Limit Exercise

Exercise:

Show that , $\underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt[3]{x}-\sqrt{x}}{1-x}=\frac{1}{6}$

Solution: Let $x={{t}^{6}}$ as $x\to 1,t\to 1$

So $\underset{x\to 1}{\mathop{\lim }}\,\frac{\sqrt[3]{x}-\sqrt{x}}{1-x}=\underset{t\to 1}{\mathop{\lim }}\,\frac{\sqrt[3]{{{t}^{6}}}-\sqrt{{{t}^{6}}}}{1-{{t}^{6}}}=\underset{t\to 1}{\mathop{\lim }}\,\frac{{{t}^{2}}-{{t}^{3}}}{1-{{t}^{6}}}=\underset{t\to 1}{\mathop{\lim }}\,\frac{{{t}^{2}}\left( 1-t \right)}{1-{{t}^{6}}}$

But $1-{{t}^{6}}=1-{{\left( {{t}^{2}} \right)}^{3}}=\left( 1-{{t}^{2}} \right)\left( {{t}^{4}}+{{t}^{2}}+1 \right)$

So $\underset{t\to 1}{\mathop{\lim }}\,\frac{{{t}^{2}}\left( 1-t \right)}{1-{{t}^{6}}}=\underset{t\to 1}{\mathop{\lim }}\,\frac{{{t}^{2}}\left( 1-t \right)}{\left( 1-{{t}^{2}} \right)\left( {{t}^{4}}+{{t}^{2}}+1 \right)}=\underset{t\to 1}{\mathop{\lim }}\,\frac{{{t}^{2}}\left( 1-t \right)}{\left( 1-t \right)\left( 1+t \right)\left( {{t}^{4}}+{{t}^{2}}+1 \right)}$

$=\underset{t\to 1}{\mathop{\lim }}\,\frac{{{t}^{2}}}{\left( 1+t \right)\left( {{t}^{4}}+{{t}^{2}}+1 \right)}=\frac{1}{2\times 3}=\frac{1}{6}$