This note shows how to find limits of fractions that give an indeterminate form such as 0/0 or ∞/∞ when you substitute. l'Hôpital's rule replaces the fraction by a quotient of derivatives, which is often much simpler. It is a quick way to handle limits involving exponentials, logarithms and trigonometric functions.
Key ideas
Some limits cannot be found by direct substitution. For example, in \(\lim_{x\to0}\dfrac{\sin x}{x}\), both the numerator and the denominator tend to 0. The expression \(\tfrac00\) has no value by itself, so it is called an indeterminate form. The limit might be any number, or might not exist. Another indeterminate form is \(\tfrac{\infty}{\infty}\), where both parts grow without bound.
l'Hôpital's rule. Suppose \(f\) and \(g\) are differentiable near \(a\) and \(\lim_{x\to a}\dfrac{f(x)}{g(x)}\) is of the form \(\tfrac00\) or \(\tfrac\infty\infty\). Then \[\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)},\] provided the limit on the right exists. The rule also works for \(x\to\infty\) and \(x\to-\infty\).
How to use it.
- First substitute. Write down what the numerator and the denominator tend to. Only use the rule if you really get \(\tfrac00\) or \(\tfrac\infty\infty\).
- Differentiate the top and the bottom separately. This is not the quotient rule.
- Find the new limit. If it is still \(\tfrac00\) or \(\tfrac\infty\infty\), apply the rule again. Repeat until the limit can be found directly.
Other forms. If a limit is a product, such as \(x\ln x\) as \(x\to0^+\), it has the form \(0\times\infty\). Rewrite it as a fraction first: \(\dfrac{\ln x}{1/x}\), which is of type \(\tfrac{-\infty}{\infty}\). Now the rule applies: the derivatives give \(\dfrac{1/x}{-1/x^2}=-x\), which tends to 0. So \(x\ln x\to0\). A difference such as \(\dfrac1x-\dfrac1{\sin x}\) can be combined into a single fraction before you start.
Growth rates. Repeated use of the rule on \(\dfrac{x^n}{e^x}\) shows that, for large \(x\), the exponential \(e^x\) beats every power of \(x\), so \(\dfrac{x^n}{e^x}\to0\). In the same way, powers of \(x\) beat \(\ln x\), so \(\dfrac{\ln x}{x^n}\to 0\) for \(n\gt0\).
Why it works (idea). Near \(x=a\), when \(f(a)=g(a)=0\), each function is close to its tangent line, so \(\dfrac{f(x)}{g(x)}\approx\dfrac{f'(a)(x-a)}{g'(a)(x-a)}=\dfrac{f'(a)}{g'(a)}\). That is why the ratio of gradients decides the limit.
A longer chain. Some limits need several applications. Take \(\lim_{x\to0}\dfrac{x-\sin x}{x^3}\). At \(x=0\) it is \(\tfrac00\). The first application gives \(\dfrac{1-\cos x}{3x^2}\), still \(\tfrac00\). The second gives \(\dfrac{\sin x}{6x}\), still \(\tfrac00\). The third gives \(\dfrac{\cos x}{6}\), which tends to \(\dfrac16\). Notice that you could also have used the standard limit \(\dfrac{\sin x}{x}\to1\) at the second stage.
Checking numerically. A quick check on your GDC is to make a table of values of the fraction for \(x\) very close to \(a\), from both sides. If the values settle towards your answer, you can be more confident. This does not replace the algebra, but it catches slips such as differentiating wrongly.
When it does not apply. If the form is not \(\tfrac00\) or \(\tfrac\infty\infty\), do not use the rule. For example, \(\lim_{x\to0}\dfrac{x+1}{x}\) has numerator tending to 1 and denominator tending to 0. Applying the rule would wrongly give 1. In fact, this limit does not exist, as the fraction grows without bound with one sign from the right and the other from the left.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Indeterminate form \(\tfrac00\), \(\tfrac\infty\infty\) | Substitution gives no answer. The rule may be used. |
| \(\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}\) | l'Hôpital's rule. Differentiate top and bottom separately. |
| \(\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1\) | A standard limit, with \(x\) in radians. It also follows from the rule. |
| \(\displaystyle\lim_{x\to\infty}\frac{x^n}{e^x}=0\) | Exponential growth beats power growth. |
| \(\displaystyle\lim_{x\to\infty}\frac{\ln x}{x^n}=0,\ n\gt0\) | Power growth beats logarithmic growth. |
Worked examples
Example 1 Find \(\displaystyle\lim_{x\to0}\frac{e^{3x}-1}{\sin 2x}\).
At \(x=0\): numerator \(=e^0-1=0\) and denominator \(=\sin0=0\). The form is \(\tfrac00\).
Differentiate top and bottom separately: \(\dfrac{3e^{3x}}{2\cos2x}\).
Now substitute \(x=0\): \(\dfrac{3\cdot1}{2\cdot1}=\dfrac32\).
Answer: \(\tfrac32\).
Example 2 Find \(\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}\).
At \(x=0\): \(1-\cos0=0\) and \(0^2=0\). The form is \(\tfrac00\).
First application: \(\dfrac{\sin x}{2x}\). At \(x=0\) this is again \(\tfrac00\).
Second application: \(\dfrac{\cos x}{2}\). This is not indeterminate. At \(x=0\) it equals \(\dfrac12\).
Answer: \(\tfrac12\).
Example 3 A signal is modelled by \(f(t)=\dfrac{t^2}{e^{t}}\). Find the limit of \(f(t)\) as \(t\to\infty\).
As \(t\to\infty\), \(t^2\to\infty\) and \(e^t\to\infty\). The form is \(\tfrac\infty\infty\).
First application: \(\dfrac{2t}{e^t}\). It is still \(\tfrac\infty\infty\).
Second application: \(\dfrac{2}{e^t}\). The numerator is fixed at 2 and the denominator tends to \(\infty\), so the fraction tends to 0.
Answer: \(0\). The signal dies away.
Common errors
- Common error: using the quotient rule on \(\dfrac{f}{g}\). In l'Hôpital's rule you differentiate \(f\) and \(g\) separately.
- Common error: applying the rule when the form is not \(\tfrac00\) or \(\tfrac\infty\infty\), so the answer is wrong. Always test by substitution first.
- Common error: stopping after one application when the new fraction is still indeterminate.
- Common error: forgetting that trigonometric limits need radians.
- Common error: applying the rule to a product or a difference without first rewriting it as a single fraction.
Exam tip: Write the form (such as "\(\tfrac00\)") after each substitution. This shows why the rule is allowed, and tells you whether another application is needed.
Check yourself
- Find \(\displaystyle\lim_{x\to0}\frac{\sin5x}{x}\).
Answer
The form is \(\tfrac00\). Differentiating gives \(\frac{5\cos5x}{1}\to5\). - Find \(\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}\).
Answer
The form is \(\tfrac\infty\infty\). Differentiating gives \(\frac{1/x}{1}=\frac1x\to0\). - Find \(\displaystyle\lim_{x\to0}\frac{e^x-1-x}{x^2}\).
Answer
The form is \(\tfrac00\). First: \(\frac{e^x-1}{2x}\), still \(\tfrac00\). Second: \(\frac{e^x}{2}\to\frac12\). - Explain why l'Hôpital's rule cannot be used for \(\displaystyle\lim_{x\to0}\frac{x+1}{x}\).
Answer
The numerator tends to 1, not 0 or \(\infty\), so the form is not indeterminate. The rule would wrongly give 1. The limit does not exist.