Question Number 192969 by mustafazaheen last updated on 01/Jun/23
Commented by mustafazaheen last updated on 01/Jun/23
$$\mathrm{how}\:\mathrm{is}\:\mathrm{solution}? \\ $$
Answered by MM42 last updated on 01/Jun/23
$${lim}_{{x}\rightarrow\mathrm{1}^{+} } \:\left(\frac{\mathrm{3}}{{x}−\mathrm{1}}\right)^{{ln}\frac{\mathrm{4}}{\mathrm{3}}} =\left(\frac{\mathrm{3}}{\mathrm{0}^{+} }\right)^{+} =\left(+\infty\right)^{+} =+\infty \\ $$
Answered by aba last updated on 01/Jun/23
$$\left(\frac{\mathrm{x}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} =\mathrm{e}^{\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\mathrm{ln}\left(\mathrm{x}−\mathrm{1}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\left(\frac{\mathrm{x}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} =\mathrm{e}^{+\infty} =+\infty \\ $$