Question Number 182978 by TUN last updated on 18/Dec/22
Answered by dumitrel last updated on 18/Dec/22
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Commented by dumitrel last updated on 18/Dec/22
Answered by cortano1 last updated on 18/Dec/22
![L= lim_(n→∞) ((√(n^2 +n)) .((n^3 −n^2 ))^(1/3) −n^2 ) L= lim_(n→∞) n^2 (√(1+(1/n))) .((1−(1/n)))^(1/3) −n^2 [ let (1/n) = x ] L = lim_(x→0) (((√(1+x)) .((1−x))^(1/3) −1)/x^2 ) = lim_(x→0) (((((1+x)^3 ))^(1/6) .(((1−x)^2 ))^(1/6) −1)/x^2 ) = lim_(x→0) (((((1+x)(1−x^2 )^2 ))^(1/6) −1)/x^2 ) = lim_(x→0) (((((1+x)(1−2x^2 +x^4 )))^(1/6) −1)/x^2 ) = lim_(x→0) ((((1+x^5 +x^4 −2x^3 −2x^2 +x))^(1/6) −1)/x^2 ) = lim_(x→0) ((1+(((x^5 +x^4 −2x^3 −2x^2 +x)/6) )−1)/x^2 )= ∞](https://www.tinkutara.com/question/Q182989.png)
$$\:{L}=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{n}^{\mathrm{2}} +{n}}\:.\sqrt[{\mathrm{3}}]{{n}^{\mathrm{3}} −{n}^{\mathrm{2}} }−{n}^{\mathrm{2}} \:\right) \\ $$$$\:{L}=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}^{\mathrm{2}} \:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\:.\sqrt[{\mathrm{3}}]{\mathrm{1}−\frac{\mathrm{1}}{{n}}}−{n}^{\mathrm{2}} \\ $$$$\:\left[\:{let}\:\frac{\mathrm{1}}{{n}}\:=\:{x}\:\right] \\ $$$$\:{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{x}}\:.\sqrt[{\mathrm{3}}]{\mathrm{1}−{x}}−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{6}}]{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }\:.\sqrt[{\mathrm{6}}]{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{6}}]{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\sqrt[{\mathrm{6}}]{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +{x}^{\mathrm{4}} \right)}−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{6}}]{\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}}−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\left(\frac{{x}^{\mathrm{5}} +{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}}{\mathrm{6}}\:\right)−\mathrm{1}}{{x}^{\mathrm{2}} }=\:\infty \\ $$