Question Number 153352 by liberty last updated on 06/Sep/21
$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{27}^{{x}} +\mathrm{9}^{{x}} }\:−\sqrt{\mathrm{9}^{{x}} +\mathrm{3}^{{x}} }\:=? \\ $$
Answered by MJS_new last updated on 06/Sep/21
![lim_(x→+∞) (27^x +9^x )^(1/3) −(9^x +3^x )^(1/2) = [x=−((ln t)/(ln 3)) → t→0^+ ] =lim_(t→0^+ ) (((t+1)^(1/3) −(t+1)^(1/2) )/t) = [L′Ho^ pital] =lim_(t→0^+ ) ((2−3(t+1)^(1/6) )/(6(t+1)^(2/3) )) =−(1/6)](https://www.tinkutara.com/question/Q153361.png)
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\left(\mathrm{27}^{{x}} +\mathrm{9}^{{x}} \right)^{\mathrm{1}/\mathrm{3}} −\left(\mathrm{9}^{{x}} +\mathrm{3}^{{x}} \right)^{\mathrm{1}/\mathrm{2}} \:= \\ $$$$\:\:\:\:\:\left[{x}=−\frac{\mathrm{ln}\:{t}}{\mathrm{ln}\:\mathrm{3}}\:\rightarrow\:{t}\rightarrow\mathrm{0}^{+} \right] \\ $$$$=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\left({t}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} −\left({t}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} }{{t}}\:= \\ $$$$\:\:\:\:\:\left[\mathrm{L}'\mathrm{H}\hat {\mathrm{o}pital}\right] \\ $$$$=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{2}−\mathrm{3}\left({t}+\mathrm{1}\right)^{\mathrm{1}/\mathrm{6}} }{\mathrm{6}\left({t}+\mathrm{1}\right)^{\mathrm{2}/\mathrm{3}} }\:=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by liberty last updated on 06/Sep/21
$${yes}\:{sir} \\ $$
Commented by MJS_new last updated on 06/Sep/21
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Commented by liberty last updated on 06/Sep/21
$$\:{how}\:{about}\:{for}\:{this}\: \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{3}}]{\mathrm{8}^{{x}} +\mathrm{3}^{{x}} }−\sqrt{\mathrm{4}^{{x}} −\mathrm{2}^{{x}} }\:=? \\ $$$$\:{result}\:\mathrm{0}\:{or}\:−\infty\:? \\ $$
Commented by MJS_new last updated on 07/Sep/21
$$\mathrm{it}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by liberty last updated on 07/Sep/21
$${yes}\:{sir} \\ $$
Answered by liberty last updated on 06/Sep/21
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{9}^{{x}} \left(\mathrm{3}^{{x}} +\mathrm{1}\right)}−\sqrt{\mathrm{3}^{{x}} \left(\mathrm{3}^{{x}} +\mathrm{1}\right)} \\ $$$${let}\:\mathrm{3}^{{x}} +\mathrm{1}\:=\:{u}^{\mathrm{6}} \:\Rightarrow\mathrm{3}^{{x}} ={u}^{\mathrm{6}} −\mathrm{1} \\ $$$$\:\underset{{u}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{3}}]{\left(\mathrm{3}^{{x}} \right)^{\mathrm{2}} }\:{u}^{\mathrm{2}} −\sqrt{\mathrm{3}^{{x}} }\:{u}^{\mathrm{3}} \:= \\ $$$$\underset{{u}\rightarrow\infty} {\mathrm{lim}}{u}^{\mathrm{2}} \:\sqrt[{\mathrm{3}}]{\left({u}^{\mathrm{6}} −\mathrm{1}\right)^{\mathrm{2}} }−{u}^{\mathrm{3}} \:\sqrt{{u}^{\mathrm{6}} −\mathrm{1}}\:= \\ $$$$\underset{{u}\rightarrow\infty} {\mathrm{lim}}\:{u}^{\mathrm{6}} \:\sqrt[{\mathrm{3}}]{\left(\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{6}} }\right)^{\mathrm{2}} }−{u}^{\mathrm{6}} \:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{6}} }}\:= \\ $$$${let}\:\frac{\mathrm{1}}{{u}^{\mathrm{6}} }\:=\:{t}\wedge{t}\rightarrow\mathrm{0} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }−\sqrt{\mathrm{1}−{t}}}{{t}}\:= \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−{t}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} +\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−{t}}}\:= \\ $$$$−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{4}}{\mathrm{6}}+\frac{\mathrm{3}}{\mathrm{6}}=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Answered by Aras last updated on 06/Sep/21
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Answered by bramlexs22 last updated on 07/Sep/21
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}3}^{\mathrm{x}} \left(\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{x}} }}−\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{x}} }}\right) \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{+\mathrm{x}}−\sqrt{\mathrm{1}+\mathrm{x}}}{\mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$