Finding a limit quickly

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Lorena_Santoro
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We might recognize the shape of a derivative.
$$\lim_{x\to\infty} x(e^{\frac 1x}-1)
=\lim_{x\to 0} \frac 1x(e^x-1)
=\lim_{x\to 0} \frac{e^x-e^0}{x-0}$$

This is the derivative of $e^x$ at $x=0$, which is $e^0=1$.