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Given $f(x) = x -1 $ for $ x \leq 1$ and $\sin(\pi x)$ for $x > 1$, is it derivable in 1? I did the left and right limits of f in 1, and computed $f(1)$ and they are equal, therefore the function is continuous in 1 and derivable. I then tried to compute $\lim_{x->1^+} \frac{f(x) - 0}{x -1} = \lim_{x->1} \frac{\sin \pi}{x-1}$ but then I got stuck. How do I get rid of the $x-1$ from the denominator so that I can compute the limit?

I know that the answer is $-1$ and can be computed using l'Hopital's rule, but I was wondering how to do it without?

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andreas.vitikan
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    Using de l'Hopital:$\lim_{x\rightarrow1}\frac{\sin\pi x}{x-1}=\lim_{x\rightarrow1}\frac{\pi\cos\pi x}{1}=\pi\cos\pi=-\pi$. So the mentioned function is not derivable in $x_{0}=1$. – drhab Feb 25 '14 at 11:06
  • I understand what you mean, but one normally says the function is "differentiable at" $1$ rather than "derivable in" $1$. – MPW Feb 25 '14 at 12:26

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In the knowledge that $\sin\pi1=0$ we recognize $\lim_{x\rightarrow1}\frac{\sin\pi x}{x-1}=\frac{\sin\pi x-\sin\pi1}{x-1}$ as the computation of $g'\left(1\right)$ where $g\left(x\right)=\sin\pi x$. It is wellknown that $g'\left(x\right)=\pi\cos\pi x$ here, so $g'\left(1\right)=\pi\cos\pi=-\pi$.

Did I use de l'Hôpital here? Well, I was not aware of it :-)

addendum:

For a small $x>0$ we find:$\frac{\sin\pi\left(1+x\right)-\sin\pi}{\left(1+x\right)-1}=\frac{\sin\pi\cos\pi x+\cos\pi\sin\pi x}{x}=-\frac{\sin\pi x}{x}=-\pi\times\frac{\sin\pi x}{\pi x}$.

So if you can prove that $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$ then you are ready. Have a look here for that.

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drhab
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  • This uses derivatives and right now we're using the limit definition to computer derivatives and not actually deriving functions, so I can't use this. Even though it is correct – andreas.vitikan Feb 25 '14 at 12:20
  • I edited something. Especially the link in it can help you out. – drhab Feb 25 '14 at 13:05