Jan 4, 2022 #2 I like Serena Science Advisor Homework Helper MHB Messages 16,335 Reaction score 258 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 $e^0=1$.
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 $e^0=1$.