Answer to Question #223100 in Calculus for Jay

Question

Evaluate the limit as x turns to 1^-

Lim [sin(x²-1) / e^(Inx) - 1]

Answer 1

$\frac{sin(x^2-1)}{e^{lnx}-1}=\frac{sin(x^2-1)}{x-1}$

using L'hopital's rule:

$(sin(x^2-1))'=2xcos(x^2-1)$

$(x-1)'=1$

$\displaystyle \lim_{x\to 1^-}\frac{sin(x^2-1)}{x-1}=\displaystyle \lim_{x\to 1^-}2xcos(x^2-1)=2$