Question

$\text{The value of}{lim}_{x\to \infty }{\left(\frac{x^2-1}{x^2+1}\right)}^{x^2}is$

• A. 1
• B. $e^{-1}$
• C. $e^{-2}$
• D. $e^{-3}$

Answer 1

The correct option is C

$e^{-2}$

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