Can anyone give me some hints on starting to prove the following
$$\lim_{x \rightarrow 0} \frac{e^x-\sin x}{x-\sin x}=1$$
Solving limit problems which include "e"
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limits
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2The right limit is $+ \infty$ and the left limit is $- \infty$. It does not approach to $1$. – 2017-01-11
1 Answers
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$\lim\limits_{x\to 0^{\pm}} x-\sin x = [0^{\pm}]=0\,$, so $$\lim\limits_{x\to 0^{\pm}} \frac{ e^x - \sin x}{x-\sin x} =\left[\frac{1}{0^{\pm}}\right] = \pm \infty $$
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0So the question is incorrect right? – 2017-01-11
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1You asked if anyone can give you hints on starting to prove your equation... So there is a hint - don't start :) – 2017-01-11