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I am required to find $\lim\limits_{x \to 0^+} \frac{xe^x}{e^x-1}$.


My attempt:

$\lim\limits_{x \to 0^+} \frac{xe^x}{e^x-1}$ = $\lim\limits_{x \to 0^+} e^x$ $\cdot$ $\lim\limits_{x \to 0^+} \frac{x}{e^x-1}$

$=1\cdot\lim\limits_{x \to 0^+} \frac{x}{e^x-1}$

$=\lim\limits_{x \to 0^+} e^{-x}$ (L'Hopital's Rule)

$= 1$ (which is the correct answer)


My question is: Why am I not able to apply the rule to the equation right from the beginning [since substituting $0$ we get $\frac{0}{0}$]?

$\lim\limits_{x \to 0^+} \frac{xe^x}{e^x-1}$

$=\lim\limits_{x \to 0^+} \frac{xe^x}{e^x}$

$=\lim\limits_{x \to 0^+}x$

$=0$ (wrong)

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    $\begingroup$ It is applicable, you're just computing the derivative wrong. $$(xe^x)^\prime = xe^x + e^x.$$ $\endgroup$ – Philip Hoskins Feb 14 '16 at 7:13
  • $\begingroup$ But why use L'Hopital's rule? $\endgroup$ – gniourf_gniourf Feb 14 '16 at 7:32
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If you want to apply L'Hospital's rule right from the beginning, you may write $$ \begin{align} \lim\limits_{x \to 0^+}\frac{xe^x}{e^x-1}&=\lim\limits_{x \to 0^+}\frac{e^x+xe^x}{e^x}=\frac{1+0}1=1 \end{align} $$ since $$ (xe^x)'=e^x+xe^x. $$

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