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 118247 by bemath last updated on 16/Oct/20

   lim_(x→∞)  ((1+x^4 ))^(1/4)  − ((1+x^5 ))^(1/5)  =?

$$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}^{\mathrm{4}} }\:−\:\sqrt[{\mathrm{5}}]{\mathrm{1}+{x}^{\mathrm{5}} }\:=? \\ $$

Answered by bobhans last updated on 16/Oct/20

solve lim_(x→∞)  ((1+x^4 ))^(1/4)  − ((1+x^5 ))^(1/(5 ))  .  solution:   let x = (1/z) ; z→0    lim_(z→0)  ((1+(1/z^4 )))^(1/4) −((1+(1/z^5 ) ))^(1/5)  =      lim_(z→0)  ((((z^4 +1))^(1/4) −((z^5 +1))^(1/5) )/z) =      lim_(z→0)  ((((z^4 /4)+1)−((z^5 /5)+1))/z) = lim_(z→0)  ((((z^4 /4) −(z^5 /5))/z) )       lim_(z→0)  (z^3 /4) − (z^4 /5) = 0

$${solve}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}^{\mathrm{4}} }\:−\:\sqrt[{\mathrm{5}\:}]{\mathrm{1}+{x}^{\mathrm{5}} }\:. \\ $$$${solution}:\: \\ $$$${let}\:{x}\:=\:\frac{\mathrm{1}}{{z}}\:;\:{z}\rightarrow\mathrm{0} \\ $$$$\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{\mathrm{4}}]{\mathrm{1}+\frac{\mathrm{1}}{{z}^{\mathrm{4}} }}−\sqrt[{\mathrm{5}}]{\mathrm{1}+\frac{\mathrm{1}}{{z}^{\mathrm{5}} }\:}\:=\: \\ $$$$\:\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}}]{{z}^{\mathrm{4}} +\mathrm{1}}−\sqrt[{\mathrm{5}}]{{z}^{\mathrm{5}} +\mathrm{1}}}{{z}}\:=\: \\ $$$$\:\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\frac{{z}^{\mathrm{4}} }{\mathrm{4}}+\mathrm{1}\right)−\left(\frac{{z}^{\mathrm{5}} }{\mathrm{5}}+\mathrm{1}\right)}{{z}}\:=\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\frac{{z}^{\mathrm{4}} }{\mathrm{4}}\:−\frac{{z}^{\mathrm{5}} }{\mathrm{5}}}{{z}}\:\right)\:\: \\ $$$$\:\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{z}^{\mathrm{3}} }{\mathrm{4}}\:−\:\frac{{z}^{\mathrm{4}} }{\mathrm{5}}\:=\:\mathrm{0} \\ $$$$ \\ $$

Commented by bemath last updated on 16/Oct/20

gave kudos man

$${gave}\:{kudos}\:{man}\: \\ $$

Answered by mathmax by abdo last updated on 16/Oct/20

let f(x)=(1+x^4 )^(1/4) −(1+x^5 )^(1/5)  ⇒  f(x)=x(1+(1/x^4 ))^(1/4) −x(1+(1/x^5 ))^(1/5)  ⇒f(x)∼x{ 1+(1/(4x^4 ))−1−(1/(5x^5 ))}  ⇒f(x)∼(1/(4x^3 ))−(1/(5x^4 )) ⇒lim_(x→∞) f(x)=0

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\left(\mathrm{1}+\mathrm{x}^{\mathrm{5}} \right)^{\frac{\mathrm{1}}{\mathrm{5}}} \:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }\right)^{\frac{\mathrm{1}}{\mathrm{5}}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\mathrm{x}\left\{\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{4}} }−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5x}^{\mathrm{5}} }\right\} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{1}}{\mathrm{4x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{5x}^{\mathrm{4}} }\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com