I've been learning Squeeze Theorem (and limits in general), but am having problems understanding how to apply it. I understand the basics of the theorem (I think), but I've come across a problem that I'm not even sure how to start solving. I realize that Squeeze Theorem is the way to solve it, but beyond that, I'm clueless.

I've search around the site, and a few other places online, but I can't seem to find a similar problem.

So, here's my problem:

$$\lim_{x\to 0} \frac{3 - \sin(e^x)}{\sqrt{x^2 + 2}}$$

Looks easy enough, but I'm clearly missing something obvious. How do we approach a problem like this? I'd love to show my work, but I'm not sure where to start.

  • 2
    $\begingroup$ The limit is $\frac {3-\sin (1)} {\sqrt 2}$. Why does one apply squeeze theorem in this case? $\endgroup$ – Kabo Murphy Aug 8 '18 at 11:44
  • $\begingroup$ There was a problem in my formatting, which @pointguardo kindly corrected. $\endgroup$ – plt279 Aug 8 '18 at 11:46
  • $\begingroup$ the answer is still $\frac{3 - \sin(1)}{\sqrt{2}}$ though $\endgroup$ – pointguard0 Aug 8 '18 at 11:48
  • $\begingroup$ Ah, I get what you're saying. Thanks. The answer makes sense, but I have no clue how to get to it. Could this be solved at all with Squeeze Theorem? Or am I totally on the wrong track? $\endgroup$ – plt279 Aug 8 '18 at 11:54
  • $\begingroup$ just plug in $x = 0$. $\endgroup$ – pointguard0 Aug 8 '18 at 12:00

After my best attempts, I believe this is the most appropriate answer to my question.

The problem cannot be solved by Squeeze Theorem, because:

$$\frac{2}{\sqrt{x^{2} + 2}} \leq \frac{3 - \sin{(e^{x})}}{\sqrt{x^{2} + 2}} \leq \frac{4}{\sqrt{x^{2} + 2}}$$


$$\lim_{x\to0} \frac{2}{\sqrt{x^{2} + 2}} \neq \lim_{x\to0} \frac{4}{\sqrt{x^{2} + 2}}$$


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