Question Number 222425 by klipto last updated on 26/Jun/25
$$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\infty} \left(\mathrm{4}\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{\mathrm{x}}}\right) \\ $$$$\boldsymbol{\mathrm{ans}}:\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Answered by Frix last updated on 26/Jun/25
![lim_(x→−∞) (4x+(√(16x^2 +3x))) = =lim_(x→0^− ) ((4−(√(16−3x)))/x) =^([l′Ho^ pital]) =lim_(x→0^− ) ((3/(2(√(16−3x))))/1) =(3/8) 4x+(√(16x^2 −3x))=L (√(16x^2 −3x))=L−4x 16x^2 −3x=(L−4x)^2 x=(L^2 /(8L−3)) which is not defined for L=(3/8) ⇒ Limit =(3/8)](https://www.tinkutara.com/question/Q222430.png)
$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\mathrm{4}{x}+\sqrt{\mathrm{16}{x}^{\mathrm{2}} +\mathrm{3}{x}}\right)\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\frac{\mathrm{4}−\sqrt{\mathrm{16}−\mathrm{3}{x}}}{{x}}\:\overset{\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right]} {=} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\frac{\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{16}−\mathrm{3}{x}}}}{\mathrm{1}}\:=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$ \\ $$$$\mathrm{4}{x}+\sqrt{\mathrm{16}{x}^{\mathrm{2}} −\mathrm{3}{x}}={L} \\ $$$$\sqrt{\mathrm{16}{x}^{\mathrm{2}} −\mathrm{3}{x}}={L}−\mathrm{4}{x} \\ $$$$\mathrm{16}{x}^{\mathrm{2}} −\mathrm{3}{x}=\left({L}−\mathrm{4}{x}\right)^{\mathrm{2}} \\ $$$${x}=\frac{{L}^{\mathrm{2}} }{\mathrm{8}{L}−\mathrm{3}}\:\mathrm{which}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\:{L}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\Rightarrow\:\mathrm{Limit}\:=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Answered by mr W last updated on 26/Jun/25
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{4}{x}+\sqrt{\mathrm{16}{x}^{\mathrm{2}} −\mathrm{3}{x}}\right)=+\infty \\ $$$$ \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{4}{x}+\sqrt{\mathrm{16}{x}^{\mathrm{2}} −\mathrm{3}{x}}\right) \\ $$$$=\underset{{t}\rightarrow+\infty} {\mathrm{lim}}\left(−\mathrm{4}{t}+\sqrt{\mathrm{16}{t}^{\mathrm{2}} +\mathrm{3}{t}}\right) \\ $$$$=\underset{{t}\rightarrow+\infty} {\mathrm{lim}}\frac{\left(\sqrt{\mathrm{16}{t}^{\mathrm{2}} +\mathrm{3}{t}}−\mathrm{4}{t}\right)\left(\sqrt{\mathrm{16}{t}^{\mathrm{2}} +\mathrm{3}{t}}+\mathrm{4}{t}\right)}{\:\sqrt{\mathrm{16}{t}^{\mathrm{2}} +\mathrm{3}{t}}+\mathrm{4}{t}} \\ $$$$=\underset{{t}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{3}{t}}{\:\mathrm{4}{t}\sqrt{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{16}{t}}}+\mathrm{4}{t}} \\ $$$$=\underset{{t}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{3}}{\:\mathrm{4}\sqrt{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{16}{t}}}+\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}+\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Answered by Frix last updated on 26/Jun/25
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\left(…\right)\:=+\infty \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\mathrm{4}{x}+\sqrt{\mathrm{16}{x}^{\mathrm{2}} −\mathrm{3}{x}}\right)\:=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Commented by klipto last updated on 26/Jun/25
$$\mathrm{break}\:\mathrm{it}\:\mathrm{down}… \\ $$