Question

Evaluate $\underset{x\to 0}{lim}\frac{(1+x{)}^6-1}{(1+x{)}^5-1}$

Answer 1

Put (1+x)=y, so that when $x\to 0$ then $y\to 1.$

$\because \underset{x\to 0}{lim}\frac{(1+x{)}^6-1}{(1+x{)}^5-1}$

$=\underset{y\to 1}{lim}\left(\frac{y^6-1}{y^5-1}\right)=\frac{\underset{y\to 1}{lim}\left(\frac{y^6-1}{y-1}\right)}{\underset{y\to 1}{lim}\left(\frac{y^5-1}{y-1}\right)}=\frac{\underset{y\to 1}{lim}\left(\frac{y^6-1^6}{y-1}\right)}{\underset{y\to 1}{lim}\left(\frac{y^5-1^5}{y^5-1}\right)}=\frac{6\times 1^{(6-1)}}{5\times 1^{(5-1)}}$

$=\frac{6\times 1^5}{5\times 1^4}=\frac{6}{5}[\because \underset{x\to a}{lim}\left(\frac{x^n-a^n}{x-a}\right)=n{a}^{n-1}]$