Understanding Limits

What you will learn

• Understand a limit as the value a function approaches.
• Find limits by substitution, factorising and at infinity.
• Know when a limit does not exist.

Before you start: you should be comfortable with functions and factorising. See Functions and Graphs.

The idea

Consider the function

$f(x) = \frac{x^2 - 9}{x - 3}$

Try to evaluate it at $x = 3$ and you get $\frac{0}{0}$, which is meaningless. But look at values near $3$:

As $x$ gets closer and closer to $3$ from either side, $f(x)$ gets closer and closer to $6$. We say the limit of $f(x)$ as $x$ approaches $3$ is $6$, and write

$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6$

The limit cares about where the function is heading, not what happens exactly at the point.

One-sided limits

Sometimes a function heads to different places depending on the direction of approach. We write $\lim_{x \to a^-}$ for approaching from the left and $\lim_{x \to a^+}$ from the right. The ordinary (two-sided) limit exists only when both one-sided limits exist and agree.

Limits at infinity

We can also ask what happens as $x$ grows without bound. As $x \to \infty$, $\frac{1}{x}$ shrinks towards $0$, so $\lim_{x \to \infty} \frac{1}{x} = 0$. This describes horizontal asymptotes, the lines a graph approaches far out to the left or right.

Worked examples

Example 1: Factorise and cancel

Evaluate $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ algebraically.

1. Direct substitution gives $\frac{0}{0}$, so we need to simplify.
2. Factorise the numerator as a difference of two squares: $\dfrac{(x - 3)(x + 3)}{x - 3}$.
3. For $x \neq 3$ we can cancel: the expression equals $x + 3$.
4. Now substitute: $3 + 3 = 6$.

Answer: $6$, matching the table.

Example 2: Limit at infinity

Find $\displaystyle\lim_{x \to \infty} \frac{4x^2 + 1}{2x^2 - 5x}$.

1. Divide every term by the highest power of $x$ in the denominator, $x^2$: $\dfrac{4 + \frac{1}{x^2}}{2 - \frac{5}{x}}$.
2. As $x \to \infty$, $\frac{1}{x^2} \to 0$ and $\frac{5}{x} \to 0$.
3. The limit becomes $\frac{4}{2} = 2$.

Answer: $2$, so the graph has a horizontal asymptote at $y = 2$.

Example 3: A limit that does not exist

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

1. For $x > 0$, $|x| = x$, so the expression is $1$. The right-hand limit is $1$.
2. For $x < 0$, $|x| = -x$, so the expression is $-1$. The left-hand limit is $-1$.
3. The one-sided limits disagree.

Answer: the limit does not exist.

Getting $\frac{0}{0}$ on substitution does not mean the limit is $0$ or undefined. It means “do more work”, such as factorising, rationalising or using a known limit.

Why it matters

Limits are the foundation stone of calculus. The derivative is defined as a limit of slopes, and the integral as a limit of sums. Outside calculus, limits describe long-term behaviour: the maximum speed a falling skydiver approaches, the value an investment tends towards, or how a drug concentration settles over time.