For example I need to find $\lim_{x\to 0}f(x)$. I know that $\lim_{x\to 0}\frac x{\sin x} = 1$. So I think it is reasonable to assume that $x \approx \sin x$ when $x$ approaches $0$. Then can I say that $\lim_{x\to 0}f(x)$ = $\lim_{x\to 0}f(\sin x)$ ?
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Assume that $\lim_{x\rightarrow 0}f(x)=L$. Given $\epsilon>0$, there exists some $\delta>0$ such that $0<|x|<\delta$ implies $|f(x)-L|<\epsilon$. Use the fact that for small neighborhood of $0$, $\sin$ has only one zero and closed to zero. So there exists some $\delta'>0$ such that $0<|x|<\delta'$ implies $0<|\sin x|<\delta$, then $|f(\sin x)-L|<\epsilon$, this proves $\lim_{x\rightarrow 0}f(\sin x)=L$.
The other direction is similar since we can find for some $\eta'>0$ such that $0<|x|<\delta'$ implies that $0<|\sin^{-1}x|<\eta$ where $\eta$ is such that $0<|x|<\eta$ implies $|f(\sin x)-L|<\epsilon$.
user284331
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If f is continuous
$$\lim_{x\to 0}f(\sin x)=f(\lim_{x\to 0}\sin x))=\lim_{x\to 0}f(x)$$
user
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2Not tacitly, I’ve written it! – user Mar 04 '18 at 18:53
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2I’ve written indeed if f is continuous, of course this is not a general case. – user Mar 04 '18 at 19:01
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Indeed of course it is a particular case, the general gas been proved by the accepted answer, do you think I should delete mine? – user Mar 04 '18 at 19:34
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Apology, but I overlooked your qualifying statement. This is fine. – Mark Viola Mar 04 '18 at 19:36
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1No apology necessary. I just had somehow overlooked your preface. (+1) – Mark Viola Mar 04 '18 at 20:09