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Using the Squeeze Theorem, how do I find: $$\lim_{x\to 3} (x^2-2x-3)^2\cos\left(\pi \over x-3\right)$$ I thought I knew the Squeeze Theorem, but I haven't encountered anything like this yet, so I honestly have no idea how and where to start.

I would appreciate any help I get because I really want to be able to understand these types of questions!

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    $\begingroup$ $|\cos\theta|\le 1$ and $|\sin\theta|\le1$ for any $\theta$. These facts will serve you well when solving typical Squeeze Theorem problems (and other problems). $\endgroup$ – David Mitra Oct 8 '13 at 18:35
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By these inequalities

$$0\leq\lim_{x\to 3} |(x^2-2x-3)^2\cos\left(\pi \over x-3\right)|\leq\lim_{x\to 3} |(x^2-2x-3)^2|=0$$ we have $$\lim_{x\to 3} (x^2-2x-3)^2\cos\left(\pi \over x-3\right)=0$$

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  • $\begingroup$ So the limit is 0? Why did we have to split off the $\cos\left(\pi\over x-3\right)$ part? I thought I was supposed to be given two functions to solve a Squeeze Theorem problem? $\endgroup$ – Redfoxpuppy22 Oct 9 '13 at 16:07
  • $\begingroup$ The function $x\mapsto \frac{\pi}{x-3}$ goes to $\infty$ when $x\to3$ and $\lim_{y\to\infty}\cos y$ is undefined so to get the desired limit we should split off the $\cos$ part using the fact that this function is bounded $\endgroup$ – user63181 Oct 9 '13 at 16:36
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Hint: The absolute value is between $0$ and $(x^2-2x-3)^2$.

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