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Evaluate $$L=\lim _{x \to 0}\left(\frac{e^x}{e^x-1}-\frac{1}{x}\right)$$ I tried in two methods

Method $1$. $$L=\lim_{x \to 0}\left(\frac{\frac{e^x}{x}}{\frac{e^x-1}{x}}-\frac{1}{x}\right)$$

But $$\lim_{x \to 0}\frac{e^x-1}{x}=1$$ So

$$L=\lim_{x \to 0}\left(\frac{e^x}{x}-\frac{1}{x}\right)=\lim_{x \to 0}\frac{e^x-1}{x}=1$$

Method $2$. By taking LCM we get

$$L=\lim_{x \to 0}\frac{xe^x-e^x+1}{xe^x-x}$$ Since the limit is in $\frac{0}{0}$ form using L'Hopital's Rule twice we get

$$L=\lim_{x \to 0}\frac{xe^x}{xe^x+e^x-1}$$ i.e.,

$$L=\lim_{x \to 0}\frac{xe^x+e^x}{xe^x+2e^x}=\frac{1}{2}$$

I am sure that second method is correct, but i want to know whats is wrong in first method

Felix Marin
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Umesh shankar
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3 Answers3

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The first method is not correct. You either take the limit of the whole expression or you manipulate it. You cannot take the limit of only half of it and take the limit of the rest again later.

Momo
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If limits exist for both functions then you can certainly separate the limits as $$\lim_{x\to0}f(x)g(x) = \left(\lim_{x\to0}f(x)\right)\left(\lim_{x\to0}g(x)\right)$$

But this, doesn't holds true

$$\lim_{x\to0}f(x)g(x) \neq \lim_{x\to0}\left(f(x)\left(\lim_{x\to0}g(x)\right)\right)$$

Therefore, in your method I, the following is not true

$$L=\lim_{x \to 0}\frac{\frac{e^x}{x}}{\frac{e^x-1}{x}}-\frac{1}{x} \neq \lim_{x \to 0}\frac{e^x}{x}-\frac{1}{x}$$

kishlaya
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    This doesn't relate to the issue. Rather, the OP tried to use $\lim\left( \frac{f}{g}-h\right)=\lim \left(\frac{f}{\lim g}-h\right)$. – Mark Viola Jan 18 '17 at 05:56
  • I guess it does because I was trying to point out something general and not the particular limit, which I explained about later in the answer. – kishlaya Jan 18 '17 at 08:00
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    I don't think there is a problem with the second equation you have written. It is just another way of saying the same thing as the first equation. And this is not the reason why the first approach in the question fails. – Paramanand Singh Jan 18 '17 at 13:08
  • Oh ok. I got it now. While writing the solution, the first equation in particular, I wasn't trying to point out the error in OP's solution but just attempting to make a point. But, I guess, then that doesn't qualifies for a good answer.

    Thanks for the help.

     – kishlaya Jan 18 '17 at 14:15
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Note that $e^x -1 =x+\frac12 x^2+O(x^3)$. Hence, we see that

$$\begin{align} \frac{e^x/x}{(e^x-1)/x}-\frac1x&=\frac{e^x-\left(1+\frac12 x+O(x^2)\right)}{x(1+\frac12 x+O(x^2))}\\\\ &=\frac{\frac{e^x-1}{x}-\frac12+O(x)}{1+\frac12 x+O(x^2)} \end{align}$$

whereupon passing to the limit yields $1/2$ as expected!

Mark Viola
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