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I am having problem in solving this indetermination. $\lim_{x\to 0} \dfrac{(e^x+3x-1)}{(e^{2x}-1)}$

I tried to do two limits $\dfrac {(e^x+3x-1)}{x}$ times $\dfrac {x}{(e^{2x}-1)}$ but I don't know how to get rid of $e^{\dfrac y3}$ in the first limit after doing the first variable exchange.

Could you guys give a hint, please? Thank you

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    L'Hopital's Rule? – 2017-02-26
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    Taylor series ? – 2017-02-26
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    $${e^x+3x-1\over e^{2x}-1}={1\over e^x+1}\left(1+3{x\over e^x-1}\right)\to{1\over2}\left(1+{3\over 1}\right)=\ldots$$ – 2017-02-26

2 Answers 2

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Recall that $\lim_{h\to0}\frac{e^h-1}h=1$. Thus,

$$\lim_{x\to0}\frac x{e^{2x}-1}=\frac12\lim_{x\to0}\frac{2x}{e^{2x}-1}=\frac12$$

$$\lim_{x\to0}\frac{e^x+3x-1}x=\lim_{x\to0}\frac{e^x-1}x+\frac{3x}x=1+3=4$$

Thus,

$$\lim_{x\to0} {e^x+3x-1\over e^{2x}-1}=4\times\frac12=2$$

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    Now, I can do. :) I just have to separate the (3x)/x Thank you! – 2017-02-26
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Well, we can apply l'Hôpital's rule:

$$\lim_{x\to0}\space\frac{e^x+3x-1}{e^{2x}-1}=\lim_{x\to0}\space\frac{e^x+3}{2e^{2x}}=\frac{1}{2}\cdot\lim_{x\to0}\space\frac{e^x+3}{e^{2x}}=\frac{1}{2}\cdot\frac{e^0+3}{e^{2\cdot0}}=\frac{1}{2}\cdot\frac{1+3}{1}=2$$