Why does $\lim_{x \rightarrow 0}\frac{e^x -1}{x} = \lim_{x \rightarrow 0}\frac{x}{e^x -1}=1 $ ?
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2Possible duplicate of Proof of $ f(x) = (e^x-1)/x = 1 \text{ as } x\to 0$ using epsilon-delta definition of a limit – Nov 11 '15 at 16:01
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or of Proving that $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$. Please search before asking. – Nov 11 '15 at 16:03
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L-hopitals rule can also work. – Jasser Nov 11 '15 at 16:04
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Use L'Hopital's Rule:
$$\lim_{x\rightarrow 0}\frac{e^{x}-1}{x} = \lim_{x\rightarrow 0}\frac{\frac{d}{dx}e^{x}-1}{\frac{d}{dx}x} = \lim_{x\rightarrow 0}e^{x} \rightarrow 1$$
$$\lim_{x\rightarrow 0}\frac{x}{e^{x}-1} = \lim_{x\rightarrow 0}\frac{\frac{d}{dx}x}{\frac{d}{dx}e^{x}-1} = \lim_{x\rightarrow 0}\frac{1}{e^{x}} \rightarrow 1$$
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