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A consequence of continuity is the following fact:
if $f(x)$ is continuous at $x=b$ and $\lim\limits_{x \to a} g(x)=b$, then,
$\lim\limits_{x \to a} f(g(x)) = f(\lim\limits_{x \to a}g(x))$

with this fact we can solve the following:

$\lim\limits_{x \to 0} e^{\sin x}= e^{\lim\limits_{x \to a}\sin x} = e^0 = 1$

so the fact is saying that in this case $f(\lim\limits_{x \to a}g(x)) = f(b).$
I don't understand how this fact helps solve the problem, is it because $\exp$ and $\sin$ are both continuous everywhere, and if they're continuous at the same places, you can use this fact?

Please elaborate. thanks!

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  • $\begingroup$ Any one of f or g does not continuous cannot make sure that f.g is continuous. Then, you cannot say the the limit of a certain point is the function value on that point. $\endgroup$ – eccstartup Jun 16 '13 at 1:22
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This solve your problem because they are both continuous function every where, you can pass the limit like this $\lim_{x\to x_0}f(g(x))=f(\lim_{x\to x_0}g(x))$ two times.

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