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In my assignment I have to find the Classification of discontinuities of the following function: $$f(x)=\frac{\sin^2(x)}{x|x(\pi-x)|}$$

I wanted to look what happens with the value $x=\pi$ because the function doesn't exist in that value.

I have to check if some $L \in \Bbb R$ exists such that $$\lim _{x \to \pi} f(x)=L $$

I didn't have much success in making the argument simpler, so I thought to make a little 'trick', and I know it works for sequences. I am not sure if it's "legal" to do in function. Here it is:

$$\frac{\sin^2(x)}{x|x(\pi-x)|} < \frac{\sin^2(x)}{x|x(\pi )|}$$

Now find the limit for the "bigger" function:

$$\lim _{x \to \pi} \frac {sin^2(x)}{x|x(\pi )|} = \frac{0}{\pi}=0$$

Is my solution "legal" or valid?

Thanks,

Alan

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  • $\begingroup$ Please let me know how I can improve my answer. I just want to give you the best answer I can. $\endgroup$ – Mark Viola May 22 '15 at 16:27
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$|\pi -x|<\pi \implies \frac{1}{|\pi-x|}>\pi$. So, we need a different approach.

Note that $\sin^2 x=\sin^2(\pi-x)$ and that

$$\lim_{ \pi -x \to 0}\frac{\sin(\pi-x)}{\pi-x}=1.$$

Can you finish now?

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  • $\begingroup$ I think I can. I just have to make sure that $|\pi-x|$ is positive. Is it becuase ${x \to \pi}$? $\endgroup$ – Alan May 22 '15 at 17:31
  • $\begingroup$ @Alan That is good thinking to make sure of that absolute value issue. However, things might just be simpler since you have an extra $\sin(\pi-x)$ in the numerator. Do you know how to proceed now? If not, I am more than happy to add to my answer to provide further support to you. $\endgroup$ – Mark Viola May 22 '15 at 18:52
  • $\begingroup$ You're welcome. My pleasure. And thank you for the best vote. Someone actually down voted for no reason. $\endgroup$ – Mark Viola May 23 '15 at 13:51
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Near $\pi$, you have made the denominator bigger, not smaller, so your inequality goes the other way, which does you no good. You can show, for instance using the mean value theorem on $\sin$ and then squaring the result, that $0 \leq \sin(x)^2 \leq (\pi-x)^2$. Plug that inequality in to see what you get.

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  • $\begingroup$ Oh, I'm awful. Thank you. But generaly, If I make the denumerator smaller, will my solution work? $\endgroup$ – Alan May 22 '15 at 16:17
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    $\begingroup$ @Alan In principle yes. But I don't think that will actually work out, because the denominator is not bounded away from zero. The point is that the numerator decays faster than the denominator as $x \to \pi$. $\endgroup$ – Ian May 22 '15 at 16:17
  • $\begingroup$ Thank you, I Will have to try another way. $\endgroup$ – Alan May 22 '15 at 16:18
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Hint: I think you will be more comfortable if you make the substitution $\pi-x=t$. So $x=\pi-t$, and $\sin(\pi-t)=\sin t$. Now we are looking at behaviour near $t=0$, familiar territory.

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Since $\sin x=\sin (\pi -x)$ then \begin{eqnarray*} \lim_{x\rightarrow \pi }\frac{\sin ^{2}(x)}{x\left\vert x(\pi -x)\right\vert } &=&\lim_{x\rightarrow \pi }\left( \frac{\sin (\pi -x)}{(\pi -x)}\right) ^{2}\frac{\left\vert \pi -x\right\vert }{x\left\vert x\right\vert } \\ &=&\lim_{x\rightarrow \pi }\left( \frac{\sin (\pi -x)}{(\pi -x)}\right) ^{2}\cdot \lim_{x\rightarrow \pi }\frac{\left\vert \pi -x\right\vert }{% x\left\vert x\right\vert } \\ &=&1^{2}\cdot \frac{\left\vert \pi -\pi \right\vert }{\pi \left\vert \pi \right\vert } \\ &=&1\cdot \frac{0}{\pi ^{2}}=0. \end{eqnarray*} I have used the standard limit \begin{equation*} \lim_{u\rightarrow 0}\frac{\sin (u)}{(u)}=1. \end{equation*}

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