Question

Evaluate : ${lim}_{x\to 0}\frac{(1+x{)}^6-1}{(1+x{)}^2-1}$

Answer 1

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

Applying L-Hospital rule

$⟹\underset{x\to 0}{lim}\frac{6(1+x{)}^5}{2(1+x)}=3$