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Question Number 203650 by Davidtim last updated on 24/Jan/24
lim_(x→0) (((2^x +3^x )/2))^(2/x) =?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}^{{x}} +\mathrm{3}^{{x}} }{\mathrm{2}}\right)^{\frac{\mathrm{2}}{{x}}} =? \\ $$
Answered by witcher3 last updated on 25/Jan/24
(2/x)ln(((2^x +3^x )/2)) 2^x =e^(xln(2)) =1+xln(2)+o(x) 3^x =e^(xln(3)) =1+xln(3)+o(x) 2^x +3^x =2+xln(6)+o(x) (2/x)ln(((2^x +3^x )/2))=(2/x)ln(1+(x/2)ln(6)+o(x))=(2/x)((x/2)ln(6))+o(1) →ln(6) lim_(x→0) (((2^x +3^x )/2))^(2/x) =e^(ln(6)) =6
$$\: \\ $$$$\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\frac{\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} }{\mathrm{2}}\right) \\ $$$$\mathrm{2}^{\mathrm{x}} =\mathrm{e}^{\mathrm{xln}\left(\mathrm{2}\right)} =\mathrm{1}+\mathrm{xln}\left(\mathrm{2}\right)+\mathrm{o}\left(\mathrm{x}\right) \\ $$$$\mathrm{3}^{\mathrm{x}} =\mathrm{e}^{\mathrm{xln}\left(\mathrm{3}\right)} =\mathrm{1}+\mathrm{xln}\left(\mathrm{3}\right)+\mathrm{o}\left(\mathrm{x}\right) \\ $$$$\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} =\mathrm{2}+\mathrm{xln}\left(\mathrm{6}\right)+\mathrm{o}\left(\mathrm{x}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\frac{\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} }{\mathrm{2}}\right)=\frac{\mathrm{2}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{6}\right)+\mathrm{o}\left(\mathrm{x}\right)\right)=\frac{\mathrm{2}}{\mathrm{x}}\left(\frac{\mathrm{x}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{6}\right)\right)+\mathrm{o}\left(\mathrm{1}\right) \\ $$$$\rightarrow\mathrm{ln}\left(\mathrm{6}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} }{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{x}}} =\mathrm{e}^{\mathrm{ln}\left(\mathrm{6}\right)} =\mathrm{6} \\ $$
Answered by Calculusboy last updated on 26/Jan/24
Solution: by sub directly,we get (1^∞ ) let y=(((2^x +3^x )/2))^(2/x) (apply log to both sides) logy=lim_(x→0) log(((2^x +3^x )/2))^(2/x) ⇒ logy=lim_(x→0) (2/x)log(((2^x +3^x )/2)) logy=lim_(x→0) ((2(d/dx)[log(((2^x +3^x )/2))])/((d/dx)(x))) ⇒ logy=lim_(x→0) ((2{(((1/2)[2^x In(2)+3^x In(3)])/((((2^x +3^x )/2))))})/1) logy=lim_(x→0) ((2^x In(2)+3^x In(3))/((2^x +3^x )/2)) ⇒ logy=2∙((lim_(x→0) [2^x In(2)+3^x In(3)])/(lim_(x→0) (2^x +3^x ))) logy=2∙(([In(2)+In(3)])/((1+1))) logy=In(6) y=e^(In(6)) y=6 ∴lim_(x→0) (((2^x +3^x )/2))^(2/x) =6
$$\boldsymbol{{Solution}}:\:\boldsymbol{{by}}\:\boldsymbol{{sub}}\:\boldsymbol{{directly}},\boldsymbol{{we}}\:\boldsymbol{{get}}\:\left(\mathrm{1}^{\infty} \right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{y}}=\left(\frac{\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{3}^{\boldsymbol{{x}}} }{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\boldsymbol{{x}}}} \left(\boldsymbol{{apply}}\:\boldsymbol{{log}}\:\boldsymbol{{to}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}}\right) \\ $$$$\boldsymbol{{logy}}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}log}}\left(\frac{\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{3}^{\boldsymbol{{x}}} }{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\boldsymbol{{x}}}} \:\:\Rightarrow\:\:\boldsymbol{{logy}}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{2}}{\boldsymbol{{x}}}\boldsymbol{{log}}\left(\frac{\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{3}^{\boldsymbol{{x}}} }{\mathrm{2}}\right) \\ $$$$\boldsymbol{{logy}}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{2}\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\left[\boldsymbol{{log}}\left(\frac{\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{3}^{\boldsymbol{{x}}} }{\mathrm{2}}\right)\right]}{\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\left(\boldsymbol{{x}}\right)}\:\:\:\Rightarrow\:\:\boldsymbol{{logy}}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{2}\left\{\frac{\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}^{\boldsymbol{{x}}} \boldsymbol{{In}}\left(\mathrm{2}\right)+\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{In}}\left(\mathrm{3}\right)\right]}{\left(\frac{\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{3}^{\boldsymbol{{x}}} }{\mathrm{2}}\right)}\right\}}{\mathrm{1}} \\ $$$$\boldsymbol{{logy}}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{2}^{\boldsymbol{{x}}} \boldsymbol{{In}}\left(\mathrm{2}\right)+\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{In}}\left(\mathrm{3}\right)}{\frac{\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{3}^{\boldsymbol{{x}}} }{\mathrm{2}}}\:\:\:\Rightarrow\:\:\boldsymbol{{logy}}=\mathrm{2}\centerdot\frac{\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\left[\mathrm{2}^{\boldsymbol{{x}}} \boldsymbol{{In}}\left(\mathrm{2}\right)+\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{In}}\left(\mathrm{3}\right)\right]}{\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\left(\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{3}^{\boldsymbol{{x}}} \right)} \\ $$$$\boldsymbol{{logy}}=\mathrm{2}\centerdot\frac{\left[\boldsymbol{{In}}\left(\mathrm{2}\right)+\boldsymbol{{In}}\left(\mathrm{3}\right)\right]}{\left(\mathrm{1}+\mathrm{1}\right)} \\ $$$$\boldsymbol{{logy}}=\boldsymbol{{In}}\left(\mathrm{6}\right) \\ $$$$\boldsymbol{{y}}=\boldsymbol{{e}}^{\boldsymbol{{In}}\left(\mathrm{6}\right)} \\ $$$$\boldsymbol{{y}}=\mathrm{6} \\ $$$$\therefore\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\left(\frac{\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{3}^{\boldsymbol{{x}}} }{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\boldsymbol{{x}}}} =\mathrm{6} \\ $$

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