Question Number 149569 by mathdanisur last updated on 06/Aug/21
$$\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{5x}\:+\:\mathrm{6}}{\mathrm{2x}\:-\:\mathrm{9}}\right)^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } =\:? \\ $$
Answered by Ar Brandon last updated on 06/Aug/21
$$\mathscr{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{5}{x}+\mathrm{6}}{\mathrm{2}{x}−\mathrm{9}}\right)^{{x}^{\mathrm{2}} } =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{5}{x}\left(\mathrm{1}+\frac{\mathrm{6}}{\mathrm{5}{x}}\right)}{\mathrm{2}{x}\left(\mathrm{1}−\frac{\mathrm{9}}{\mathrm{2}{x}}\right)}\right)^{{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}5}{e}^{{x}^{\mathrm{2}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{6}}{\mathrm{5}{x}}\right)} \centerdot\mathrm{2}{e}^{−{x}^{\mathrm{2}} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{9}}{\mathrm{5}{x}}\right)} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}5}{e}^{{x}^{\mathrm{2}} \left(\frac{\mathrm{6}}{\mathrm{5}{x}}\right)} \centerdot\mathrm{2}{e}^{−{x}^{\mathrm{2}} \left(−\frac{\mathrm{9}}{\mathrm{5}{x}}\right)} \\ $$$$\:\:\:\:\:=\mathrm{10}\underset{{x}\rightarrow\infty} {\mathrm{lim}}{e}^{\mathrm{3}{x}} \approx+\infty \\ $$
Commented by mathdanisur last updated on 06/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$