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Question Number 72905 by Tony Lin last updated on 04/Nov/19

f(x)≥0, and lim_(x→a) f(x)=0,lim_(x→a) g(x)=∞ then lim_(x→a) f(x)^(g(x)) =?

$${f}\left({x}\right)\geqslant\mathrm{0},\:{and}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0},\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)=\infty \\ $$$${then}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)^{{g}\left({x}\right)} =? \\ $$

Commented by mathmax by abdo last updated on 04/Nov/19

we have by taylor f(x)=f(a)+(x−a)f^′ (a)+o(x−a) ⇒f(x)∼(x−a)f^′ (a) because lim_(x→a) f(x)=0 (we suppose f derivable at point a) ⇒ f(x)^(g(x)) ∼{(x−a)f^′ (a)}^(g(x)) =e^(g(x)ln{(x−a)f^′ (a)}) if lim_(x→a) g(x)=+∞ we have lim_(x→a^+ ) ln{(x−a)f^′ (a)}=−∞ ⇒ lim_(x→a^+ ) g(x)ln{(x−a)f^′ (a)}=−∞ ⇒ lim_(x→a^+ ) {f(x)}^(g(x)) =0 if lim_(x→a^+ ) g(x)=−∞ ⇒lim_(x→a^+ ) g(x)ln{(x−a)f^′ (a)}=+∞ ⇒ lim_(x→a^+ ) {f(x)}^(g(x)) =+∞ and if you have a clear answer post it..

$${we}\:{have}\:{by}\:{taylor}\:{f}\left({x}\right)={f}\left({a}\right)+\left({x}−{a}\right){f}^{'} \left({a}\right)+{o}\left({x}−{a}\right) \\ $$$$\Rightarrow{f}\left({x}\right)\sim\left({x}−{a}\right){f}^{'} \left({a}\right)\:\:{because}\:{lim}_{{x}\rightarrow{a}} {f}\left({x}\right)=\mathrm{0} \\ $$$$\left({we}\:{suppose}\:{f}\:{derivable}\:{at}\:{point}\:{a}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)^{{g}\left({x}\right)} \sim\left\{\left({x}−{a}\right){f}^{'} \left({a}\right)\right\}^{{g}\left({x}\right)} \:={e}^{{g}\left({x}\right){ln}\left\{\left({x}−{a}\right){f}^{'} \left({a}\right)\right\}} \\ $$$${if}\:{lim}_{{x}\rightarrow{a}} {g}\left({x}\right)=+\infty\:\:{we}\:{have}\:{lim}_{{x}\rightarrow{a}^{+} } {ln}\left\{\left({x}−{a}\right){f}^{'} \left({a}\right)\right\}=−\infty\:\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow{a}^{+} } \:\:{g}\left({x}\right){ln}\left\{\left({x}−{a}\right){f}^{'} \left({a}\right)\right\}=−\infty\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow{a}^{+} } \:\:\:\left\{{f}\left({x}\right)\right\}^{{g}\left({x}\right)} =\mathrm{0} \\ $$$${if}\:{lim}_{{x}\rightarrow{a}^{+} } \:\:{g}\left({x}\right)=−\infty\:\:\:\Rightarrow{lim}_{{x}\rightarrow{a}^{+} } \:\:{g}\left({x}\right){ln}\left\{\left({x}−{a}\right){f}^{'} \left({a}\right)\right\}=+\infty\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow{a}^{+} } \:\:\:\left\{{f}\left({x}\right)\right\}^{{g}\left({x}\right)} \:=+\infty\:\:{and}\:{if}\:{you}\:{have}\:{a}\:{clear}\:{answer}\:{post}\:{it}.. \\ $$

Commented by Tony Lin last updated on 05/Nov/19

thanks sir

$${thanks}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 05/Nov/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

Answered by mind is power last updated on 04/Nov/19

f(x)^(g(x)) =e^(g(x)ln(f(x))) lim g(x)=+∞ lim[f(x)→0 ⇒lim ln(f(x))→−∞ ⇒g(x)ln(f)→−∞ ⇒e^(g(x)ln(f(x))) →0

$$\mathrm{f}\left(\mathrm{x}\right)^{\mathrm{g}\left(\mathrm{x}\right)} =\mathrm{e}^{\mathrm{g}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)} \\ $$$$\mathrm{lim}\:\mathrm{g}\left(\mathrm{x}\right)=+\infty \\ $$$$\mathrm{lim}\left[\mathrm{f}\left(\mathrm{x}\right)\rightarrow\mathrm{0}\right. \\ $$$$\Rightarrow\mathrm{lim}\:\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\rightarrow−\infty \\ $$$$\Rightarrow\mathrm{g}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{f}\right)\rightarrow−\infty \\ $$$$\Rightarrow\mathrm{e}^{\mathrm{g}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)} \rightarrow\mathrm{0} \\ $$

Commented by Tony Lin last updated on 05/Nov/19

thanks sir

$${thanks}\:{sir} \\ $$

Answered by Tony Lin last updated on 05/Nov/19

lim_(x→a) f(x)^(g(x)) =0

$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)^{{g}\left({x}\right)} =\mathrm{0} \\ $$

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