Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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)) =+∞

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{16x}^{\mathrm{4}} \:+\mathrm{16x}^{\mathrm{3}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:+\mathrm{4x}^{\mathrm{2}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{2x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\mathrm{4x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\sim\:\mathrm{2x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{16}}\right)\right)+\mathrm{4x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\sim\:\mathrm{2x}\left\{\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2x}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{32}}\right\}\:+\mathrm{4x}^{\mathrm{2}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\sim\mathrm{2x}+\mathrm{1}\:−\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{16}}\:+\mathrm{4x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow−\infty} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{lim}_{\mathrm{x}\rightarrow−\infty} −\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{16}}\:=+\infty \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com