Limits Explained: Substitution, Factoring and L'Hôpital's Rule

A limit is the value a function approaches as $x$ gets close to some number, whether or not the function is defined there. Try direct substitution first. If that gives $\frac{0}{0}$, factor and cancel or use L'Hôpital's rule. For limits at infinity, compare the highest powers.

Step 1: Try direct substitution

If the function is continuous at the point, just plug it in:

Polynomials, $\sin$, $\cos$ and $e^x$ are continuous everywhere, so substitution always works for them.

Step 2: 0/0 means "do more work"

Substituting gives $\frac{0}{0}$, which is indeterminate: it does not mean the limit fails to exist, only that you need another method.

Factor and cancel:

So the limit is $2 + 2 = 4$. The graph is the line $y = x + 2$ with a hole at $x = 2$.

Step 3: L'Hôpital's rule

If substitution gives $\frac{0}{0}$ (or $\frac{\infty}{\infty}$), differentiate the top and bottom separately:

The famous one:

Applying it twice:

Limits at infinity

For a rational function, compare the highest powers of $x$:

When a limit does not exist

$\lim_{x \to 0}\frac{1}{x}$ does not exist: from the right the values shoot up to $+\infty$, from the left down to $-\infty$. A two-sided limit only exists if both sides agree.

Common mistakes

• Saying "the limit is undefined" as soon as you see $\frac{0}{0}$.
• Using L'Hôpital's rule when substitution already gives a number.
• Forgetting to check both sides for piecewise or $\frac{1}{x}$-type functions.

Practice questions

1. $\lim_{x \to 1}\frac{x^2 - 1}{x - 1}$
2. $\lim_{x \to \infty}\frac{5x + 2}{x^2}$
3. $\lim_{x \to 0}\frac{e^x - 1}{x}$

Answers: 1) $2$ 2) $0$ 3) $1$

The limit calculator tries substitution, then L'Hôpital's rule, and checks the answer by approaching from both sides. It uses the same differentiation engine as the derivative calculator. See also derivatives for beginners.

Frequently asked questions

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