L'Hopital's Rule failure due to derivative of denominator equaling 0

Question

I'm working on a problem demonstrating that the hypothesis in l’Hopital’s rule that g′(x)= 0 is really needed. I'm asked to show that if

(a) $f(x) = x+\cos(x)\sin(x)$

(b) $g(x) = e^{\sin(x)}(x+\cos(x)\sin(x))$

Then

(1)$\lim_{x→∞} f′(x)/g′(x) = 0$

(2)$\lim_{x→∞} f(x)/g(x)$ does not exist

I proved (2), as this was quite trivial. I'm having trouble proving (1). So far I've computed the derivatives of f and g

$f′(x)=1+\cos^2(x)-sin^2(x)$

$g′(x)=e^{\sin(x)}(1+\cos^2(x)-\sin^2(x))+\cos(x)e^{\sin(x)}(x+\cos(x)\sin(x))$

I know that $|f′(x)|\leq 2$ due to the triangle inequality.

Thanks!

You have x in the denominator so as x goes to infinity you have infinity in the denominator (since all the other terms are trivially bounded) — Avrham Aton, Jan 13 '15 at 17:58
@AvrhamAton: Not necessarily -- the $\cos x$ factor will periodically make the denominator cross zero. — hmakholm left over Monica, Jan 13 '15 at 18:00

score 2 · Accepted Answer · answered Jan 13 '15 at 18:11

It becomes less horrible once you realize that $1+\cos^2x-\sin^2x = 2\cos^2x $. Then

$$ \frac{f'(x)}{g'(x)}=\frac{2\cos^2x}{2\cos^2x+\cos(x)(x+\cos x \sin x)}e^{-\sin x}$$

Cancel $\cos x$ in the fraction to get

$$ \frac{f'(x)}{g'(x)}=\frac{2\cos x}{2\cos x+x+\cos x \sin x}e^{-\sin x}$$

Now the denominator does go to $+\infty$, while the numerator (and the exponential factor) stays bounded. So the whole thing goes to $0$.

1 Answers1