Question

Solve;
$\underset{x\to 0}{lim}\frac{{(1+x)}^6-1}{{(1+x)}^2-1}$

Answer 1

Now,

$\underset{x\to 0}{lim}\frac{{(1+x)}^6-1}{{(1+x)}^2-1}$ $\left(\right)$ form

Now applying L'Hospital's rule we get,

$=\underset{x\to 0}{lim}\frac{6{(1+x)}^5}{2(1+x)}$

$=\frac{6}{2}$

$=3$.