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Question Number 101615 by john santu last updated on 03/Jul/20
Commented by bramlex last updated on 03/Jul/20
i got −∞ sir
$$\mathrm{i}\:\mathrm{got}\:−\infty\:\mathrm{sir}\: \\ $$
Commented by john santu last updated on 04/Jul/20
lim_(x→−∞) ((16x^3 −1)/( (√(16x^4 +16x^3 −1))−4x^2 )) lim_(x→−∞) ((16x^3 −1)/(−4x^2 (√(1+(1/x)−(1/(16x^2 ))))−4x^2 )) lim_(x→−∞) ((16x^3 −1)/(−4x^2 ((√(1+(1/x)−(1/(16x^2 ))))+1))) lim_(x→−∞) ((−x^3 (−16+(1/x^3 )))/(−4x^2 ((√(1+(1/x)−(1/(16x^2 ))))+1))) lim_(x→−∞) ((x(−16+(1/x^3 )))/( (√(1+(1/x)−(1/(16x^2 ))))+1)) = ∞ (JS⊛) it correct ?
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\:\frac{\mathrm{16x}^{\mathrm{3}} −\mathrm{1}}{\:\sqrt{\mathrm{16x}^{\mathrm{4}} +\mathrm{16x}^{\mathrm{3}} −\mathrm{1}}−\mathrm{4x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{16x}^{\mathrm{3}} −\mathrm{1}}{−\mathrm{4x}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{16x}^{\mathrm{2}} }}−\mathrm{4x}^{\mathrm{2}} }\: \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{16x}^{\mathrm{3}} −\mathrm{1}}{−\mathrm{4x}^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{16x}^{\mathrm{2}} }}+\mathrm{1}\right)} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{−\mathrm{x}^{\mathrm{3}} \left(−\mathrm{16}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)}{−\mathrm{4x}^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{16x}^{\mathrm{2}} }}+\mathrm{1}\right)} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{x}\left(−\mathrm{16}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\right)}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{16x}^{\mathrm{2}} }}+\mathrm{1}}\:=\:\infty\:\left(\mathrm{JS}\circledast\right) \\ $$$$\mathrm{it}\:\mathrm{correct}\:? \\ $$
Commented by john santu last updated on 04/Jul/20
sorry i mistake write question.
$$\mathrm{sorry}\:\mathrm{i}\:\mathrm{mistake}\:\mathrm{write}\:\mathrm{question}.\: \\ $$
Commented by john santu last updated on 04/Jul/20
it should be lim_(x→−∞) (√(16x^4 +16x^3 −1)) + 4x^2
$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\: \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt{\mathrm{16x}^{\mathrm{4}} +\mathrm{16x}^{\mathrm{3}} −\mathrm{1}}\:+\:\mathrm{4x}^{\mathrm{2}} \: \\ $$
Answered by bramlex last updated on 04/Jul/20
lim_(x→−∞) (((16x^4 +16x^3 −1)−16x^4 )/( ((16x^4 +16x^3 −1))^(1/ ) −4x^2 )) = lim_(x→−∞) ((16x^3 −1)/( (√((2x)^4 (1+(1/x)−(1/(16x^4 )))))−4x^2 )) = lim_(x→−∞) ((−2x^2 (−8x+(1/(2x^2 ))))/(−2x^2 ((√(1+(1/x)−(1/(16x^4 ))))+2)))= lim_(x→−∞) ((−8x+(1/(2x^2 )))/( (√(1+(1/x)−(1/(16x^4 ))))+2)) = ∞
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{16x}^{\mathrm{4}} +\mathrm{16x}^{\mathrm{3}} −\mathrm{1}\right)−\mathrm{16x}^{\mathrm{4}} }{\:\sqrt[{\:}]{\mathrm{16x}^{\mathrm{4}} +\mathrm{16x}^{\mathrm{3}} −\mathrm{1}}\:−\mathrm{4x}^{\mathrm{2}} }\:= \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{16x}^{\mathrm{3}} −\mathrm{1}}{\:\sqrt{\left(\mathrm{2x}\right)^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{16x}^{\mathrm{4}} }\right)}−\mathrm{4x}^{\mathrm{2}} }\:= \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{−\mathrm{2x}^{\mathrm{2}} \left(−\mathrm{8x}+\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }\right)}{−\mathrm{2x}^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{16x}^{\mathrm{4}} }}+\mathrm{2}\right)}= \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{−\mathrm{8x}+\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{16x}^{\mathrm{4}} }}+\mathrm{2}}\:=\:\infty \\ $$
Commented by bramlex last updated on 03/Jul/20
oo sorry sir. you are right
$$\mathrm{oo}\:\mathrm{sorry}\:\mathrm{sir}.\:\mathrm{you}\:\mathrm{are}\:\mathrm{right} \\ $$
Answered by Dwaipayan Shikari last updated on 04/Jul/20
lim_(x→−∞) (x(16+((16)/x)−(1/x^4 ))^(1/4) +4x^2 )=2x+4x^2 →∞
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left({x}\left(\mathrm{16}+\frac{\mathrm{16}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{4}{x}^{\mathrm{2}} \right)=\mathrm{2}{x}+\mathrm{4}{x}^{\mathrm{2}} \rightarrow\infty \\ $$
Answered by mathmax by abdo last updated on 05/Jul/20
let f(x) =(16x^4 +16x^3 −1)^(1/4) +4x^2 ⇒f(x) =2x(1+(1/x)−(1/(16))x^4 )^(1/2) +4x^2 ⇒f(x) ∼ 2x(1+(1/2)((1/x)−(x^4 /(16))))+4x^2 ⇒f(x) ∼ 2x{ 1+(1/(2x))−(x^4 /(32))} +4x^2 ⇒f(x) ∼2x+1 −(x^5 /(16)) +4x^2 ⇒ lim_(x→−∞) f(x) =lim_(x→−∞) −(x^5 /(16)) =+∞
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{16x}^{\mathrm{4}} \:+\mathrm{16x}^{\mathrm{3}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:+\mathrm{4x}^{\mathrm{2}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{2x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\mathrm{4x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\sim\:\mathrm{2x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{16}}\right)\right)+\mathrm{4x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\sim\:\mathrm{2x}\left\{\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2x}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{32}}\right\}\:+\mathrm{4x}^{\mathrm{2}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\sim\mathrm{2x}+\mathrm{1}\:−\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{16}}\:+\mathrm{4x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow−\infty} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{lim}_{\mathrm{x}\rightarrow−\infty} −\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{16}}\:=+\infty \\ $$$$ \\ $$

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