Question

Evaluate:

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

Answer 1

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

$=\frac{(1+1{)}^6-1}{(1+1{)}^2-1}$

$=\frac{(2{)}^6-1}{(2{)}^2-1}$

$=\frac{64-1}{4-1}$

$=\frac{63}{3}$

$=21$