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Question Number 127421 by Study last updated on 29/Dec/20

lim_(x→−∞) ((√(x^2 +5x−3))/(10x−3))=???

$${li}\underset{{x}\rightarrow−\infty} {{m}}\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{3}}}{\mathrm{10}{x}−\mathrm{3}}=??? \\ $$

Answered by ebi last updated on 29/Dec/20

lim_(x→−∞) ((√(x^2 +5x−3))/(10x−3))  =lim_(x→−∞) ((√(x^2 +5x−3))/(10x−3))=lim_(x→−∞) ((√(x^2 (1+(5/x)−(3/x^2 ))))/(x(10−(3/x))))  =lim_(x→−∞) ((∣x∣(√((1+(5/x)−(3/x^2 )))))/(x(10−(3/x))))  =lim_(x→−∞) ((−x(√((1+(5/x)−(3/x^2 )))))/(x(10−(3/x)))), ∣x∣= { ((x, x≥0)),((−x, x<0)) :}  =lim_(x→−∞) ((−(√((1+(5/x)−(3/x^2 )))))/((10−(3/x))))=((−(√(1+0−0)))/(10))=−(1/(10))

$$\underset{{x}\rightarrow−\infty} {{lim}}\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{3}}}{\mathrm{10}{x}−\mathrm{3}} \\ $$$$=\underset{{x}\rightarrow−\infty} {{lim}}\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{3}}}{\mathrm{10}{x}−\mathrm{3}}=\underset{{x}\rightarrow−\infty} {{lim}}\frac{\sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{5}}{{x}}−\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\right)}}{{x}\left(\mathrm{10}−\frac{\mathrm{3}}{{x}}\right)} \\ $$$$=\underset{{x}\rightarrow−\infty} {{lim}}\frac{\mid{x}\mid\sqrt{\left(\mathrm{1}+\frac{\mathrm{5}}{{x}}−\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\right)}}{{x}\left(\mathrm{10}−\frac{\mathrm{3}}{{x}}\right)} \\ $$$$=\underset{{x}\rightarrow−\infty} {{lim}}\frac{−{x}\sqrt{\left(\mathrm{1}+\frac{\mathrm{5}}{{x}}−\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\right)}}{{x}\left(\mathrm{10}−\frac{\mathrm{3}}{{x}}\right)},\:\mid{x}\mid=\begin{cases}{{x},\:{x}\geqslant\mathrm{0}}\\{−{x},\:{x}<\mathrm{0}}\end{cases} \\ $$$$=\underset{{x}\rightarrow−\infty} {{lim}}\frac{−\sqrt{\left(\mathrm{1}+\frac{\mathrm{5}}{{x}}−\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\right)}}{\left(\mathrm{10}−\frac{\mathrm{3}}{{x}}\right)}=\frac{−\sqrt{\mathrm{1}+\mathrm{0}−\mathrm{0}}}{\mathrm{10}}=−\frac{\mathrm{1}}{\mathrm{10}} \\ $$

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