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Question Number 5343 by sanusihammed last updated on 09/May/16

[((1−x^(−1) −6x^(−2) )/(1−5x^(−1) −24x^(−2) ))]^x Limit x → ∞ please help.

$$\left[\frac{\mathrm{1}−{x}^{−\mathrm{1}} −\mathrm{6}{x}^{−\mathrm{2}} }{\mathrm{1}−\mathrm{5}{x}^{−\mathrm{1}} −\mathrm{24}{x}^{−\mathrm{2}} }\right]^{{x}} \\ $$$$ \\ $$$${Limit}\:{x}\:\rightarrow\:\infty \\ $$$$ \\ $$$${please}\:{help}. \\ $$

Commented by Yozzii last updated on 09/May/16

Let u=x^(−1) l=lim_(x→∞) (((1−x^(−1) −6x^(−2) )/(1−5x^(−1) −24x^(−2) )))^x =lim_(u→0) (((1−u−6u^2 )/(1−5u−24u^2 )))^(1/u) l=lim_(u→0) exp((1/u){ln(1+u(−1−6u)−ln(1+u(−5−24u)}) l=exp(lim_(u→0) (1/u)ln(1+u(−1−6u))−lim_(u→0) (1/u)ln(1+u(−5−24u))) Since ln(1+x)=x−(x^2 /2)+(x^3 /3)−(x^4 /4)+.... ⇒lim_(u→0) (1/u)ln(1+u(−1−6u))=lim_(u→0) ((1/u)u(−1−6u)−((u(1+6u)^2 )/2)+((u^2 (−1−6u)^3 )/3)−....)=−1−0+0−0+...=−1.= Similarly, lim_(u→0) (1/u)ln(1+u(−5−24u))=−5. ∴ l=exp(−1−(−5))=e^4 .

$${Let}\:{u}={x}^{−\mathrm{1}} \\ $$$${l}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}−{x}^{−\mathrm{1}} −\mathrm{6}{x}^{−\mathrm{2}} }{\mathrm{1}−\mathrm{5}{x}^{−\mathrm{1}} −\mathrm{24}{x}^{−\mathrm{2}} }\right)^{{x}} =\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−{u}−\mathrm{6}{u}^{\mathrm{2}} }{\mathrm{1}−\mathrm{5}{u}−\mathrm{24}{u}^{\mathrm{2}} }\right)^{\mathrm{1}/{u}} \\ $$$${l}=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}{exp}\left(\frac{\mathrm{1}}{{u}}\left\{{ln}\left(\mathrm{1}+{u}\left(−\mathrm{1}−\mathrm{6}{u}\right)−{ln}\left(\mathrm{1}+{u}\left(−\mathrm{5}−\mathrm{24}{u}\right)\right\}\right)\right.\right. \\ $$$${l}={exp}\left(\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{u}}{ln}\left(\mathrm{1}+{u}\left(−\mathrm{1}−\mathrm{6}{u}\right)\right)−\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{u}}{ln}\left(\mathrm{1}+{u}\left(−\mathrm{5}−\mathrm{24}{u}\right)\right)\right) \\ $$$${Since}\:{ln}\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+.... \\ $$$$\Rightarrow\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{u}}{ln}\left(\mathrm{1}+{u}\left(−\mathrm{1}−\mathrm{6}{u}\right)\right)=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{u}}{u}\left(−\mathrm{1}−\mathrm{6}{u}\right)−\frac{{u}\left(\mathrm{1}+\mathrm{6}{u}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{{u}^{\mathrm{2}} \left(−\mathrm{1}−\mathrm{6}{u}\right)^{\mathrm{3}} }{\mathrm{3}}−....\right)=−\mathrm{1}−\mathrm{0}+\mathrm{0}−\mathrm{0}+...=−\mathrm{1}.= \\ $$$${Similarly},\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{u}}{ln}\left(\mathrm{1}+{u}\left(−\mathrm{5}−\mathrm{24}{u}\right)\right)=−\mathrm{5}. \\ $$$$\therefore\:{l}={exp}\left(−\mathrm{1}−\left(−\mathrm{5}\right)\right)={e}^{\mathrm{4}} . \\ $$$$ \\ $$

Commented by sanusihammed last updated on 09/May/16

Thanks so much

$${Thanks}\:{so}\:{much} \\ $$

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