Question Number 116252 by bemath last updated on 02/Oct/20
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }\:=\:? \\ $$
Answered by MJS_new last updated on 02/Oct/20
$$\frac{\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} }{\mathrm{e}^{{x}} −\mathrm{e}^{−{x}} }=\frac{\mathrm{1}}{\mathrm{tanh}\:{x}}\:\Rightarrow\:\mathrm{limit}=−\mathrm{1} \\ $$$$\mathrm{or} \\ $$$$\frac{\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} }{\mathrm{e}^{{x}} −\mathrm{e}^{−{x}} }=\frac{\mathrm{e}^{\mathrm{2}{x}} +\mathrm{1}}{\mathrm{e}^{\mathrm{2}{x}} −\mathrm{1}}\:\Rightarrow\:\mathrm{limit}=−\mathrm{1} \\ $$
Commented by MJS_new last updated on 02/Oct/20
$$\mathrm{sorry}\:\mathrm{took}\:\infty\:\mathrm{instead}\:\mathrm{of}\:−\infty \\ $$
Commented by bemath last updated on 02/Oct/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by Dwaipayan Shikari last updated on 02/Oct/20
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{e}^{−\mathrm{x}} +\mathrm{e}^{\mathrm{x}} }{\mathrm{e}^{−\mathrm{x}} −\mathrm{e}^{\mathrm{x}} }=−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}e}^{−\mathrm{x}} =\mathrm{0}\right) \\ $$