Question

Compute :$\underset{x\to \infty }{lim}\frac{(\sqrt{x^2}+x-x)}{x}$

Answer 1

$\underset{x\to \infty }{lim}(\sqrt{x^2}+x-x)$

$=(x+x-x)$

$=(2x-x)$

$=x$

$\underset{x\to \infty }{lim}=\frac{x}{x}=1$.