Question

Evaluate the limit:
$\underset{x\to \infty }{lim}x[\sqrt{x^2+1}-\sqrt{x^2-1}]$

Answer 1

We have,
$\underset{x\to \infty }{lim}x[\sqrt{x^2+1}-\sqrt{x^2-1}]$

On rationalising numerator ,we get

$=\underset{x\to \infty }{lim}\frac{x[\sqrt{x^2+1}-\sqrt{x^2-1}][\sqrt{x^2+1}+\sqrt{x^2-1}]}{[\sqrt{x^2+1}+\sqrt{x^2-1}]}$

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

$[\because (a-b\left)\right(a+b)=a^2-b^2]$

$=\underset{x\to \infty }{lim}\frac{x[x^2+1-x^2+1]}{[\sqrt{x^2+1}+\sqrt{x^2-1}]}$

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

Dividing numerator and denominator by $x$, we get

$=\underset{x\to \infty }{lim}\frac{\frac{2x}{x}}{[\sqrt{\frac{x^2+1}{x^2}}+\sqrt{\frac{x^2-1}{x^2}}]}$

$=\underset{x\to \infty }{lim}\frac{2}{[\sqrt{1+\frac{1}{x^2}}+\sqrt{1-\frac{1}{x^2}}]}$

When $x\to \infty$, then $\frac{1}{x}\to 0$

$=\frac{2}{[\sqrt{1+0}+\sqrt{1-0}]}=\frac{2}{2}=1$

$\therefore \underset{x\to \infty }{lim}x[\sqrt{x^2+1}-\sqrt{x^2-1}]=1$