Question

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

Answer 1

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

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

$\log y=\frac{1}{x^2}\log \left(\frac{1+x^2}{1-x^2}\right)$

$y=e^{\underset{x\to 0}{lim}\frac{1}{x^2}[\frac{1+x^2}{1-x^2}-1]}$

$y=e^{\underset{x\to 0}{lim}\frac{1}{x^2}\left[\frac{2x^2}{1-x^2}\right]}.$ $\{\because \underset{x\to a}{lim}f(x{)}^{2\left(x\right)}$

$\Rightarrow 1^{\infty }$

$=e^{\underset{x\to 0}{lim}\left(\frac{2}{1-x^2}\right)}=e^2$ then $=e^{\underset{x\to a}{lim}g\left(x\right)\left[f\right(x)-1]}$

$y=e^2$