Question Number 117417 by bobhans last updated on 11/Oct/20
$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left(\mathrm{tanh}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)−\frac{\mathrm{1}}{\mathrm{cosh}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} =? \\ $$
Answered by Lordose last updated on 11/Oct/20
$$\mathrm{1} \\ $$
Answered by MJS_new last updated on 11/Oct/20
$$\mathrm{let}\:{x}=\frac{\mathrm{1}}{{t}} \\ $$$$\mathrm{tanh}\:{t}\:=\frac{\mathrm{e}^{\mathrm{2}{t}} −\mathrm{1}}{\mathrm{e}^{\mathrm{2}{t}} +\mathrm{1}} \\ $$$$\mathrm{cosh}\:{t}\:=\frac{\mathrm{e}^{\mathrm{2}{t}} +\mathrm{1}}{\mathrm{2e}^{{t}} } \\ $$$$\Rightarrow \\ $$$$\underset{{t}\rightarrow+\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{e}^{\mathrm{2}{t}} −\mathrm{2e}^{{t}} −\mathrm{1}}{\mathrm{e}^{\mathrm{2}{t}} +\mathrm{1}}\right)^{{t}} = \\ $$$$=\underset{{t}\rightarrow+\infty} {\mathrm{lim}}\:\left(\mathrm{1}−\frac{\mathrm{2}\left(\mathrm{e}^{{t}} +\mathrm{1}\right)}{\mathrm{e}^{\mathrm{2}{t}} +\mathrm{1}}\right)^{{t}} =\mathrm{1} \\ $$
Commented by bemath last updated on 12/Oct/20
$$\mathrm{the}\:\mathrm{last}\:\mathrm{line}\:\mathrm{why}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1}\:\mathrm{sir} \\ $$
Commented by MJS_new last updated on 12/Oct/20
$$\mathrm{1}−\frac{\mathrm{2}\left(\mathrm{e}^{{t}} +\mathrm{1}\right)}{\mathrm{e}^{\mathrm{2}{t}} +\mathrm{1}}\sim\mathrm{1}−\frac{\mathrm{2e}^{{t}} }{\mathrm{e}^{\mathrm{2}{t}} }=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{e}^{{t}} }\sim\mathrm{1} \\ $$