Question

$\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^4-1}{{\left(1+x\right)}^3-1}$

Answer 1

Now,

$\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^4-1}{{\left(1+x\right)}^3-1}$ $\left(\frac{0}{0}\right)$ form.

Now applying L'Hospital's rule we get,

$\underset{x\to 0}{lim}\frac{4{\left(1+x\right)}^3}{3{\left(1+x\right)}^2}$

$=\frac{4}{3}$.