Question Number 25026 by NECx last updated on 02/Dec/17
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{{x}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{1}/\mathrm{4}} −\mathrm{1}}\right) \\ $$$${Evaluate}\:{this} \\ $$$$ \\ $$
Answered by nnnavendu last updated on 02/Dec/17
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{{x}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{1}/\mathrm{4}} −\mathrm{1}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\left({x}^{\mathrm{1}/\mathrm{4}} \right)^{\mathrm{4}/\mathrm{3}} −\mathrm{1}^{\mathrm{4}/\mathrm{3}} }{{x}^{\mathrm{1}/\mathrm{4}} −\mathrm{1}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\mathrm{4}/\mathrm{3}\:\mathrm{1}^{\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}} \right) \\ $$$$\mathrm{4}/\mathrm{3} \\ $$$$ \\ $$$${f}\mathrm{armula}− \\ $$$$\underset{{x}\rightarrow\mathrm{n}} {\mathrm{lim}}\left(\frac{{x}^{\mathrm{n}} −\mathrm{a}^{\mathrm{n}} }{{x}−\mathrm{a}}\right)={na}^{{n}−\mathrm{1}} \\ $$$$ \\ $$
Commented by moxhix last updated on 02/Dec/17
$${I}\:{think}… \\ $$$$\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{{x}^{{n}} −{a}^{{n}} }{{x}−{a}}\right)=\left(\frac{{d}}{{dx}}{x}^{{n}} \right)_{{x}={a}} =\:{na}^{{n}−\mathrm{1}} \\ $$
Answered by jota+ last updated on 03/Dec/17
$$=\frac{\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:{x}^{−\mathrm{2}/\mathrm{3}} /\mathrm{3}}{\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:{x}^{−\mathrm{3}/\mathrm{4}} /\mathrm{4}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${Use}\:{Hopital}\:{rule} \\ $$$$ \\ $$
Commented by NECx last updated on 02/Dec/17
$${thanks}\:{bosses}….\:{I}\:{really}\:{appreciate}\:{this} \\ $$