Question

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

Answer 1

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

Applying L-Hospitals rule

$=\underset{x\to 0}{lt}\frac{5(1+x{)}^4}{1}=5(1+0{)}^4$

$=5$.