We’re given the one-sided limit, [tex]\displaystyle\lim_{x\to1^+}\frac{\sin(1-x)-e^{x-1}+1}{\ln(x)}[/tex] 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 [tex]\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}[/tex] Log in to Reply

We’re given the one-sided limit,

[tex]\displaystyle\lim_{x\to1^+}\frac{\sin(1-x)-e^{x-1}+1}{\ln(x)}[/tex]

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

[tex]\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}[/tex]