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Question Number 5593 by sanusihammed last updated on 21/May/16

Find the limit limit [1 + mt]^(2/t) t →∞

$${Find}\:{the}\:{limit} \\ $$$$ \\ $$$${limit}\:\:\:\:\:\:\:\left[\mathrm{1}\:+\:{mt}\right]^{\frac{\mathrm{2}}{{t}}} \\ $$$${t}\:\rightarrow\infty \\ $$

Answered by FilupSmith last updated on 21/May/16

L=lim_(t→∞) (1+mt)^(2/t) =lim_(t→∞) e^(ln((1+mt)^(2/t) )) =lim_(t→∞) e^((2/t)ln(1+mt)) e^x =exp(x) change of notation for easier writing =lim_(t→∞) exp(((2ln(1+mt))/t)) =exp(lim_(t→∞) ((2ln(1+mt))/t)) use L′Hopital′s Law L=exp(lim_(t→∞) ((2×k)/1)) k=(d/dt)(ln(1+mt)) let u=1+mt ⇒ du=mdt k=(d/dt)mln(u) k=(m/(1+mt)) ∴L=exp(lim_(t→∞) ((2(m/(1+mt)))/1)) L=exp(lim_(t→∞) 2(m/(1+mt))) L=exp(2×0) L=1 ∴lim_(t→∞) (1+mt)^(2/t) =1

$${L}=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+{mt}\right)^{\mathrm{2}/{t}} \\ $$$$=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:{e}^{\mathrm{ln}\left(\left(\mathrm{1}+{mt}\right)^{\mathrm{2}/{t}} \right)} \\ $$$$=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:{e}^{\frac{\mathrm{2}}{{t}}\mathrm{ln}\left(\mathrm{1}+{mt}\right)} \\ $$$${e}^{{x}} =\mathrm{exp}\left({x}\right)\:\:\:\:\:\:{change}\:{of}\:{notation}\:{for} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{easier}\:{writing} \\ $$$$=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\frac{\mathrm{2ln}\left(\mathrm{1}+{mt}\right)}{{t}}\right) \\ $$$$=\mathrm{exp}\left(\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2ln}\left(\mathrm{1}+{mt}\right)}{{t}}\right) \\ $$$$\mathrm{use}\:{L}'{Hopital}'{s}\:{Law} \\ $$$${L}=\mathrm{exp}\left(\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}×{k}}{\mathrm{1}}\right) \\ $$$${k}=\frac{{d}}{{dt}}\left(\mathrm{ln}\left(\mathrm{1}+{mt}\right)\right) \\ $$$${let}\:{u}=\mathrm{1}+{mt}\:\:\Rightarrow\:\:{du}={mdt} \\ $$$${k}=\frac{{d}}{{dt}}{m}\mathrm{ln}\left({u}\right) \\ $$$${k}=\frac{{m}}{\mathrm{1}+{mt}} \\ $$$$\therefore{L}=\mathrm{exp}\left(\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}\frac{{m}}{\mathrm{1}+{mt}}}{\mathrm{1}}\right) \\ $$$${L}=\mathrm{exp}\left(\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{2}\frac{{m}}{\mathrm{1}+{mt}}\right) \\ $$$${L}=\mathrm{exp}\left(\mathrm{2}×\mathrm{0}\right) \\ $$$${L}=\mathrm{1} \\ $$$$ \\ $$$$\therefore\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+{mt}\right)^{\mathrm{2}/{t}} =\mathrm{1} \\ $$

Commented by FilupSmith last updated on 22/May/16

Not that I can work out

$$\mathrm{Not}\:\mathrm{that}\:\mathrm{I}\:\mathrm{can}\:\mathrm{work}\:\mathrm{out} \\ $$

Commented by FilupSmith last updated on 22/May/16

No problem

$$\mathrm{No}\:\mathrm{problem} \\ $$

Commented by sanusihammed last updated on 21/May/16

Thanks so much. God bless you

$${Thanks}\:{so}\:{much}.\:{God}\:{bless}\:{you} \\ $$

Commented by Rasheed Soomro last updated on 22/May/16

Could the problem be solved without using calculus?

$$\mathrm{Could}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{without} \\ $$$$\mathrm{using}\:\mathrm{calculus}? \\ $$

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