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Question Number 10830 by Saham last updated on 26/Feb/17

lim_(x→∞)   ((3^x  − 3^(−x) )/(3^x  + 3^(−x) ))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}^{\mathrm{x}} \:−\:\mathrm{3}^{−\mathrm{x}} }{\mathrm{3}^{\mathrm{x}} \:+\:\mathrm{3}^{−\mathrm{x}} } \\ $$

Answered by bahmanfeshki last updated on 26/Feb/17

lim_(x→∞) ((3^x (1−(1/3^(2x) )))/(3^x (1+(1/3^(2x) ))))=1

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}^{{x}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}{x}} }\right)}{\mathrm{3}^{{x}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}{x}} }\right)}=\mathrm{1} \\ $$

Answered by FilupS last updated on 27/Feb/17

L=lim_(x→∞) ((3^x −3^(−x) )/(3^x +3^(−x) ))  L=lim_(x→∞) ((1−3^(−2x) )/(1+3^(−2x) ))=((1−0)/(1+0))  L=1

$${L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} } \\ $$$${L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{3}^{−\mathrm{2}{x}} }{\mathrm{1}+\mathrm{3}^{−\mathrm{2}{x}} }=\frac{\mathrm{1}−\mathrm{0}}{\mathrm{1}+\mathrm{0}} \\ $$$${L}=\mathrm{1} \\ $$

Commented by Saham last updated on 27/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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