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Question Number 144413 by imjagoll last updated on 25/Jun/21

lim_(x→1) [ (1/(4−4(√x)))−(1/(5−5(x)^(1/5) )) ] =?

$$\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left[\:\frac{\mathrm{1}}{\mathrm{4}−\mathrm{4}\sqrt{\mathrm{x}}}−\frac{\mathrm{1}}{\mathrm{5}−\mathrm{5}\sqrt[{\mathrm{5}}]{\mathrm{x}}}\:\right]\:=? \\ $$

Answered by Olaf_Thorendsen last updated on 25/Jun/21

f(x) = (1/(4−4(√x)))−(1/(5−5(x)^(1/5) )) Let x = X^(10) f(X^(10) ) = (1/(4−4(√X^(10) )))−(1/(5−5(X^(10) )^(1/5) )) f(X^(10) ) = (1/(4−4X^5 ))−(1/(5−5X^2 )) f(X^(10) ) = (1/(4(1−X)(1+X+X^2 +X^3 +X^4 )))−(1/(5(1−X)(1+X))) f(X^(10) ) = ((5(1+X)−4(1+X+X^2 +X^3 +X^4 ))/(20(1−X)(1+X)(1+X+X^2 +X^3 +X^4 ))) f(X^(10) ) = ((1+X−4X^2 (1+X+X^2 ))/(20(1−X)(1+X)(1+X+X^2 +X^3 +X^4 ))) lim_(x→1) f(x) = lim_(X→1) f(X^(10) ) = −((10)/(200×0^(+/−) )) lim_(x→1) f(x) = ±∞

$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}−\mathrm{4}\sqrt{{x}}}−\frac{\mathrm{1}}{\mathrm{5}−\mathrm{5}\sqrt[{\mathrm{5}}]{{x}}} \\ $$$$\mathrm{Let}\:{x}\:=\:{X}^{\mathrm{10}} \\ $$$${f}\left({X}^{\mathrm{10}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{4}−\mathrm{4}\sqrt{{X}^{\mathrm{10}} }}−\frac{\mathrm{1}}{\mathrm{5}−\mathrm{5}\sqrt[{\mathrm{5}}]{{X}^{\mathrm{10}} }} \\ $$$${f}\left({X}^{\mathrm{10}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{4}−\mathrm{4}{X}^{\mathrm{5}} }−\frac{\mathrm{1}}{\mathrm{5}−\mathrm{5}{X}^{\mathrm{2}} } \\ $$$${f}\left({X}^{\mathrm{10}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}−{X}\right)\left(\mathrm{1}+{X}+{X}^{\mathrm{2}} +{X}^{\mathrm{3}} +{X}^{\mathrm{4}} \right)}−\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{1}−{X}\right)\left(\mathrm{1}+{X}\right)} \\ $$$${f}\left({X}^{\mathrm{10}} \right)\:=\:\frac{\mathrm{5}\left(\mathrm{1}+{X}\right)−\mathrm{4}\left(\mathrm{1}+{X}+{X}^{\mathrm{2}} +{X}^{\mathrm{3}} +{X}^{\mathrm{4}} \right)}{\mathrm{20}\left(\mathrm{1}−{X}\right)\left(\mathrm{1}+{X}\right)\left(\mathrm{1}+{X}+{X}^{\mathrm{2}} +{X}^{\mathrm{3}} +{X}^{\mathrm{4}} \right)} \\ $$$${f}\left({X}^{\mathrm{10}} \right)\:=\:\frac{\mathrm{1}+{X}−\mathrm{4}{X}^{\mathrm{2}} \left(\mathrm{1}+{X}+{X}^{\mathrm{2}} \right)}{\mathrm{20}\left(\mathrm{1}−{X}\right)\left(\mathrm{1}+{X}\right)\left(\mathrm{1}+{X}+{X}^{\mathrm{2}} +{X}^{\mathrm{3}} +{X}^{\mathrm{4}} \right)} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)\:=\:\underset{{X}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({X}^{\mathrm{10}} \right)\:=\:−\frac{\mathrm{10}}{\mathrm{200}×\mathrm{0}^{+/−} } \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)\:=\:\pm\infty \\ $$

Answered by mathmax by abdo last updated on 25/Jun/21

f(x)=(1/(4−4(√x)))−(1/(5−5(^5 (√x)))) ⇒f(x)=(1/(4(1−(√x))))−(1/(5(1−x^(1/5) ))) changement x−1=t give f(x)=g(t)=(1/(4(1−(√(1+t)))))−(1/(5(1−(1+t)^(1/5) ))) (t→0) ⇒g(t)∼(1/(4(1−(1+(t/2)))))−(1/(5(1−(1+(t/5))))) ⇒g(t)∼(1/(−2t))−(1/(−t))=(1/t)−(1/(2t))=(1/(2t)) ⇒lim_(t→0) g(t)=∞ =lim_(x→1) g(x)

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{4}−\mathrm{4}\sqrt{\mathrm{x}}}−\frac{\mathrm{1}}{\mathrm{5}−\mathrm{5}\left(^{\mathrm{5}} \sqrt{\mathrm{x}}\right)}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}−\sqrt{\mathrm{x}}\right)}−\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{1}−\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{5}}} \right)} \\ $$$$\mathrm{changement}\:\mathrm{x}−\mathrm{1}=\mathrm{t}\:\mathrm{give}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{t}\right)=\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{t}}\right)}−\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{1}−\left(\mathrm{1}+\mathrm{t}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} \right)} \\ $$$$\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow\mathrm{g}\left(\mathrm{t}\right)\sim\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}−\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{2}}\right)\right)}−\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{1}−\left(\mathrm{1}+\frac{\mathrm{t}}{\mathrm{5}}\right)\right)} \\ $$$$\Rightarrow\mathrm{g}\left(\mathrm{t}\right)\sim\frac{\mathrm{1}}{−\mathrm{2t}}−\frac{\mathrm{1}}{−\mathrm{t}}=\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{1}}{\mathrm{2t}}=\frac{\mathrm{1}}{\mathrm{2t}}\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{g}\left(\mathrm{t}\right)=\infty\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}} \mathrm{g}\left(\mathrm{x}\right) \\ $$

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