Question Number 215831 by golsendro last updated on 19/Jan/25
$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }−\sqrt{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} }}{\:\sqrt{\mathrm{x}}}\:=? \\ $$
Answered by efronzo1 last updated on 19/Jan/25
$$\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6x}^{\mathrm{2}} +\mathrm{2}}{\:\sqrt{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }+\sqrt{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} }}\: \\ $$$$\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{6}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}+..}\:+\sqrt{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{x}}+...}\:\right)} \\ $$$$\:\:=\:\frac{\mathrm{6}}{\mathrm{1}+\mathrm{1}}\:=\:\mathrm{3}\: \\ $$
Answered by MrGaster last updated on 19/Jan/25
$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left({x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left({x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\:\sqrt{\pi}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left({x}^{\frac{\mathrm{3}}{\mathrm{2}}} +\frac{\mathrm{3}}{\mathrm{2}}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)−\left({x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{3}}{\mathrm{2}}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)}{\:\sqrt{\pi}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}\sqrt{{x}}}{\:\sqrt{{x}}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}3} \\ $$$$=\begin{array}{|c|}{\mathrm{3}}\\\hline\end{array} \\ $$