Question Number 163642 by cortano1 last updated on 09/Jan/22
$$\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\lfloor\mathrm{3}{x}\rfloor−\mathrm{3}{x}}{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}}\:=? \\ $$
Answered by mehdiAz last updated on 09/Jan/22
![lim_(x→(2/3)) ((⌊3x⌋−3x)/(9x^2 −4)) =? if x∈]−∞,0]: lim_(x→(2/3)) ((−3x−3x)/(9x^2 −4)) = lim_(x→(2/3)) ((−6x)/(9x^2 −4)) let x^2 =h ⇒ x=−(√h) we have x→(2/3) ⇒ h→ (4/9) lim_(h→(4/9)) ((6(√h))/(9h−4)) = ((6×(2/3))/(9×(4/9)−4)) = (4/(4−4)) = (4/0^+ ) = +∞ so lim_(x→(2/3)) ((⌊3x⌋−3x)/(9x^2 −4)) = +∞, for x ∈ ]−∞,0] now if x ∈ [0,+∞[, then lim_(x→(2/3)) ((3x−3x)/(9x^2 −4)) = lim_(x→(2/3)) (0/(9x^2 −4))=lim_(x→(2/3)) 0 = 0](https://www.tinkutara.com/question/Q163646.png)
$$\: \\ $$$$\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\lfloor\mathrm{3}{x}\rfloor−\mathrm{3}{x}}{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}}\:=? \\ $$$$\left.{i}\left.{f}\:{x}\in\right]−\infty,\mathrm{0}\right]:\:\underset{{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} {\mathrm{lim}}\:\frac{−\mathrm{3}{x}−\mathrm{3}{x}}{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}}\:=\:\underset{{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} {\mathrm{lim}}\:\frac{−\mathrm{6}{x}}{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}} \\ $$$${let}\:{x}^{\mathrm{2}} ={h}\:\Rightarrow\:{x}=−\sqrt{{h}}\: \\ $$$${we}\:{have}\:{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow\:{h}\rightarrow\:\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\underset{{h}\rightarrow\frac{\mathrm{4}}{\mathrm{9}}} {\mathrm{lim}}\:\frac{\mathrm{6}\sqrt{{h}}}{\mathrm{9}{h}−\mathrm{4}}\:=\:\frac{\mathrm{6}×\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{9}×\frac{\mathrm{4}}{\mathrm{9}}−\mathrm{4}}\:=\:\frac{\mathrm{4}}{\mathrm{4}−\mathrm{4}}\:=\:\frac{\mathrm{4}}{\mathrm{0}^{+} }\:=\:+\infty \\ $$$$\left.{s}\left.{o}\:\underset{{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\lfloor\mathrm{3}{x}\rfloor−\mathrm{3}{x}}{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}}\:=\:+\infty,\:{for}\:{x}\:\in\:\right]−\infty,\mathrm{0}\right] \\ $$$${now}\:{if}\:{x}\:\in\:\left[\mathrm{0},+\infty\left[,\:{then}\right.\right. \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{3}{x}−\mathrm{3}{x}}{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}}\:=\:\underset{{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{0}}{\mathrm{9}{x}^{\mathrm{2}} −\mathrm{4}}=\underset{{x}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} {\mathrm{lim}}\:\mathrm{0}\:=\:\mathrm{0} \\ $$