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

I have some function:

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.