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Evaluate $\lim_\limits{x \to 0}\ \dfrac{e^x-1}{e^{2x}-1}$ using $\delta - \varepsilon$ definition.

Attempt: I claim that $\lim_\limits{x \to 0}\ \dfrac{e^x-1}{e^{2x}-1} = \dfrac 12$. $\forall \varepsilon >0, \exists\delta>0$ such that

$$|\frac{e^x-1}{e^{2x}-1}-\frac12| = |\frac {2e^x-2-e^{2x}+1}{2e^{2x}-2}|=|\frac {2e^x-e^{2x}-1}{2e^{2x}-2}|\le |\cdot|<\varepsilon$$

I don't know how to proceed from here. I appreciate any hint.

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2 Answers 2

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HINT you can reduce the expression like, for $x \ne 0$ $$\frac{e^x-1}{e^{2x}-1}=\frac{e^x-1}{(e^{x}-1)(e^{x}+1)}=\frac{1}{e^{x}+1}$$

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Will be easier to evaluate if you first factorise the denominator followed by the delta-epsilon evaluation Hopefully you can proceed from there.

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