$$\large f(x)=\begin{cases}\left(\sin\frac\pi{2+x}\right)^{1/x^2},& x\ne0\\c,&x\ne0\end{cases}$$
I want to find the c for which this function is continuous:
I tried to find the limit of f(x) as x approches 0 to get a value which would be my C but I have trouble finding the limit.
I get infinity which would mean there is no continuous extension at 0.
How would you proceed?
Answer
Hint
Consider $$y=\sqrt[x^2]{\sin \left(\frac{\pi }{x+2}\right)}\implies \log(y)=\frac 1{x^2}\log\left(\sin \left(\frac{\pi }{x+2}\right)\right)$$ Now use Taylor series $$\frac{\pi }{x+2}=\frac{\pi }{2}-\frac{\pi x}{4}+\frac{\pi x^2}{8}+O\left(x^3\right)$$ $$\sin \left(\frac{\pi }{x+2}\right)=1-\frac{\pi ^2 x^2}{32}+O\left(x^3\right)$$ $$\log\left(\sin \left(\frac{\pi }{x+2}\right)\right)=\log\left( 1-\frac{\pi ^2 x^2}{32}+O\left(x^3\right)\right)=-\frac{\pi ^2 x^2}{32}+O\left(x^3\right)$$
I am sure that you can take it from here.
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