Limits in Calculus: A Practical Guide

The idea of a limit

Consider $f(x) = \frac{x^2 - 9}{x - 3}$. At $x = 3$ the formula gives $\frac{0}{0}$, so $f(3)$ is undefined. Yet if you evaluate $f$ at 2.9, 2.99, 3.01 and 3.1, you get 5.9, 5.99, 6.01 and 6.1. The outputs close in on 6. We say the limit of $f(x)$ as $x$ approaches 3 is 6, and write:

The value of the function at the point itself is irrelevant to the limit. Only the behaviour nearby matters. Graphically, the graph of $f$ is the line $y = x + 3$ with a hole at $(3, 6)$, and the limit is the height of the hole.

Informally, $\lim_{x \to a} f(x) = L$ means that $f(x)$ can be made as close as you like to $L$ by taking $x$ close enough to $a$, but not equal to it. The precise version of that sentence, the epsilon-delta definition, turns this into a statement that can be proved.

Method 1: direct substitution

Polynomials, rational functions where the denominator is not zero, roots, exponentials, logarithms and trig functions are continuous on their domains. For continuous functions, the limit is simply the function's value, so substitute.

For example, $\lim_{x \to 2}(x^2 + 3x) = 4 + 6 = 10$, and $\lim_{x \to 0}\frac{\cos x}{x + 1} = \frac{1}{1} = 1$. Always try substitution first. If it gives a real number, you are done. If it gives a nonzero number over zero, the limit is infinite or does not exist. Only $\frac{0}{0}$ and similar indeterminate forms need more work.

Method 2: factor and cancel

If substitution gives $\frac{0}{0}$ for a rational function, the numerator and denominator share a factor of $(x - a)$. Factor both, cancel, and substitute again.

Cancelling is allowed because the limit never uses the value at $x = a$ itself. The factoring guide covers the techniques you need here.

Method 3: multiply by the conjugate

When a square root causes the $\frac{0}{0}$, multiply the numerator and denominator by the conjugate expression, which removes the root through the difference of squares.

One-sided limits and when a limit does not exist

A one-sided limit looks at inputs from only one direction: $x \to a^-$ from the left and $x \to a^+$ from the right. The two-sided limit exists only when both one-sided limits exist and are equal.

For $f(x) = \frac{|x|}{x}$, every positive input gives 1 and every negative input gives $-1$. So $\lim_{x \to 0^+} f(x) = 1$ and $\lim_{x \to 0^-} f(x) = -1$. They disagree, so $\lim_{x \to 0} f(x)$ does not exist. Piecewise functions are the most common source of this behaviour, so always check both sides at a boundary between pieces.

A limit can also fail to exist because the function oscillates. As $x \to 0$, $\sin\frac{1}{x}$ swings between $-1$ and 1 infinitely often and settles on no value.

Infinite limits and limits at infinity

If a function grows without bound near $a$, we write the limit as $\infty$, as in $\lim_{x \to 0}\frac{1}{x^2} = \infty$. Strictly the limit does not exist as a number, but the notation records how it fails, and it signals a vertical asymptote at $x = a$.

Limits as $x \to \infty$ describe long-run behaviour and horizontal asymptotes. For rational functions, divide every term by the highest power of $x$ in the denominator:

If the denominator has the higher degree, the limit is 0; for instance $\lim_{x \to \infty}\frac{2x + 1}{x^2} = 0$. If the numerator has the higher degree, the function grows without bound. One famous limit at infinity defines a constant: $\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e \approx 2.71828$.

Special trigonometric limits and the squeeze theorem

Two limits underpin all the trigonometric derivatives:

Substitution gives $\frac{0}{0}$ in both, and algebra alone cannot remove it. Numerically, $\frac{\sin 0.1}{0.1} \approx 0.99833$ and $\frac{\sin 0.01}{0.01} \approx 0.99998$, which suggests the answer, and a geometric argument with the squeeze theorem proves it. The angle must be in radians.

The squeeze theorem says that if $g(x) \le f(x) \le h(x)$ near $a$ and both $g$ and $h$ approach $L$, then $f$ approaches $L$ too. For example, $-x^2 \le x^2\sin\frac{1}{x} \le x^2$, and both bounds approach 0, so $\lim_{x \to 0} x^2\sin\frac{1}{x} = 0$ even though $\sin\frac{1}{x}$ alone has no limit.

L'Hopital's rule

When a limit has the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, and the functions are differentiable near the point, you may replace the numerator and denominator by their derivatives: $\lim\frac{f}{g} = \lim\frac{f'}{g'}$, provided the new limit exists.

Differentiate the numerator and denominator separately. L'Hopital's rule is not the quotient rule, and it only applies to indeterminate forms.

Limits, continuity and derivatives

A function is continuous at $a$ when $\lim_{x \to a} f(x) = f(a)$: the limit exists, the function is defined there, and the two agree. The derivative itself is a limit, $f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$, which always starts as $\frac{0}{0}$; see what is a derivative. Integrals and series convergence are defined through limits as well.

The limit calculator evaluates one-sided and two-sided limits, including limits at infinity, and shows which method it used. For checking a limit visually, the graphing calculator lets you zoom in on the point.

The limit laws

Limits behave well with arithmetic. If $\lim f(x)$ and $\lim g(x)$ both exist, then the limit of the sum is the sum of the limits, the limit of the product is the product of the limits, and the limit of the quotient is the quotient of the limits as long as the denominator's limit is not zero. Constant multiples and powers pass through in the same way. These laws are why direct substitution works for continuous functions: a polynomial is built from sums and products of $x$ and constants, so its limit is built the same way.

The laws need both individual limits to exist. The limit of $x\sin\frac{1}{x}$ at 0 is 0, but you cannot get that by multiplying limits, because $\sin\frac{1}{x}$ has no limit there. Cases like that need the squeeze theorem instead.

A strategy that works for most problems

1. Substitute. A real number is the answer.
2. Nonzero over zero: check the sign from each side to decide between $\infty$, $-\infty$ or no limit.
3. Zero over zero with polynomials: factor and cancel.
4. Zero over zero with square roots: multiply by the conjugate.
5. Zero over zero with trig functions: rewrite using $\frac{\sin x}{x} \to 1$.
6. Limits at infinity: divide by the highest power in the denominator.
7. If algebra stalls, use L'Hopital's rule, or check numerically with values on both sides.

Practice with answers: $\lim_{x \to 2}\frac{x^3 - 8}{x - 2} = 12$, by factoring the difference of cubes. $\lim_{x \to 0}\frac{\sin 3x}{x} = 3$, by writing it as $3\cdot\frac{\sin 3x}{3x}$. $\lim_{x \to \infty}\frac{4x - 7}{2x + 9} = 2$, by dividing through by $x$.

Frequently asked questions