Question Number 187859 by TUN last updated on 23/Feb/23
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}−{x}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 23/Feb/23
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}−{x}\right) \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}−{x}\right)\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}\:+{x}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}\:+{x}} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\:{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}−{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}\:+{x}} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\:{x}\left(\mathrm{3}+\frac{\mathrm{2}}{{x}}\right)}{\:{x}\left(\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}\:+\mathrm{1}\right)} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\:\mathrm{3}+\frac{\mathrm{2}}{{x}}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}\:+\mathrm{1}} \\ $$$$\:\frac{\:\mathrm{3}+\mathrm{0}}{\:\sqrt{\mathrm{1}+\mathrm{0}+\mathrm{0}}\:+\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Frix last updated on 23/Feb/23
$$\mathrm{The}\:\mathrm{mistake}\:\mathrm{is}\:\mathrm{for}\:{x}<\mathrm{0}: \\ $$$$\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}}{\mid{x}\mid}=−\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}}{{x}} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}+\frac{\mathrm{2}}{{x}}}{\mathrm{1}−\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }}}\:=\frac{\mathrm{3}−\mathrm{0}}{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{0}+\mathrm{0}}}\:\mathrm{undefined} \\ $$
Commented by Rasheed.Sindhi last updated on 23/Feb/23
$$\mathcal{T}{hanks}\:{sir},\:{I}\:{learnt}! \\ $$
Answered by cortano12 last updated on 23/Feb/23
![L=lim_(x→−∞) (√(x^2 (1+(3/x)+(2/x^2 ))))−x = lim_(x→−∞) −x[(√(1+(3/x)+(2/x^2 ))) +1 ] = lim_(x→0) (((√(1−3x−2x^2 ))+1)/x)=∞](https://www.tinkutara.com/question/Q187862.png)
$$\:\mathrm{L}=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\right)}−\mathrm{x} \\ $$$$\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:−\mathrm{x}\left[\sqrt{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{x}}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }}\:+\mathrm{1}\:\right] \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}−\mathrm{3x}−\mathrm{2x}^{\mathrm{2}} }+\mathrm{1}}{\mathrm{x}}=\infty \\ $$
Answered by aba last updated on 23/Feb/23
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{3x}+\mathrm{2}}=+\infty\:\wedge\underset{{x}\rightarrow−\infty} {\mathrm{lim}x}=−\infty \\ $$$$\Rightarrow\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{3x}+\mathrm{2}}−\mathrm{x}\right)=+\infty \\ $$