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Question Number 187598 by horsebrand11 last updated on 19/Feb/23

 lim_(x→0^+ ) (6/x)−(√(((36)/x^2 )+(4/x)+9)) =?

$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{6}}{{x}}−\sqrt{\frac{\mathrm{36}}{{x}^{\mathrm{2}} }+\frac{\mathrm{4}}{{x}}+\mathrm{9}}\:=? \\ $$

Answered by cortano12 last updated on 19/Feb/23

  L=lim_(x→0^+ )  (6/x) −(1/x)(√(36+4x+9x^2 ))     L=lim_(x→0^+ )  ((6−(√(9x^2 +4x+36)))/x)   L=lim_(x→0^+ )  ((−x(9x+4))/(x(6+(√(9x^2 +4x+36)) )))   L=−(1/3)

$$\:\:{L}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{6}}{{x}}\:−\frac{\mathrm{1}}{{x}}\sqrt{\mathrm{36}+\mathrm{4}{x}+\mathrm{9}{x}^{\mathrm{2}} }\: \\ $$$$\:\:{L}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{6}−\sqrt{\mathrm{9}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{36}}}{{x}} \\ $$$$\:{L}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{−{x}\left(\mathrm{9}{x}+\mathrm{4}\right)}{{x}\left(\mathrm{6}+\sqrt{\mathrm{9}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{36}}\:\right)} \\ $$$$\:{L}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Answered by a.lgnaoui last updated on 19/Feb/23

L=(6/x)−(√(((6/x))^2 +(2/3)((6/x))+9))  posons  X=(6/x)⇒X−(√(X^2 +(2/3)X+9))    X−(√((X+(1/3))^2 +((80)/9))) =X−(X+(1/3))(√(1+((80)/(9(X+(1/3))^2 ))))  x→0^(0+) ⇒X→+∞et (√(1+((80)/(9(x+(1/3))^2 )))) →1    lim_(x→0^+ ) (6/x)−(√(((36)/x^2 ) +(4/x)+ 9)) =(−(1/3))

$$\mathrm{L}=\frac{\mathrm{6}}{{x}}−\sqrt{\left(\frac{\mathrm{6}}{{x}}\right)^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{6}}{{x}}\right)+\mathrm{9}} \\ $$$${posons}\:\:\mathrm{X}=\frac{\mathrm{6}}{{x}}\Rightarrow\mathrm{X}−\sqrt{\mathrm{X}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}\mathrm{X}+\mathrm{9}} \\ $$$$\:\:\mathrm{X}−\sqrt{\left(\mathrm{X}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{80}}{\mathrm{9}}}\:=\mathrm{X}−\left(\mathrm{X}+\frac{\mathrm{1}}{\mathrm{3}}\right)\sqrt{\mathrm{1}+\frac{\mathrm{80}}{\mathrm{9}\left(\mathrm{X}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} }} \\ $$$${x}\rightarrow\mathrm{0}^{\mathrm{0}+} \Rightarrow\mathrm{X}\rightarrow+\infty\mathrm{et}\:\sqrt{\mathrm{1}+\frac{\mathrm{80}}{\mathrm{9}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} }}\:\rightarrow\mathrm{1} \\ $$$$ \\ $$$$\mathrm{lim}_{{x}\rightarrow\mathrm{0}^{+} } \frac{\mathrm{6}}{{x}}−\sqrt{\frac{\mathrm{36}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{4}}{{x}}+\:\mathrm{9}}\:=\left(−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$

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