Question Number 154586 by Dadoubleny last updated on 19/Sep/21 | ||
Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21 | ||
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\left({n}+\mathrm{1}\right)^{\alpha} −{n}^{\alpha} \right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({n}^{\alpha} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{\alpha} −{n}^{\alpha} \right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left({n}^{\alpha} \left(\mathrm{1}+\frac{\alpha}{{n}}\right)−{n}^{\alpha} \right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\alpha}{{n}}\right)=\mathrm{0} \\ $$ | ||