Question

$\underset{x\to \infty }{lt}\frac{e^{x^2}-1}{e^{x^2}+1}=$

Answer 1

$\underset{x\to \infty }{Lt}\frac{e^{x^2}-1}{e^{x^2}+1}$
$=\underset{x\to \infty }{Lt}\frac{1-{\left(\frac{1}{e}\right)}^{x^2}}{1+{\left(\frac{1}{e}\right)}^{x^2}}=\frac{1-0}{1+0}=1.$
$\because e>1\therefore \frac{1}{e}<1$ and ${\left(\frac{1}{e}\right)}^{x^2}\to 0$ as $x\to \infty$