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Question-103366

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Question Number 103366 by Study last updated on 14/Jul/20
Answered by mathmax by abdo last updated on 14/Jul/20
let f(x) =((x^(1/4) −5^(1/{) )/((lnx)^(1/8) −(ln5)^(1/8) )) let find lim_(x→5) f(x) we do the changement x =5−t ⇒f(x) =g(t) =(((5−t)^(1/4) −5^(1/4) )/((ln(5−t))^(1/8) −(ln5)^(1/8) )) (t→0) =((5^(1/4) {(1−(t/5))^(1/4) −1})/((ln(5)+ln(1−(t/5)))^(1/8) −(ln5)^(1/8) )) ⇒g(t) ∼((5^(1/4) {1−(t/(20))−1})/((ln(5)−(t/5))^(1/8) −(ln5)^(1/8) )) =((5^(1/4) {−(t/(20))})/((ln5)^(1/8) {(1−(t/(5ln5)))^(1/8) −1})) ∼((5^(1/4) {−(t/(20))})/((ln5)^(1/8) {1−(t/(40ln5))−1})) →(5^(1/4) /((ln5)^(1/8) ))×((40ln5)/(20)) =((2 ln5 ×5^(1/4) )/((ln5)^(1/8) )) =2.5^(1/4) (ln5)^(7/8) ⇒lim_(x→5) f(x)= 2(^4 (√5))(ln5)^(7/8)
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{5}^{\frac{\mathrm{1}}{\left\{\right.}} }{\left(\mathrm{lnx}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} −\left(\mathrm{ln5}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} }\:\:\mathrm{let}\:\mathrm{find}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{5}} \mathrm{f}\left(\mathrm{x}\right)\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement} \\ $$$$\mathrm{x}\:=\mathrm{5}−\mathrm{t}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{g}\left(\mathrm{t}\right)\:=\frac{\left(\mathrm{5}−\mathrm{t}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\left(\mathrm{ln}\left(\mathrm{5}−\mathrm{t}\right)\right)^{\frac{\mathrm{1}}{\mathrm{8}}} −\left(\mathrm{ln5}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} }\:\:\:\left(\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$=\frac{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} \left\{\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{5}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{1}\right\}}{\left(\mathrm{ln}\left(\mathrm{5}\right)+\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{5}}\right)\right)^{\frac{\mathrm{1}}{\mathrm{8}}} −\left(\mathrm{ln5}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} } \\ $$$$\Rightarrow\mathrm{g}\left(\mathrm{t}\right)\:\sim\frac{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} \left\{\mathrm{1}−\frac{\mathrm{t}}{\mathrm{20}}−\mathrm{1}\right\}}{\left(\mathrm{ln}\left(\mathrm{5}\right)−\frac{\mathrm{t}}{\mathrm{5}}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} −\left(\mathrm{ln5}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} }\:=\frac{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} \left\{−\frac{\mathrm{t}}{\mathrm{20}}\right\}}{\left(\mathrm{ln5}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} \left\{\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{5ln5}}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} −\mathrm{1}\right\}} \\ $$$$\sim\frac{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} \left\{−\frac{\mathrm{t}}{\mathrm{20}}\right\}}{\left(\mathrm{ln5}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} \left\{\mathrm{1}−\frac{\mathrm{t}}{\mathrm{40ln5}}−\mathrm{1}\right\}}\:\rightarrow\frac{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\left(\mathrm{ln5}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} }×\frac{\mathrm{40ln5}}{\mathrm{20}}\:=\frac{\mathrm{2}\:\mathrm{ln5}\:×\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\left(\mathrm{ln5}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} } \\ $$$$=\mathrm{2}.\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\left(\mathrm{ln5}\right)^{\frac{\mathrm{7}}{\mathrm{8}}} \:\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{5}} \mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{2}\left(^{\mathrm{4}} \sqrt{\mathrm{5}}\right)\left(\mathrm{ln5}\right)^{\frac{\mathrm{7}}{\mathrm{8}}} \\ $$
Answered by bemath last updated on 14/Jul/20
(1) lim_(x→5) (((x)^(1/4) −(5)^(1/4) )/( ((lnx))^(1/8) −((ln 5))^(1/8) )) = lim_(x→5) ((1/(4 x^(3/4) ))/(1/(8x (ln x)^(7/8) ))) = lim_(x→5) ((8x (ln x)^(7/8) )/(4x^(3/4) )) = ((10.(ln 5)^(7/8) )/5^(3/4) ) ■
$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\:\:\frac{\sqrt[{\mathrm{4}}]{{x}}−\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\:\sqrt[{\mathrm{8}}]{\mathrm{ln}{x}}\:−\sqrt[{\mathrm{8}}]{\mathrm{ln}\:\mathrm{5}}}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{4}\:{x}^{\mathrm{3}/\mathrm{4}} }}{\frac{\mathrm{1}}{\mathrm{8}{x}\:\left(\mathrm{ln}\:{x}\right)^{\mathrm{7}/\mathrm{8}} }}\:=\:\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\:\frac{\mathrm{8}{x}\:\left(\mathrm{ln}\:{x}\right)^{\mathrm{7}/\mathrm{8}} }{\mathrm{4}{x}^{\mathrm{3}/\mathrm{4}} } \\ $$$$=\:\frac{\mathrm{10}.\left(\mathrm{ln}\:\mathrm{5}\right)^{\mathrm{7}/\mathrm{8}} }{\mathrm{5}^{\mathrm{3}/\mathrm{4}} }\:\blacksquare\: \\ $$
Answered by bemath last updated on 14/Jul/20
(2) lim_(x→3) (((e^x )^(1/6) −(e^3 )^(1/6) )/( (x)^(1/3) −(3)^(1/3) )) = lim_(x→3) ((((e^x )^(−5/6) .e^x )/6)/(1/(3 (x^2 )^(1/3) ))) = lim_(x→3) ((3 (x^2 )^(1/3) )/6) ×e^(x/6) = (((9)^(1/3) .e^(1/2) )/2) = (((√e) .(9)^(1/3) )/2) ■
$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{6}}]{{e}^{{x}} }−\sqrt[{\mathrm{6}}]{{e}^{\mathrm{3}} }}{\:\sqrt[{\mathrm{3}}]{{x}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\frac{\left({e}^{{x}} \right)^{−\mathrm{5}/\mathrm{6}} .{e}^{{x}} }{\mathrm{6}}}{\frac{\mathrm{1}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }\:}}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{3}\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}{\mathrm{6}}\:×{e}^{\frac{{x}}{\mathrm{6}}} \:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{9}}\:.{e}^{\mathrm{1}/\mathrm{2}} }{\mathrm{2}}\: \\ $$$$=\:\frac{\sqrt{{e}}\:.\sqrt[{\mathrm{3}}]{\mathrm{9}}}{\mathrm{2}}\:\blacksquare \\ $$
Answered by mathmax by abdo last updated on 15/Jul/20
let g(x) =((e^(x/6) −e^(1/2) )/(x^(1/3) −3^(1/3) )) we do the cha7gement t =x−3 (so t→0) ⇒ g(x) =((e^((t+3)/6) −e^(1/2) )/((t+3)^(1/3) −3^(1/3) )) =((e^(1/2) {e^(t/6) −1})/(3^(1/3) { (1+(t/3))^(1/3) −1})) =ϕ(t) ⇒ ϕ(t) ∼(e^(1/2) /3^(1/3) ){ ((1+(t/6)−1)/(1+(t/9)−1))} →(e^(1/2) /3^(1/3) )×(9/6) =(3/2)×((√e)/((^3 (√3))))(t→0) ⇒ lim _(x→3) g(x) =((3(√e))/(2(^3 (√3))))
$$\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{6}}} −\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} }\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{cha7gement}\:\mathrm{t}\:=\mathrm{x}−\mathrm{3}\:\left(\mathrm{so}\:\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{e}^{\frac{\mathrm{t}+\mathrm{3}}{\mathrm{6}}} −\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{t}+\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} }\:=\frac{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} \left\{\mathrm{e}^{\frac{\mathrm{t}}{\mathrm{6}}} −\mathrm{1}\right\}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} \left\{\:\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}\right\}}\:=\varphi\left(\mathrm{t}\right)\:\Rightarrow \\ $$$$\varphi\left(\mathrm{t}\right)\:\sim\frac{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} }\left\{\:\frac{\mathrm{1}+\frac{\mathrm{t}}{\mathrm{6}}−\mathrm{1}}{\mathrm{1}+\frac{\mathrm{t}}{\mathrm{9}}−\mathrm{1}}\right\}\:\rightarrow\frac{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}}} }×\frac{\mathrm{9}}{\mathrm{6}}\:=\frac{\mathrm{3}}{\mathrm{2}}×\frac{\sqrt{\mathrm{e}}}{\left(^{\mathrm{3}} \sqrt{\mathrm{3}}\right)}\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{lim}\:_{\mathrm{x}\rightarrow\mathrm{3}} \mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{3}\sqrt{\mathrm{e}}}{\mathrm{2}\left(^{\mathrm{3}} \sqrt{\mathrm{3}}\right)} \\ $$

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