Menu Close

lim-x-3-x-3-x-3-x-3-x-

Tinku Tara 0 Comments
Pin




Question Number 137737 by bemath last updated on 06/Apr/21
lim_(x→−∞) ((3^x −3^(−x) )/(3^x +3^(−x) )) =?
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }\:=? \\ $$$$ \\ $$
Answered by greg_ed last updated on 06/Apr/21
L′Hopital′s rule lim_(x→−∞) ((3^x −3^(−x) )/(3^x +3^(−x) )) = lim_(x→−∞) (((d/dx)(3^x −3^(−x) ))/((d/dx)(3^x +3^(−x) ))) = lim_(x→−∞) (((ln(3)×9^x +ln(3))/3^x )/((ln(3)×9^x −ln(3))/3^x )) = lim_(x→−∞) ((ln(3)×9^x +ln(3))/(ln(3)×9^x −ln(3))) = lim_(x→−∞) ((9^x +1)/(9^x −1)) since lim_(x→−∞) 9^x = 0, then : lim_(x→−∞) ((3^x −3^(−x) )/(3^x +3^(−x) )) = −1.
$$\boldsymbol{\mathrm{L}}'\boldsymbol{\mathrm{Hopital}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{rule}}\: \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\:\frac{\frac{{d}}{{dx}}\left(\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} \right)}{\frac{{d}}{{dx}}\left(\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} \right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\:\frac{\frac{{ln}\left(\mathrm{3}\right)×\mathrm{9}^{{x}} +{ln}\left(\mathrm{3}\right)}{\mathrm{3}^{\boldsymbol{{x}}} }}{\frac{{ln}\left(\mathrm{3}\right)×\mathrm{9}^{{x}} −{ln}\left(\mathrm{3}\right)}{\mathrm{3}^{\boldsymbol{{x}}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\:\frac{\boldsymbol{{ln}}\left(\mathrm{3}\right)×\mathrm{9}^{{x}} +\boldsymbol{{ln}}\left(\mathrm{3}\right)}{\boldsymbol{{ln}}\left(\mathrm{3}\right)×\mathrm{9}^{{x}} −\boldsymbol{{ln}}\left(\mathrm{3}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\:\frac{\mathrm{9}^{{x}} +\mathrm{1}}{\mathrm{9}^{{x}} −\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{since}}\:\underset{\boldsymbol{{x}}\rightarrow−\infty} {\boldsymbol{\mathrm{lim}}}\mathrm{9}^{\boldsymbol{{x}}} \:=\:\mathrm{0},\:\:\:\boldsymbol{\mathrm{then}}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow−\infty} {\boldsymbol{\mathrm{lim}}}\:\:\frac{\mathrm{3}^{\boldsymbol{{x}}} −\mathrm{3}^{−\boldsymbol{{x}}} }{\mathrm{3}^{\boldsymbol{{x}}} +\mathrm{3}^{−\boldsymbol{{x}}} }\:=\:−\mathrm{1}. \\ $$
Answered by bemath last updated on 06/Apr/21

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *