FP1 2.4 - L'Hospital's rule | Edexcel A-Level Further Mathematics

Evaluate indeterminate limits by differentiating the numerator and denominator separately. You will check when the rule applies, repeat it when necessary and turn differences, products and powers into suitable quotients.

Check the form before differentiating

What happens to $(1+1/x)^x$ as $x$ grows without bound? Its base approaches $1$ , but its exponent grows. The answer is $\mathrm{e}$ , rather than $1$ . To explain that balance, we will first turn the power into a quotient whose limit can be found by differentiation.

You already know

A limit concerns values near a point, so the expression need not be defined at the point itself. Taylor series can reveal the first surviving terms of a quotient. Recall the usual chain, product and quotient rules for differentiation, and that $\ln(a^b)=b\ln a$ for a positive real base $a$ .

Direct substitution into $(\mathrm{e}^x-1)/x$ at $x=0$ produces $0/0$ . This is an indeterminate form, not the number zero and not a value of the quotient. Both functions approach zero, and their relative rates decide the limit.

The derivative ratio compares those rates. For example, $\mathrm{e}^x/1\to1$ , which gives the correct limit of $(\mathrm{e}^x-1)/x$ . The rule needs conditions; it is not an instruction to differentiate every fraction you encounter.

L'Hospital's rule

Suppose $f$ and $g$ are differentiable in a punctured interval approaching $a$ , with $g(x)\ne0$ and $g'(x)\ne0$ there. If $f/g$ has form $0/0$ or $\infty/\infty$ , and

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

exists, then

\lim_{x\to a}\frac{f(x)}{g(x)}=L

Here an infinite limit $L$ is allowed. The rule also applies to a one-sided limit or to a limit as $x\to\pm\infty$ , with the same conditions on the relevant interval. For the infinite quotient form, signs may differ: it is the magnitudes of both numerator and denominator that grow without bound.

Not in the formula booklet: learn it.

The equality is between limits. It does not say that $f(x)/g(x)=f'(x)/g'(x)$ for ordinary values of $x$ . Nor are you taking the derivative of the quotient: differentiate its numerator and denominator separately.

Key idea

L'Hospital compares the limits of two quotients after the entry conditions are met. It does not assert that the original quotient equals the derivative ratio at ordinary points.

For a useful non-example, $(x+1)/(x+2)\to1/2$ as $x\to0$ . Differentiating top and bottom would give $1$ , but the original form is neither $0/0$ nor $\infty/\infty$ , so the rule is unavailable. If the derivative ratio has no limit, the rule gives no conclusion about the original limit.

Watch out

Wrong: use L'Hospital immediately on a product, a difference or a power. Right: first produce a quotient, check its indeterminate form and check again before every further application.

Choose a quotient that meets the rule's entry condition.

Repeat only while the form is indeterminate

Some quotients remain $0/0$ after one differentiation. Recheck the form, then apply the rule again. Once the derivative ratio has an ordinary finite value, stop and evaluate it.

Keep limit notation at each stage. For these trigonometric examples, the denominator derivatives are non-zero sufficiently close to $0$ , apart from $0$ itself where appropriate, so the rule works on both sides.

Worked example 1: two applications

Question. Use L'Hospital's rule to find $\displaystyle\lim_{x\to0}\frac{1-\cos2x}{x\sin3x}$ .

Plan. Check $0/0$ , differentiate top and bottom separately, and recheck before differentiating again.

\begin{aligned} \lim_{x\to0}\frac{1-\cos2x}{x\sin3x} &=\lim_{x\to0}\frac{2\sin2x}{\sin3x+3x\cos3x} && \text{first application: }0/0 \\ &=\lim_{x\to0}\frac{4\cos2x}{6\cos3x-9x\sin3x} && \text{second application: still }0/0 \\ &=\frac46=\frac23 && \text{denominator now tends to }6 \end{aligned}

Why this step? The second denominator derivative contains two contributions of $3\cos3x$ : one from differentiating $\sin3x$ , the other from the product $3x\cos3x$ .

Answer. The limit is $2/3$ .

Check. The leading terms are $1-\cos2x\sim2x^2$ and $x\sin3x\sim3x^2$ , agreeing with the derivative calculation. Here $f\sim g$ means $f/g\to1$ .

Change only the angle multiplier in the numerator, from $2$ to $4$ . Its two chain-rule factors change the limit by a factor of $4$ .

Worked example 2: the chain-rule factors matter

Question. Use L'Hospital's rule to find $\displaystyle\lim_{x\to0}\frac{1-\cos4x}{x\sin3x}$ .

Plan. Use the same two-stage calculation with the new numerator.

\begin{aligned} \lim_{x\to0}\frac{1-\cos4x}{x\sin3x} &=\lim_{x\to0}\frac{4\sin4x}{\sin3x+3x\cos3x} && 0/0 \\ &=\lim_{x\to0}\frac{16\cos4x}{6\cos3x-9x\sin3x} && \text{still }0/0 \\ &=\frac{16}{6}=\frac83 && \text{direct evaluation now applies} \end{aligned}

Why this step? Differentiating $4\sin4x$ gives $16\cos4x$ . Retaining only the first factor of $4$ would produce an incorrect limit.

Answer. The limit is $8/3$ .

Check. The leading numerator is $8x^2$ , while the denominator remains $3x^2$ .

The series check is useful, but a question requesting L'Hospital needs the derivative route and its indeterminate-form checks. Neither particular result is printed in the booklet.

Watch out

Wrong: differentiate $x\sin3x$ as $3x\cos3x$ alone. The product rule also contributes $\sin3x$ . Right: use $\sin3x+3x\cos3x$ , then recheck the form before another application. Write each differentiated denominator in full before taking its limit. A final correct number cannot repair missing product-rule terms in the method.

Vary the numerator angle once more.

Combine a difference before applying the rule

A difference of two divergent terms can have a finite limit. Put the terms over a common denominator first: the cancellation then becomes part of the numerator that L'Hospital can handle.

For example, $1/\sin x$ and $1/x$ both become large in magnitude near $0$ , but their difference is much smaller than either term. Subtracting separately evaluated infinite quantities would not determine that difference.

Worked example 3: a difference of reciprocals

Question. Use L'Hospital's rule to find $\displaystyle\lim_{x\to0}\left(\frac1{\sin x}-\frac1x\right)$ .

Plan. Combine the fractions, then use the rule twice.

\begin{aligned} \frac1{\sin x}-\frac1x&=\frac{x-\sin x}{x\sin x} && \text{common denominator} \\ \lim_{x\to0}\frac{x-\sin x}{x\sin x} &=\lim_{x\to0}\frac{1-\cos x}{\sin x+x\cos x} && 0/0 \\ &=\lim_{x\to0}\frac{\sin x}{2\cos x-x\sin x} && \text{still }0/0 \\ &=0 && \text{numerator tends to }0,\text{ denominator to }2 \end{aligned}

Why this step? L'Hospital applies to the single quotient after combining, not to the subtraction of two infinite limits.

Answer. The limit is $0$ .

Check. Series give $x-\sin x\sim x^3/6$ and $x\sin x\sim x^2$ ; their quotient tends to $0$ .

Change only the first reciprocal from $1/\sin x$ to $1/\tan x$ . Combining the fractions still creates a quotient of form $0/0$ .

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FP1 2.4 - L'Hospital's rule

Check the form before differentiating

Repeat only while the form is indeterminate

Combine a difference before applying the rule

Continue this lesson.

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