Question

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

Answer 1

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

Put 1+x =y

as $x\to 0,y\to 1$

${lim}_{x\to 0}\frac{y^6-1}{y^2-1}$

Dividing numerator and denominator by y -1

${lim}_{y\to 1}\frac{\frac{y^6-1}{y-1}}{\frac{y^2-1}{y-1}}$

$=\frac{{lim}_{y\to 1}\frac{y^6-(1{)}^6}{y-1}}{{lim}_{y\to 1}\frac{y^2-(1{)}^2}{y-1}}$

$=\frac{6(1{)}^{6-1}}{2(1{)}^{2-1}}=\frac{6}{2}=3$