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Question Number 116252 by bemath last updated on 02/Oct/20

lim_(x→−∞) ((e^x +e^(−x) )/(e^x −e^(−x) )) = ?

$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }\:=\:? \\ $$

Answered by MJS_new last updated on 02/Oct/20

((e^x +e^(−x) )/(e^x −e^(−x) ))=(1/(tanh x)) ⇒ limit=−1 or ((e^x +e^(−x) )/(e^x −e^(−x) ))=((e^(2x) +1)/(e^(2x) −1)) ⇒ limit=−1

$$\frac{\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} }{\mathrm{e}^{{x}} −\mathrm{e}^{−{x}} }=\frac{\mathrm{1}}{\mathrm{tanh}\:{x}}\:\Rightarrow\:\mathrm{limit}=−\mathrm{1} \\ $$$$\mathrm{or} \\ $$$$\frac{\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} }{\mathrm{e}^{{x}} −\mathrm{e}^{−{x}} }=\frac{\mathrm{e}^{\mathrm{2}{x}} +\mathrm{1}}{\mathrm{e}^{\mathrm{2}{x}} −\mathrm{1}}\:\Rightarrow\:\mathrm{limit}=−\mathrm{1} \\ $$

Commented by MJS_new last updated on 02/Oct/20

sorry took ∞ instead of −∞

$$\mathrm{sorry}\:\mathrm{took}\:\infty\:\mathrm{instead}\:\mathrm{of}\:−\infty \\ $$

Commented by bemath last updated on 02/Oct/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Oct/20

lim_(x→−∞) ((e^x +e^(−x) )/(e^x −e^(−x) ))=lim_(x→∞) ((e^(−x) +e^x )/(e^(−x) −e^x ))=−1 (lim_(x→∞) e^(−x) =0)

$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{e}^{−\mathrm{x}} +\mathrm{e}^{\mathrm{x}} }{\mathrm{e}^{−\mathrm{x}} −\mathrm{e}^{\mathrm{x}} }=−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}e}^{−\mathrm{x}} =\mathrm{0}\right) \\ $$

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