D7 - L’Hôpital’s rule | AQA A-Level Further Mathematics

Differentiation offers another way to evaluate certain indeterminate limits. You will recognise when l’Hôpital’s rule applies, differentiate the numerator and denominator separately, and check the conditions again before repeating the rule.

Recognising indeterminate quotient forms

Why does a quotient's limit sometimes depend on the derivatives of its two parts? Near zero, the changes in $\mathrm{e}^x-1$ and $x$ are both controlled first by their gradients. The series method found $(\mathrm{e}^x-1)/x\to1$ ; l’Hôpital’s rule turns this idea into a precise limit rule.

First identify what happens to the numerator and denominator separately. The symbols $0/0$ and $\infty/\infty$ describe indeterminate forms, not arithmetic with zero or infinity. They do not decide a limit: both $x/x\to1$ and $x^2/x\to0$ have the form $0/0$ at zero.

Quotient and approach	Separate behaviour	What it tells us
$(\mathrm{e}^x-1)/x$ , $x\to0$	$0/0$	A possible l’Hôpital calculation
$\ln x/x$ , $x\to\infty$	$\infty/\infty$	A possible l’Hôpital calculation
$(1+\sin x)/x$ , $x\to0$	$1/0$	Not an indeterminate quotient form
$\mathrm{e}^x/(1+x)$ , $x\to0$	$1/1$	Direct substitution gives 1

For the third row, the numerator is positive near zero. The quotient grows positively from the right and negatively from the left. Differentiating both parts would wrongly give $\cos x/1\to1$ ; that calculation is invalid because the original form is $1/0$ .

L’Hôpital’s rule: conditions matter

Suppose, on an interval approaching the point under consideration:

$f$ and $g$ are differentiable away from the limiting point, and $g(x)\ne0$ and $g'(x)\ne0$ ;
$f/g$ has the form $0/0$ , or numerator and denominator both tend to signed infinities;
$f'(x)/g'(x)$ has a limit $L$ , finite or infinite.

Then

\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}=L

The same principle applies to one-sided limits and limits at infinity. For a two-sided limit, the hypotheses and conclusion must work on both sides, with the same result.

The rule compares limits; it does not say the two quotient functions are equal. The point itself can be excluded from the domain, as it is in $\sin x/x$ at zero. If the derivative quotient has no limit, the rule in this form gives no conclusion; that alone does not prove the original limit fails to exist.

For limits at infinity, $\infty/\infty$ includes negative signs: inspect them in the actual functions. For example, $-\ln x/x$ as $x\to\infty$ has numerator tending to $-\infty$ and denominator to $+\infty$ , still an eligible form if the other hypotheses hold.

Differentiating numerator and denominator separately

Once the conditions hold, form $f'/g'$ . This is not the quotient-rule derivative $(f'g-fg')/g^2$ . You are finding a new quotient with the same limit, not differentiating the original quotient as a whole.

Worked example 1: a trigonometric zero-over-zero limit

Question. Use l’Hôpital’s rule to find $\displaystyle\lim_{x\to0}\frac{\sin(3x)}{2x}$ .

Plan. Verify $0/0$ , then differentiate the two parts separately using radians.

The numerator and denominator both tend to zero. They are differentiable near zero and the denominator derivative is $2\ne0$ .

\begin{aligned} \lim_{x\to0}\frac{\sin(3x)}{2x} &=\lim_{x\to0}\frac{3\cos(3x)}{2} && \text{l’Hôpital's rule}\\ &=\frac32 && \text{substitute in the continuous quotient} \end{aligned}

Answer. $3/2$ .

Check. The sine series begins $\sin(3x)=3x-(9/2)x^3+\cdots$ . Dividing by $2x$ independently gives $3/2-(9/4)x^2+\cdots$ , with limit $3/2$ .

Add a quadratic term only to the denominator. Its derivative still tends to 2, so the leading limit should stay unchanged.

Worked example 2: a changed denominator

Question. Use l’Hôpital’s rule to find $\displaystyle\lim_{x\to0}\frac{\sin(3x)}{2x+x^2}$ .

Plan. Recheck the original form and use the derivative of the whole denominator.

Both parts tend to zero. The denominator derivative $2+2x$ is nonzero sufficiently near zero, for example when $|x|<1/2$ .

\begin{aligned} \lim_{x\to0}\frac{\sin(3x)}{2x+x^2} &=\lim_{x\to0}\frac{3\cos(3x)}{2+2x}\\ &=\frac32 \end{aligned}

Answer. $3/2$ .

Check. Factor the original denominator: the quotient equals $[\sin(3x)/x]/(2+x)$ for $x\ne0$ . The series gives the numerator limit 3 and the denominator limit 2, agreeing with $3/2$ .

The rule also handles growth at infinity; derivatives need not keep growing.

Worked example 3: comparing logarithmic and linear growth

Question. Find $\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}$ .

Plan. Check the infinity-over-infinity form and compare the derivatives.

As $x\to\infty$ , both parts tend to $+\infty$ . For $x>0$ , both are differentiable and the denominator derivative is 1.

\lim_{x\to\infty}\frac{\ln x}{x} =\lim_{x\to\infty}\frac{1/x}{1}=0

Answer. $0$ .

Check. At $x=10,100,1000$ , the original quotient is approximately $0.2303,0.04605,0.006908$ . The decreasing positive values support the exact result. The result expresses a useful connection: logarithmic growth is slower than linear growth, although both functions are unbounded.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Core Pure

Mechanics

Statistics

Discrete Mathematics

D7 - L’Hôpital’s rule

Recognising indeterminate quotient forms

Differentiating numerator and denominator separately

Continue this lesson.

Learn faster with Nova