5
$\begingroup$

Wiki provides a counterexample limit that fails with L'Hopital's Rule, because it doesn't satisfy the $g'(x) \ne 0$ rule they provide earlier.

Let $f(x)=x+\sin x \cos x$ and $g(x)=f(x)e^{\sin x}.$ Then there is no limit for $f(x)/g(x)$ as $x\to\infty$ However,

$$ \begin{align} \frac{f'(x)}{g'(x)} &= \frac{2\cos^2 x}{(2 \cos^2 x) e^{\sin x} + (x+\sin x \cos x) e^{\sin x} \cos x} \\[4pt] &= \frac{2\cos x}{2 \cos x +x+\sin x \cos x} e^{-\sin x}, \end{align} $$

which tends to 0 as $x\to\infty$.

My question: is it possible to construct a counterexample where $x$ goes to a real number, rather than infinity? Or does approaching a real number protect you from failing in the $g'(x) = 0$ case.

$\endgroup$
2
  • $\begingroup$ Maybe $\dfrac{x^2\sin(1/x)}{\sin(x)}$ when $x\to0$. $\endgroup$
    – Mikasa
    Jan 12, 2022 at 6:25
  • $\begingroup$ @Mikasa $\lim_{x\to 0} \frac{f'(x)}{g'(x)}$ doesn't exist, so it isn't a counterexample (which demonstrates why $g'(x)\ne 0$ is necessary). $\endgroup$
    – Snaw
    Jan 12, 2022 at 7:03

1 Answer 1

5
$\begingroup$

Consider $$f_1(x)\equiv f\circ h(x)\\ g_1(x)\equiv g\circ h(x)\\ h(x)\equiv (x-a)^{-2},a\in\mathbb R,$$ where $f$ and $g$ are defined as in your counterexample. Then there is no limit for $f_1(x)/g_1(x)$ as $x\rightarrow a$ but L'Hopital's rule imputes a limit of zero.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .