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Limits, and why epsilon-delta is not as scary as it looks

By Sara Lin · Last updated April 2026 · 12 min read

A limit answers the question: as $x$ approaches some value $a$, what value does $f(x)$ approach? It is the building block of every other idea in calculus — the derivative is a limit; so is the definite integral — but it is also the topic that makes the most students think they have wandered into the wrong subject.

Most of that fear comes from the formal definition (the “epsilon-delta” one), which is genuinely hard to read on the first pass. The good news is that you can compute almost every limit you will meet at school using a few mechanical rules and never write down an epsilon. This article covers those rules, and then briefly sketches what the formal definition is actually saying.

The intuitive definition

$\lim_{x \to a} f(x) = L$ means: you can make $f(x)$ as close to $L$ as you like by taking $x$ close enough to $a$ (but not equal to it). The “not equal to” part is what distinguishes a limit from plain function evaluation: the limit only cares about what happens near $a$, not at $a$ itself.

Direct substitution: when it works

For most functions you will write down at school — polynomials, rational functions away from their poles, $\sin$, $\cos$, $\exp$, $\ln$ on its domain — the function is continuous, which means you can compute the limit by simply plugging $a$ in:

$$\lim_{x \to 2} (x^2 + 3x) = 4 + 6 = 10.$$

Always try direct substitution first. If you get a number, you are done. If you get something that looks like $\tfrac{0}{0}$, $\tfrac{\infty}{\infty}$, $0 \cdot \infty$, or $\infty - \infty$, you have an indeterminate form and need a different tool.

The seven indeterminate forms

The seven cases that direct substitution cannot resolve are: $\tfrac{0}{0}$, $\tfrac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$, $0^0$, $1^\infty$, and $\infty^0$. The overwhelmingly most common one in school problems is $\tfrac{0}{0}$, and the standard treatment for it is either factoring or L’Hôpital’s rule.

Tool 1 — factor and cancel

$\displaystyle \lim_{x \to 2} \frac{x^2 - 4}{x - 2}$. Direct substitution gives $\tfrac{0}{0}$. But the numerator factors:

$$\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2 \quad \text{for } x \ne 2.$$

The cancelled form has limit $2 + 2 = 4$. The original function and the cancelled one differ only at $x = 2$, and the limit does not care about that point.

Factoring works whenever the indeterminate form arises from a common factor in the numerator and denominator, which is most of the time on school exam questions. If the numerator does not factor obviously, try multiplying by a conjugate (for limits involving square roots) or expanding (for limits involving binomial powers).

Tool 2 — L’Hôpital’s rule

For limits of the form $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$,

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)},$$

provided the right-hand limit exists. Note this is not the quotient rule; you differentiate $f$ and $g$ separately, not as a quotient. A worked example. $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}$ is $\tfrac{0}{0}$. Apply L’Hôpital:

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

This particular limit is one you should memorise: $\lim_{x \to 0} \tfrac{\sin x}{x} = 1$. It comes up everywhere.

One-sided limits

Sometimes a function approaches different values from the left and the right of $a$. Then the two-sided limit does not exist, but the one-sided limits do:

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

A two-sided limit exists only when both one-sided limits exist and are equal. This is the standard test for continuity at a point.

Limits at infinity

$\lim_{x \to \infty} f(x)$ asks where $f$ ends up as $x$ grows without bound. For rational functions, the rule is to compare the degrees of the numerator and denominator:

$\lim_{x \to \infty} \tfrac{3x^2 + 1}{2x^2 - 5} = \tfrac{3}{2}$ — same degree, ratio of leading coefficients.

The mistakes I see most often

1. Substituting first instead of last

For $\lim_{x \to 2} \tfrac{x^2 - 4}{x - 2}$, plugging $x = 2$ first gives $\tfrac{0}{0}$ and a feeling that the limit does not exist. It does; you have to factor first. The rule is: try direct substitution, and if it gives an indeterminate form, do something else before concluding anything.

2. Using L’Hôpital on a determinate form

$\lim_{x \to 0} \tfrac{x + 1}{x + 2}$ is just $\tfrac{1}{2}$ by direct substitution. Applying L’Hôpital here gives $\tfrac{1}{1} = 1$, which is wrong. L’Hôpital’s rule applies only to $\tfrac{0}{0}$ and $\tfrac{\infty}{\infty}$ forms. Always check the form first.

3. Forgetting that L’Hôpital differentiates top and bottom separately

Differentiating $\tfrac{f}{g}$ as a quotient gives the quotient rule, which has cross terms. L’Hôpital differentiates $f$ and $g$ each by themselves and takes the limit of the ratio. Different rules, different answers.

4. Treating $\infty$ as a number

$\infty - \infty$ is not $0$; $\tfrac{\infty}{\infty}$ is not $1$; $0 \cdot \infty$ is not $0$. These are all indeterminate forms whose value depends on the specific functions involved. Each requires its own analysis.

What the epsilon-delta definition is actually saying

For completeness: the formal definition is that $\lim_{x \to a} f(x) = L$ iff for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $0 < |x - a| < \delta$ implies $|f(x) - L| < \varepsilon$. That sentence translates roughly as: “no matter how strict you make the requirement that $f(x)$ be near $L$ ($\varepsilon$), I can guarantee it by making $x$ near enough to $a$ ($\delta$).” You will not need to write epsilon-delta proofs for school work, but recognising what the definition is doing — making “arbitrarily close” precise — is part of understanding why limits work.

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