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 149637 by mnjuly1970 last updated on 06/Aug/21

Answered by iloveisrael last updated on 06/Aug/21

   lim_(x→∞)  ((x^2 (√(1+x^(−1) +x^(−4) )) −x^2 (1+x^(−1) ))/x)  = lim_(x→∞)  x{(√(1+x^(−1) +x^(−4) ))−(1+x^(−1) )}  set x^(−1) = u ∧ u→0  =lim_(u→0) (((√(1+u+u^4 ))−(1+u))/u)  =lim_(u→0) (((1+(1/2)(u+u^4 ))−(1+u))/u)    =lim_(u→0) ((−(1/2)u+(1/2)u^4 )/u)  =lim_(u→0) (−(1/2)+(1/2)u^3 )=−(1/2)

$$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{1}} +\mathrm{x}^{−\mathrm{4}} }\:−\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}^{−\mathrm{1}} \right)}{\mathrm{x}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\left\{\sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{1}} +\mathrm{x}^{−\mathrm{4}} }−\left(\mathrm{1}+\mathrm{x}^{−\mathrm{1}} \right)\right\} \\ $$$$\mathrm{set}\:\mathrm{x}^{−\mathrm{1}} =\:\mathrm{u}\:\wedge\:\mathrm{u}\rightarrow\mathrm{0} \\ $$$$=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+\mathrm{u}+\mathrm{u}^{\mathrm{4}} }−\left(\mathrm{1}+\mathrm{u}\right)}{\mathrm{u}} \\ $$$$=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{u}+\mathrm{u}^{\mathrm{4}} \right)\right)−\left(\mathrm{1}+\mathrm{u}\right)}{\mathrm{u}}\:\: \\ $$$$=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{u}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{u}^{\mathrm{4}} }{\mathrm{u}} \\ $$$$=\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{u}^{\mathrm{3}} \right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 06/Aug/21

thx alot..

$${thx}\:{alot}.. \\ $$

Answered by EDWIN88 last updated on 06/Aug/21

 lim_(x→∞) (((x^4 +x^3 +1)−(x^2 +x)^2 )/(x((√(x^4 +x^3 +1)) +(x+x^2 ))))  = lim_(x→∞)  ((x^4 +x^3 +1−(x^4 +2x^3 +x^2 ))/(x(x^2 (√(1+(1/x)+(1/x^4 )))+x^2 (1+(1/x)))))  = lim_(x→∞) ((−x^3 −x^2 +1)/(x^3 ((√(1+(1/x)+(1/x^4 )))+1+(1/x))))  =lim_(x→∞)  ((−1−(1/x)+(1/x^3 ))/( (√(1+(1/x)+(1/x^4 )))+1+(1/x)))  =−(1/(1+1))=−(1/2)

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +\mathrm{1}\right)−\left({x}^{\mathrm{2}} +{x}\right)^{\mathrm{2}} }{{x}\left(\sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +\mathrm{1}}\:+\left({x}+{x}^{\mathrm{2}} \right)\right)} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +\mathrm{1}−\left({x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} \right)}{{x}\left({x}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}+{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\right)} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{−{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{3}} \left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}+\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{−\mathrm{1}−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}+\mathrm{1}+\frac{\mathrm{1}}{{x}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 06/Aug/21

  thank you so much..

$$\:\:{thank}\:{you}\:{so}\:{much}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com