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I want to know how to do a problem like this. What steps are taken and why.

I have some function:

$$ \lim_{h\to3} \ \sin{\left(\frac{1}{h-3}\right)} \ e^h \ (h-3)^2 $$

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  • $\begingroup$ If you knew the following inequality $\frac{\sin(x)}{x} \leq 1$, could you tackle this problem? $\endgroup$ – Demophilus Jun 8 '17 at 17:29
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    $\begingroup$ Try drawing a graph of @rt6's solution to see what is really going on here. $\endgroup$ – John Joy Jun 10 '17 at 1:20
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We know that $-1\leq \sin(x)\leq 1$ for every $x\in\mathbb{R}$.

It follows then for every value of $h\in \mathbb{R}$ except at $h=3$,

$$-e^h(h-3)^2\leq \sin\left(\frac{1}{h-3}\right)e^h(h-3)^2\leq e^h(h-3)^2$$

We can now apply the squeeze theorem as $h$ tends to $3$ to obtain that the desired limit is zero.

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