Question Number 5704 by sanusihammed last updated on 24/May/16
![Show that ... Limit [((3^x − 3^(−x) )/(3^(x ) + 3^(−x) ))] = − 1 x → −∞](Q5704.png)
$${Show}\:{that}\:... \\ $$$$ \\ $$$${Limit}\:\:\:\:\:\:\left[\frac{\mathrm{3}^{{x}} \:−\:\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}\:} \:+\:\mathrm{3}^{−{x}} }\right]\:=\:−\:\mathrm{1} \\ $$$${x}\:\rightarrow\:−\infty \\ $$
Answered by FilupSmith last updated on 24/May/16
$${L}=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} } \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{{x}} −\frac{\mathrm{1}}{\mathrm{3}^{{x}} }}{\mathrm{3}^{{x}} +\frac{\mathrm{1}}{\mathrm{3}^{{x}} }} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\frac{\mathrm{3}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{3}^{{x}} }}{\frac{\mathrm{3}^{\mathrm{2}{x}} +\mathrm{1}}{\mathrm{3}^{{x}} }} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{{x}} \left(\mathrm{3}^{\mathrm{2}{x}} −\mathrm{1}\right)}{\mathrm{3}^{{x}} \left(\mathrm{3}^{\mathrm{2}{x}} +\mathrm{1}\right)} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}{x}} +\mathrm{1}} \\ $$$$\mathrm{use}\:\mathrm{quotant}\:\mathrm{rule} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}\left(−\infty\right)} +\mathrm{1}} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{\mathrm{2}{x}} −\mathrm{1}}{\mathrm{1}} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\mathrm{3}^{\mathrm{2}{x}} −\mathrm{1} \\ $$$${L}=−\mathrm{1} \\ $$
Commented by Yozzii last updated on 24/May/16
$${The}\:{definition}\:{of}\:{the}\:{limit}\:{of}\:{f}\left({x}\right)\:{to}\:−\infty, \\ $$$${if}\:{it}\:{exists},{can}\:{be}\:{used}\:{to}\:{verify}\:{that}\:{the}\:{limit} \\ $$$${found}\:{is}\:{plausible}. \\ $$$$−−−−−−−−−−−−−−−−−−−−− \\ $$$${According}\:{to}\:{the}\:\epsilon−\delta\:{definition}\:{of} \\ $$$${the}\:{limit}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left({x}\right)={l},\:{for}\:{any}\:\epsilon>\mathrm{0}\:{there} \\ $$$${exists}\:{a}\:{number}\:\delta<\mathrm{0}\:{such}\:{that}\:\mid{f}\left({x}\right)−{l}\mid<\epsilon\:{whenever}\:{x}<\delta. \\ $$$${Suppose}\:{that}\:\delta=\frac{{ln}\left(\mathrm{2}\epsilon^{−\mathrm{1}} −\mathrm{1}\right)^{−\mathrm{1}} }{\mathrm{2}{ln}\mathrm{3}},\:\left(\mathrm{2}\epsilon^{−\mathrm{1}} −\mathrm{1}>\mathrm{0}\Rightarrow\epsilon<\mathrm{2}\:{which}\:{is}\:{appropriate}\:{for}\:{closeness}\right) \\ $$$${Then},\:{if}\:{x}<\delta \\ $$$$\Rightarrow{x}<\frac{{ln}\left(\mathrm{2}\epsilon^{−\mathrm{1}} −\mathrm{1}\right)^{−\mathrm{1}} }{\mathrm{2}{ln}\mathrm{3}} \\ $$$${ln}\mathrm{3}^{\mathrm{2}{x}} <−{ln}\left(\mathrm{2}\epsilon^{−\mathrm{1}} −\mathrm{1}\right) \\ $$$${ln}\mathrm{3}^{−\mathrm{2}{x}} >{ln}\left(\mathrm{2}\epsilon^{−\mathrm{1}} −\mathrm{1}\right) \\ $$$$\mathrm{3}^{−\mathrm{2}{x}} >\mathrm{2}\epsilon^{−\mathrm{1}} −\mathrm{1} \\ $$$$\mathrm{3}^{−\mathrm{2}{x}} +\mathrm{1}>\mathrm{2}\epsilon^{−\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{\mathrm{1}+\mathrm{3}^{−\mathrm{2}{x}} }<\epsilon \\ $$$$\frac{\mathrm{2}×\mathrm{3}^{{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }<\epsilon \\ $$$$\frac{\mathrm{2}×\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} −\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }<\epsilon \\ $$$$\frac{\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} +\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }<\epsilon \\ $$$$\frac{\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }+\mathrm{1}<\epsilon \\ $$$$\Rightarrow\mid\frac{\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }+\mathrm{1}\mid<\mid\epsilon\mid=\epsilon\:\:\:\:\:\because\:\epsilon>\mathrm{0} \\ $$$$\therefore\mid\frac{\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }−\left(−\mathrm{1}\right)\mid<\epsilon\:\:\:{whenever}\:\:\:{x}<\delta=\frac{{ln}\left(\mathrm{2}\epsilon^{−\mathrm{1}} −\mathrm{1}\right)}{−\mathrm{2}{ln}\mathrm{3}}. \\ $$$${So}\:{indeed}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }=−\mathrm{1}. \\ $$
Commented by sanusihammed last updated on 24/May/16
$${Thanks} \\ $$
Commented by sanusihammed last updated on 24/May/16
$${i}\:{really}\:{appreciate}\: \\ $$