5.13 Limits and L’Hôpital’s rule

Many important limits cannot be found by substituting the value that $x$ approaches. The derivative itself is defined by such a limit: in $\lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$

This subtopic collects the methods for evaluating limits of quotients $\frac{f(x)}{g(x)}$, as $x$ approaches a number or as $x$ becomes very large. Algebraic simplification works for many rational functions. L'Hôpital's rule, which replaces $f$

Limits and direct substitution

A limit describes the value that $f(x)$ approaches as $x$ gets closer and closer to some value $a$, without needing $x$ to equal $a$. It is written $\lim_{x \to a} f(x) = L$

For a function that is continuous at $a$, the limit is simply the value of the function there, so it can be found by direct substitution. Polynomials, $\sin x$, $\cos x$, $e^x$ and $\ln x$

Substitution is always the first thing to try. Only when it fails, by giving an expression with no meaning, is another method needed.

When substitution gives a non-zero number over zero

If the numerator tends to a non-zero number and the denominator tends to $0$, the quotient grows without bound in size. For example, as $x \to 0$, $\frac{1}{x^2}$ becomes larger than any number, and

Indeterminate forms

The situation is very different when the numerator and the denominator both tend to $0$. Substituting $x = 0$ into each of the three quotients

gives $\frac{0}{0}$ every time, yet their limits as $x \to 0$ are $1$, $0$ and "does not exist" respectively, since they simplify to $1$

The form $\frac{\infty}{\infty}$ arises in the same way when the numerator and the denominator both grow without bound, usually as $x \to \infty$. Here the outcome depends on which grows faster. The table compares the forms that are determined with those that are not.

Algebraic methods for limits

A quotient that gives $\frac{0}{0}$ at $x = a$ often has a common factor that is zero at $a$. Once that factor is removed, substitution works.

Factorising and cancelling

If a polynomial $p(x)$ satisfies $p(a) = 0$, then $(x - a)$ is a factor of $p(x)$, by the factor theorem. So when both the numerator and the denominator of a rational function are zero at

The graph of $y = \frac{x^2 + x - 6}{x^2 - 4}$ is the graph of $y = \frac{x + 3}{x + 2}$

Rationalising

When a square root makes the quotient indeterminate, multiplying the numerator and the denominator by the conjugate expression removes the root from the place where it causes the problem.

Limits at infinity

A limit at infinity describes what happens to $f(x)$ as $x$ grows without bound. If $f(x)$ approaches a number $L$, then $\lim_{x \to \infty} f(x) = L$

Dividing by the highest power

A rational function such as $\frac{3x^2 - 5x}{2x^2 + 7}$ has the form $\frac{\infty}{\infty}$

The three cases for rational functions

The same method gives a general rule. Let $p(x)$ have degree $m$ and leading coefficient $a$, and $q(x)$ have degree $n$ and leading coefficient $b$

The cases make sense in terms of dominant terms. For large $x$ the quotient behaves like $\frac{ax^m}{bx^n} = \frac{a}{b}x^{m - n}$

The standard limit of sin x over x

Substituting $x = 0$ into $\frac{\sin x}{x}$ gives $\frac{0}{0}$

Since $\frac{\sin(-x)}{-x} = \frac{\sin x}{x}$

A geometric argument

A table of values suggests a limit but does not prove it. A proof for small positive $x$ compares three areas in the unit circle. Let $O$ be the centre, $A = (1, 0)$ and $P$ the point on the circle at angle $x$, and let the tangent at $A$

The triangle $OAP$ lies inside the sector $OAP$, which lies inside the triangle $OAT$. Using the area of a triangle and the area of a sector, $\frac{1}{2}r^2\theta$

Multiplying by $\frac{2}{\sin x}$, which is positive, and taking reciprocals (which reverses the inequalities) gives

As $x \to 0$, $\cos x \to 1$. The quotient $\frac{\sin x}{x}$

Limits that follow from the standard limit

Many related limits reduce to $\lim_{x \to 0}\frac{\sin x}{x} = 1$ by rewriting them so that the angle inside the sine matches the denominator. The same idea, with the definition of the derivative, gives two further standard limits:

The first is the gradient of $y = e^x$ at $x = 0$, written from first principles; the second is the gradient of $y = \ln(1 + x)$

L'Hôpital's rule

Factorising and the standard limits only deal with particular functions. A single method that works for almost every $\frac{0}{0}$ or $\frac{\infty}{\infty}$ limit is L'Hôpital's rule.

The rule is named after the French mathematician Guillaume de l'Hôpital, who published it in 1696 in the first textbook of differential calculus. Its name is also spelled "L'Hospital".

Why the rule works

For the form $\frac{0}{0}$ with $f$ and $g$ differentiable at $a$, the reason is the tangent line approximation. Since $f(a) = 0$

Dividing, the factors of $(x - a)$ cancel:

and the approximation becomes exact in the limit. Near $a$, the ratio of two quantities that both vanish is decided by the ratio of the rates at which they vanish.

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Check the form before every use

The rule applies only to indeterminate forms. Applied to a limit that substitution already settles, it gives a wrong answer. For example,

by substitution, but differentiating the top and the bottom would give $\frac{1}{1} = 1$. The conditions in the definition are not a formality: without the $\frac{0}{0}$

Repeated use of L'Hôpital's rule

After one application, the new quotient $\frac{f'(x)}{g'(x)}$ may still have the form $\frac{0}{0}$

The answer to (a) gives a useful approximation: for small $x$, $1 - \cos x \approx \frac{x^2}{2}$, so $\cos x \approx 1 - \frac{x^2}{2}$

The form infinity over infinity

L'Hôpital's rule applies equally to $\frac{\infty}{\infty}$. It is particularly useful for comparing how quickly different kinds of function grow, where dividing by a power of $x$ does not help.

The same argument works for any power: applying the rule $n$ times to $\frac{x^n}{e^x}$ leaves $\frac{n!}{e^x} \to 0$

Any positive power of $x$ eventually outgrows $\ln x$, and $e^x$ eventually outgrows any power of $x$. Exponential models therefore always overtake polynomial ones in the long run, however large the polynomial's coefficients.

Rewriting a product as a quotient

A product such as $x\ln x$ as $x \to 0^+$ has the form $0 \times (-\infty)$, which is also indeterminate. Writing it as a quotient turns it into one of the two forms that L'Hôpital's rule handles.

Limits using Maclaurin series

A Maclaurin series writes a function as a polynomial with infinitely many terms. The formula booklet gives the series for $e^x$, $\ln(1 + x)$, $\sin x$, $\cos x$ and $\arctan x$

Replacing each function by its series turns a $\frac{0}{0}$ limit as $x \to 0$ into a quotient of polynomials, where the lowest powers of $x$ cancel and the limit can be read off.

The method

1. Replace each function in the numerator and the denominator by enough terms of its series.

2. Simplify, so that the lowest power of $x$ in the numerator and the denominator is visible.

3. Divide the numerator and the denominator by the lowest power of $x$ in the denominator.

4. Let $x \to 0$: every term with a positive power of $x$ vanishes.

The series also show the speed at which each quotient approaches its limit. In (a), the error is roughly $\frac{x^2}{120}$, which explains why a table of values settles quickly.

Choosing a method

Algebraic methods, L'Hôpital's rule and series are equivalent routes to the same value, and a second method is a good check on the first.

Limits involving a parameter

Some problems ask for the value of a constant that makes a limit exist, or makes it equal a given number. The key observation is that if the denominator tends to $0$ and the limit is finite, the numerator must also tend to $0$; otherwise the form would be non-zero over zero.

A limit also decides how to fill a gap in a function to make it continuous. The function $f(x) = \frac{\sin x}{x}$ is undefined at $0$, but defining $f(0) = 1$

Exam-style questions on limits

Self-check

• To do:

  I can explain why $\frac{0}{0}$ and $\frac{\infty}{\infty}$ are indeterminate, while $\frac{k}{0}$