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How can I prove:

$$\lim_{x\to \frac{\pi}{2}}\sin{\frac{x}{2}} \cdot [\sin{x}] = 0$$

I know that for every $0<x<\pi$, where $x\neq \pi/2$, $[\sin{x}]=0$

How can I go from here?

rtybase
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3 Answers3

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You said yourself that $[\sin(x)] = 0$ for all $x \in (0, \pi)$ except at $x= \pi/2$.

Then $\sin(x/2)[\sin(x)] = \begin{cases} 0 & x \in (0, \pi) \text{ and } x \neq \pi/2 \\ 1 & x = \pi/2 \end{cases}$. It follows from the definition of $\lim \limits_{x \to a} f(x) = y$ that the limit as $x \to \frac{\pi}{2}$ of this expression is $0$. Recall the $\epsilon$-$\delta$ definition:

$\lim \limits_{x \to a} f(x) = y$ means for each $\epsilon > 0$ there is a $\delta > 0$ so that $0 < |x - a| < \delta$ implies $|f(x) - y| < \epsilon$. Note the $0 < |x -a|$ part, which means we are only considering points near $a$, but not the value $a$ itself, since if $x = a$, then $|x - a| = 0$.

It's your job to now fill in the details/apply this definition based on the hint I gave above.

layman
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    The definition of limit implies $0 < | x − a | < \delta $! https://en.wikipedia.org/wiki/Limit_of_a_function#Functions_of_a_single_variable – rtybase Dec 13 '16 at 19:51
  • @EugenCovaci I didn't write it, but I think you can exclude the point $a$ from the $\epsilon$-$\delta$ definition for $x \to a$. We care about the behavior of the function near $a$, but not necessarily at $a$. – layman Dec 13 '16 at 19:52
  • @rtybase I don't understand the significance of your comment. – layman Dec 13 '16 at 19:52
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    @user46944 it's the $| x − a | < \delta $ vs $0 < | x − a | < \delta $ In this case you are right and EugenCovaci is wrong (including his -1 downvote). – rtybase Dec 13 '16 at 19:54
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    @rtybase Good point! :) Thank you!!! I was questioning what's wrong with the definition I wrote and how we can exclude $\pi/2$. It's because $x = \pi/2$ makes $|x - \pi/2| = 0$. Thanks again! :) – layman Dec 13 '16 at 19:55
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You almost answered the question yourself. The limitand is zero in a neighborhood of the point. Sine is continuous on a neighborhood of the point. What happens at $x=\pi/2$ is irrelevant; it's zero from the right and the left.

GFauxPas
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First of all, defining the domain would be helpful for those of use trying to help you. Are these defined over $\Bbb R$? Try using the properties of limits: $Lim$(fg)=lim(f)lim(g). Then, we can focus on each limit individually. Also, as stated by @GFauxPas, analyze the left and right sided limits. Good luck

wesssg
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