Prove that: $\underset{x\to \infty }{\text{lim}}\frac{x+\sin (x) \cos (x)}{e^{\sin (x)} (x+\sin (x) \cos (x))} = \text{doesn't exist}$ Prove that:
$$\underset{x\to \infty }{\text{lim}}\frac{x+\sin (x) \cos (x)}{e^{\sin (x)} (x+\sin (x) \cos (x))} = \text{doesn't exist}$$
We were showing that the limit doesn't exist with L'hopital rule: We have done L'hopital 4 times and then seen that the zero in the denominator is not isolated (the epsilon neighbourhood of the zero contained another zero).
I understand that $e^{\sin{x}}$ is oscillating and bounded,however, I do not really understand why the $e^{\sin{x}}$ is "stronger" the $x$, since it is bounded and the limit is not one.
Another question is, how valid is a proof with L'hopital rulle used to proof that the limit doesn't exist.
 A: Let me point out two misconceptions:

*

*I cannot say for sure, but I expect that the proof using de l'Hospital's rule you're mentioning is false. Simply because it's a theorem of the type if (...) then the limit exists, and not of the type if (...) the the limit doesn't exist.

*You may consider this picky, but one shouldn't write
$$\lim_{x \to \infty} f(x) = \text{doesn't exist},$$
but rather
$$\lim_{x \to \infty} f(x) \quad \text{doesn't exist}.$$
The difference lies in the fact that $\lim_{x \to \infty} f(x)$ is a priori just a symbol. If the limit exists, we can assign a value to this symbol and write $\lim_{x \to \infty} f(x) = (\ldots)$. But when the limit doesn't exist, the symbol is not related to any number, and writing an equal sign next to it is misleading.


As for the problem itself, it's helpful to note that
$$
\frac{x+\sin (x) \cos (x)}{e^{\sin (x)} (x+\sin (x) \cos (x))}
= e^{-\sin x}
$$
for all $x$ for which the first expression makes sense. This reduces the problem to the study of $e^{-\sin x}$, or even $\sin x$.
