Limits are the gateway to calculus, and unfortunately also the place where most students give up. The truth is, most limits are easy — direct substitution works. The remaining minority follow a small handful of techniques. This guide walks you through them in increasing difficulty so you can recognise on sight which method to apply.

What a limit really means

The notation $\lim_{x \to a} f(x) = L$ says: as $x$ gets arbitrarily close to $a$ (from either side), $f(x)$ gets arbitrarily close to $L$ . The function does not need to be defined at $a$ — and even if it is, $f(a)$ doesn't have to equal $L$ .

This last point is what makes limits useful. They let us discuss "approaching" behaviour where the function might be undefined or jump.

Method 1: Direct substitution (works ~70% of the time)

If $f$ is continuous at $a$ , $\lim_{x \to a} f(x) = f(a)$ . Plug in. Done.

Example: $\lim_{x \to 3}(x^2 + 2x - 1) = 9 + 6 - 1 = 14$ .

Polynomials, rational functions (where the denominator is nonzero), exp, sin, cos, ln (in domain) — all continuous, all yield to substitution.

Method 2: Factor and cancel (for 0/0 indeterminate form)

If direct substitution gives $\frac{0}{0}$ , try factoring numerator and denominator.

Example: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ .

Direct: $\frac{0}{0}$ ❌
Factor: $\frac{(x-2)(x+2)}{x-2}$ .
Cancel: $\lim_{x \to 2} (x + 2) = 4$ .

The cancelled factor caused the original $0/0$ ; once it's gone, substitute.

Method 3: Rationalise (when factoring fails on radicals)

For limits with square roots that give $0/0$ , multiply by the conjugate.

Example: $\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}$ .

Multiply by $\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}$ : numerator becomes $(x+1) - 1 = x$ .
Cancel $x$ : $\lim_{x \to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{2}$ .

Method 4: Limits at infinity

For rational functions as $x \to \infty$ , divide every term by the highest power of $x$ in the denominator.

Example: $\lim_{x \to \infty} \frac{3x^2 + 2x - 1}{2x^2 - 5}$ .

Divide top and bottom by $x^2$ : $\frac{3 + 2/x - 1/x^2}{2 - 5/x^2}$ .
As $x \to \infty$ , the $1/x$ and $1/x^2$ terms go to $0$ .
Limit: $\frac{3}{2}$ .

Rule of thumb: for $\frac{p(x)}{q(x)}$ as $x \to \infty$ :

If $\deg p < \deg q$ → limit is $0$ .
If $\deg p = \deg q$ → limit is ratio of leading coefficients.
If $\deg p > \deg q$ → limit is $\pm\infty$ .

Method 5: The fundamental trig limit

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

This is the trig version of $\frac{0}{0}$ . Combined with $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$ , it solves most introductory trig limits.

Example: $\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} 3 \cdot \frac{\sin(3x)}{3x} = 3 \cdot 1 = 3$ .

Method 6: L'Hôpital's rule

When 0/0 or ∞/∞ won't yield to algebra, L'Hôpital's rule lets you differentiate top and bottom independently:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad (\text{indeterminate forms only})$

Example: $\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1$ . ✓ (Same answer, faster derivation.)

What is continuity?

A function $f$ is continuous at $a$ if three conditions hold:

$f(a)$ is defined.
$\lim_{x \to a} f(x)$ exists.
The two are equal: $\lim_{x \to a} f(x) = f(a)$ .

Common discontinuities:

Removable (a hole): can be "fixed" by redefining $f(a)$ .
Jump: left and right limits differ.
Infinite: vertical asymptote.

Continuity is the prerequisite for calculus's most powerful theorems: Intermediate Value Theorem, Extreme Value Theorem, and the very definition of differentiability.

Common mistakes

Assuming limit equals function value. Limits and values are different concepts; $\lim_{x \to 0}\frac{\sin x}{x} = 1$ even though the function is undefined at $x = 0$ .
Applying L'Hôpital to non-indeterminate forms. $\lim_{x \to 0}\frac{\sin x + 1}{x + 1}$ is not $\frac{0}{0}$ — direct substitution gives $1$ , period.
Splitting limits incorrectly. $\lim (f + g) = \lim f + \lim g$ only if both individual limits exist.
Forgetting one-sided limits. $\lim_{x \to 0^+} \frac{1}{x} = +\infty$ but $\lim_{x \to 0^-} \frac{1}{x} = -\infty$ — the two-sided limit doesn't exist.

Try it yourself

Drop any limit into the free Limit Calculator — the AI picks the right method (substitution, factoring, conjugate, L'Hôpital) and shows every step.

Related material:

Limits and Continuity Without the Headache

A clear introduction to limits, indeterminate forms, and continuity. Six worked examples — direct substitution, factoring, conjugate, infinity, sin(x)/x, and L'Hôpital — with the standard rules.