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Show f(x) = sin(1/x), for x does not equal 0, is differentiable for nonzero real numbers.

I was wondering if this would be enough to show the the previous statement:

Let c<0 => c does not equal 0.

limit of f(x) as x approaches c from the left = lim sin(1/x) as x approaches c from the left = sin (1/c) since c does not equal 0.

The same works as x approaches c from the right.

Then, repeat this process with c>0.

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It's just a composition of differentiable functions. no proof necessary. on $\mathbf{R}$ with origin removed both $\sin$ and $1/x$ are differentiable.

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So, since sin(x) is differentiable, and 1/x for nonzero x is differentiable, it is enough to say that sin(1/x) is for nonzero x? – user99930 Feb 12 '14 at 21:46
@user99930 yes. You might even calculate the derivative by the chain rule: $f'(x) = (\sin(1/x))'\cdot(1/x)'=..$ – user127.0.0.1 Feb 12 '14 at 21:51

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