Question

Solve:-

$\underset{x\to 0}{lim}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+4}-2}$

Answer 1

Now,

$\underset{x\to 0}{lim}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+4}-2}$

Rationalize both the numerator and the denominator

$=\underset{x\to 0}{lim}\frac{\left\{\right(\sqrt{x^2+1}{)}^2-1^2\left\}\right\{\sqrt{x^2+4}+2\}}{\left(\sqrt{x^2+4}{)}^2-2^2\right\}\{\sqrt{x^2+1}+1\}}$

$=\underset{x\to 0}{lim}\frac{\left\{\right(x^2\left)\right\}\{\sqrt{x^2+4}+2\}}{\left(x^2\right)\{\sqrt{x^2+1}+1\}}$

$=\underset{x\to 0}{lim}\frac{\{\sqrt{x^2+4}+2\}}{\{\sqrt{x^2+1}+1\}}$

$=\frac{2+2}{1+1}$

$=2$.