Exercise 795Calculus › Single-variable calculus › Differentiation › Differentiable functionsL1Added 2 Nov 2026

When L'Hôpital's rule gives no answer

#L'Hôpital's rule#counterexamples#squeeze theorem

Statement

Consider $f(x)=x+\sin x$ and $g(x)=x$ .

Show that $\dfrac{f(x)}{g(x)}$ has a limit as $x\to+\infty$ and compute it.
Show that $\dfrac{f'(x)}{g'(x)}$ has no limit as $x\to+\infty$ .
What does this say about L'Hôpital's rule?

Hint

For part 1, divide by $x$ and use $|\sin x|\le1$ .

Show solutionHide solution

Solution

1. $\dfrac{x+\sin x}x=1+\dfrac{\sin x}x\to1$ , because $\left|\frac{\sin x}x\right|\le\frac1x\to0$ .

2. $\dfrac{f'}{g'}=\dfrac{1+\cos x}1=1+\cos x$ , which oscillates between $0$ and $2$ and has no limit.

3. L'Hôpital's rule says: *if* $\frac{f'}{g'}$ has a limit, *then* $\frac fg$ has the same limit. The converse is false: $\frac fg$ may have a limit while $\frac{f'}{g'}$ does not. So failure of the second limit to exist proves nothing, and one must use another method (here the squeeze theorem).

Remark. In an exam, the correct way to use the rule is to make sure the limit of the derivative quotient exists. A careless " $\frac{f'}{g'}$ doesn't converge, so neither does $\frac fg$ " is a classic mistake.

When L'Hôpital's rule gives no answer

Statement

Solution

More problems in Differentiable functions