Question

Evaluate the limit $\underset{x\to \infty }{lim}{\left(\frac{x^2+1}{x^2-1}\right)}^{x^2}=$

• A. $e$
• B. $\frac{1}{e}$
• C. $e^2$
• D. $e^{-2}$

Answer 1

The correct option is C $e^2$

$\underset{x\to \infty }{lim}{\left(\frac{x^2+1}{x^2-1}\right)}^{x^2}=1^{\infty }$

$=e^{\underset{x\to \infty }{lim}{x}^2(\frac{x^2+1}{x^2-1}-1)}=e^{\underset{x\to \infty }{lim}{x}^2\left(\frac{2}{x^2-1}\right)}$

$=e^{2\underset{x\to \infty }{lim}(1+\frac{1}{x^2-1})}=e^{2x(1+0)}=e^2$.