Question-149637

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

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

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

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

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