x) = 7$, and $f(5) = 2$. What is $\displaystyle\lim_{x \to 5} f(x)$?

Solution

Both one-sided limits are $7$, so $\displaystyle\lim_{x \to 5} f(x) = 7$. The value $f(5) = 2$ doesn’t matter.

4. (Core) Let $p(x) = \begin{cases} 3x - 2, & x \lt 1 \\ x^2, & x \ge 1 \end{cases}$. Find $\displaystyle\lim_{x \to 1} p(x)$, if it exists.

Solution

$$\lim_{x \to 1^-} (3x - 2) = 1, \qquad \lim_{x \to 1^+} x^2 = 1$$

Both sides agree, so $\displaystyle\lim_{x \to 1} p(x) = 1$.

5. (Core) Let $q(x) = \begin{cases} x + 4, & x \lt -1 \\ 2 - x^2, & x \gt -1 \end{cases}$. Find both one-sided limits at $x = -1$. Does $\displaystyle\lim_{x \to -1} q(x)$ exist?

Solution

$$\lim_{x \to -1^-} (x + 4) = 3, \qquad \lim_{x \to -1^+} (2 - x^2) = 2 - 1 = 1$$

The one-sided limits are different ($3 \ne 1$), so the limit does not exist.

6. (Core) Find the value of $k$ that makes $\displaystyle\lim_{x \to 2} r(x)$ exist, where $r(x) = \begin{cases} kx + 1, & x \lt 2 \\ x^2 - k, & x \ge 2 \end{cases}$. What is the limit?

Solution

The left-hand limit is $2k + 1$ and the right-hand limit is $4 - k$. Set them equal:

$$2k + 1 = 4 - k \quad\Rightarrow\quad 3k = 3 \quad\Rightarrow\quad k = 1$$

Then both sides approach $2(1) + 1 = 3$, so the limit is $3$.

7. (Core) A stone dropped from a cliff falls $d(t) = 4.9t^2$ metres in $t$ seconds. Find its average speed from $t = 1$ to $t = 1 + h$ for $h = 0.1$, $0.01$, and $0.001$. Use the results to estimate its speed at exactly $t = 1$.

Solution

The average speed is $\dfrac{4.9(1 + h)^2 - 4.9}{h}$.

$h$	$0.1$	$0.01$	$0.001$
average speed (m/s)	$10.29$	$9.849$	$9.8049$

The values approach $9.8$, so the speed at $t = 1$ is about $9.8$ m/s.

(Expanding shows why: $\dfrac{4.9(1 + 2h + h^2) - 4.9}{h} = 9.8 + 4.9h$, which approaches $9.8$ as $h \to 0$.)

8. (Challenge) Find $a$ and $b$ so that $\displaystyle\lim_{x \to -1} f(x)$ and $\displaystyle\lim_{x \to 1} f(x)$ both exist, where

$$f(x) = \begin{cases} x + 2, & x \lt -1 \\ ax + b, & -1 \le x \lt 1 \\ 3x^2, & x \ge 1 \end{cases}$$

Solution

At $x = -1$: the left side approaches $-1 + 2 = 1$ and the right side approaches $-a + b$. So $-a + b = 1$.

At $x = 1$: the left side approaches $a + b$ and the right side approaches $3(1)^2 = 3$. So $a + b = 3$.

Adding the equations gives $2b = 4$, so $b = 2$ and then $a = 1$.

9. (Challenge) Explain why $\displaystyle\lim_{x \to 0} \sin\left(\frac{\pi}{x}\right)$ does not exist. (Use radians.)

Solution

Look at two lists of $x$-values that both approach $0$.

When $x = 1, \tfrac{1}{2}, \tfrac{1}{3}, \dots$, we get $\dfrac{\pi}{x} = \pi, 2\pi, 3\pi, \dots$, so $\sin\left(\dfrac{\pi}{x}\right) = 0$ every time.

When $x = 2, \tfrac{2}{5}, \tfrac{2}{9}, \dots$, we get $\dfrac{\pi}{x} = \dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2}, \dots$, so $\sin\left(\dfrac{\pi}{x}\right) = 1$ every time.

No matter how close you get to $0$, the function keeps taking both values $0$ and $1$ (in fact, every value from $-1$ to $1$). It never settles on one number, so the limit does not exist.