Question Number 184960 by mathlove last updated on 14/Jan/23
Answered by cortano1 last updated on 15/Jan/23
$${L}=−\frac{\mathrm{9}}{\mathrm{4}}\left(\mathrm{ln}\:\mathrm{4}−\mathrm{1}\right) \\ $$
Commented by mathlove last updated on 15/Jan/23
$${solution}??? \\ $$
Answered by cortano1 last updated on 15/Jan/23
$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left({x}+\mathrm{1}\right)−\mathrm{ln}\:\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{\mathrm{7}+{x}}−\sqrt{\mathrm{1}+\mathrm{3}{x}}}\:×\:\underset{{x}\rightarrow\mathrm{1}\:} {\mathrm{lim}}\frac{\mathrm{3}\left(\mathrm{4}^{{x}−\mathrm{1}} −{x}\right)}{\mathrm{sin}\:\left({x}−\mathrm{1}\right)} \\ $$$${L}_{\mathrm{1}} =\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left({t}+\mathrm{2}\right)−\mathrm{ln}\:\mathrm{2}}{\:\sqrt[{\mathrm{3}}]{{t}+\mathrm{8}}−\sqrt{\mathrm{4}+\mathrm{3}{t}}}\: \\ $$$${L}_{\mathrm{1}} =\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{{t}+\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{\left({t}+\mathrm{8}\right)^{\mathrm{2}} }}−\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{4}+\mathrm{3}{t}}}}\: \\ $$$${L}_{\mathrm{1}} =\:\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{12}}−\frac{\mathrm{3}}{\mathrm{4}}}\:=\frac{\mathrm{6}}{\mathrm{1}−\mathrm{9}}\:=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${L}_{\mathrm{2}} =\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{4}^{{x}−\mathrm{1}} −{x}\right)}{\mathrm{sin}\:\left({x}−\mathrm{1}\right)}=\:\mathrm{3}.\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{4}^{{x}−\mathrm{1}} −\mathrm{1}−\left({x}−\mathrm{1}\right)}{{x}−\mathrm{1}} \\ $$$$\:{L}_{\mathrm{2}} =\:\mathrm{3}\left(\mathrm{ln}\:\mathrm{4}−\mathrm{1}\right) \\ $$$$\therefore\:{L}={L}_{\mathrm{1}} ×{L}_{\mathrm{2}} =−\frac{\mathrm{9}}{\mathrm{4}}\left(\mathrm{ln}\:\mathrm{4}−\mathrm{1}\right) \\ $$
Commented by mathlove last updated on 15/Jan/23
$${thanks}\:{dear} \\ $$