# How can I use squeeze theorm for $\lim_{x\rightarrow0^+} xe^{9\sin(1/x)}$

I am trying to use to squeeze theorem to evaluate the following limit:

$$\lim_{x\rightarrow0^+} xe^{9\sin(1/x)}$$

The main issue I had (the online assignment has already been submitted) was determining the right interval for $e^{9\sin(1/x)}$. Since the function is an exponential, I assumed that there was no max and min i.e. the interval was $(-\infty, \infty)$. Which didn't make much sense to me when I put it in context. However, I still continued and obtained the following:

\begin{align} \lim_{x\rightarrow 0^+} x * -\infty \lt &e^{9\sin(1/x)} \lt \lim_{x\rightarrow 0^+} x * \infty \\ 0 \lt &e^{9\sin(1/x)} \lt 0 \end{align}

Therefore, the limit is $0$. Now, that's what I put on the assignment and it got marked as correct but does that make sense? Are my steps correct? or did I just get lucky?

Thanks!

• Hint: Use that $-1\leq \sin(1/x)\leq 1$. – Start wearing purple Oct 23 '13 at 3:52
• @O.L. That was the thought I had at first but I tried to figure it out in relation to the problem and it didn't work out. Wait, was I suppose to do what you mention and have $-x$ and $x$ on either side? and from there I get zero as the limit? – Jeel Shah Oct 23 '13 at 3:54
• That inequality gives you a way to use squeezing via $e^{-9} \leq e^{9 \sin 1/x} \leq e^9$. – Start wearing purple Oct 23 '13 at 3:57

$-1\le \sin {1\over x}\le 1\Rightarrow -9\le 9\sin {1\over x}\le 9\Rightarrow e^{-9}\le e^{9\sin {1\over x}}\le e^9\Rightarrow xe^{-9}\le xe^{9\sin {1\over x}}\le xe^9$