Question

Solve:

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

Answer 1

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

L'Hospital's Rule,

$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

$\Rightarrow$ $\underset{x\to 0}{lim}\frac{\frac{d}{dx}(x+1{)}^5-\frac{d}{dx}(1)}{\frac{d}{dx}\left(x\right)}$

$\Rightarrow$ $\underset{x\to 0}{lim}\frac{5(x+1)(1+0)-0}{1}$

$\Rightarrow$ $\underset{x\to 0}{lim}5(x+1)$

$\Rightarrow$ $5(0+1)$

$\Rightarrow$ $5$