I want to compute $\lim_{x \to \infty} e^x \sin(1/x)$. Here what I did:
$\lim_{x \to \infty} e^x \sin(1/x) = \lim_{x \to \infty} \dfrac{\sin(1/x)}{e^{-x}}$. By using L'Hospital Rule I get $\lim_{x \to \infty} e^x \sin(1/x) = \lim_{x \to \infty} \dfrac{\cos(1/x) e^x}{x^2}$.
What can I do right now? I know I can not use the product rule for limits as
$\lim_{x \to \infty} \dfrac{\cos(1/x) e^x}{x^2} = \lim_{x \to \infty} \cos(1/x) \cdot \lim_{x \to \infty} \dfrac{e^x}{x^2}$
since the second limit is infinity. Any help would be appreciated.
Best Answer
You can indeed use the product rule for limits, since $1\cdot (+\infty)$ is not an indeterminate form. The result is $+\infty$.
If you want a solution that does not use neither L'Hospital nor Taylor, you can just observe that $$ e^x \sin (1/x) = \frac{e^x \sin(1/x)}{1/x}\frac{1}{x} = \frac{e^x}{x} \frac{\sin(1/x)}{1/x} \to (+\infty)\cdot1 = +\infty \quad \text{as } x \to +\infty $$