Question

Lim x->1+( sin(1-x)-(e^(x-1))+1)/ lnx

Answer 1

We're given the one-sided limit,

$\displaystyle\lim_{x\to1^+}\frac{\sin(1-x)-e^{x-1}+1}{\ln(x)}$

Evaluating the limand directly at x = 1 gives the indeterminate from

(sin(1 - 1) - exp(1 - 1) + 1) / ln(1) = 0/0

so we can potentially solve the limit by applying L'Hopital's rule. Doing so gives

$\displaystyle\lim_{x\to1^+}\frac{\sin(1-x)-e^{x-1}+1}{\ln(x)}=\lim_{x\to1^+}\frac{-\cos(1-x)-e^{x-1}}{\frac1x}=\frac{-\cos(0)-e^0}{\frac11}=\boxed{-2}$