Question

lim x->1 [(ln x)^2/e^(x-1)-x] ?

Answer 1

Use L'Hopital's rule 2 times

Answer 2

differentiate top and bottom
\[\rightarrow \lim_{x \rightarrow 1}\frac{\frac{2\ln(x)}{x}}{e^{x-1} -1} = \lim_{x \rightarrow 1} \frac{2\ln(x)}{xe^{x-1}-x}\]
differentiate again
\[\rightarrow \lim_{x \rightarrow 1}\frac{2/x}{xe^{x-1}+e^{x-1}-1} = 2\]

Answer 3

thanks!