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Question Number 112855 by ruwedkabeh last updated on 10/Sep/20

$$$$$$1.\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{5x+4}-\sqrt{3x+9}}{4x}=...$$$$$$$$2.\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{5-4x+3x^2}-\sqrt{4-3x+3x^2}}{2x}=...$$$$$$

Answered by bobhans last updated on 10/Sep/20

$$\left(1\right)\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{5\mathrm{x}+4}-\sqrt{3\mathrm{x}+9}}{4\mathrm{x}}=\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{\frac{5}{\mathrm{x}}+\frac{4}{{\mathrm{x}}^2}}-\sqrt{\frac{3}{\mathrm{x}}+\frac{9}{{\mathrm{x}}^2}}}{4}=0$$

Answered by bobhans last updated on 10/Sep/20

$$\left(2\right)\underset{x\to \mathrm{\infty }}{\lim }\frac{\mathrm{x}(\sqrt{\frac{5}{{\mathrm{x}}^2}-\frac{4}{\mathrm{x}}+3}-\sqrt{\frac{4}{{\mathrm{x}}^2}-\frac{3}{\mathrm{x}}+3})}{2\mathrm{x}}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{\frac{5}{{\mathrm{x}}^2}-\frac{4}{\mathrm{x}}+3}-\sqrt{\frac{4}{{\mathrm{x}}^2}-\frac{3}{{\mathrm{x}}^2}+3}}{2}=0$$

Commented by ruwedkabeh last updated on 10/Sep/20

$$thankyouverymuch.$$

Answered by mathmax by abdo last updated on 10/Sep/20

$$1)\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{5\mathrm{x}+4}-\sqrt{3\mathrm{x}+9}}{4\mathrm{x}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{5\mathrm{x}}\sqrt{1+\frac{4}{5\mathrm{x}}}-\sqrt{3\mathrm{x}}\sqrt{1+\frac{9}{3\mathrm{x}}}}{4\mathrm{x}}$$$$\sim \frac{\sqrt{5\mathrm{x}}(1+\frac{2}{5\mathrm{x}})-\sqrt{3\mathrm{x}}(1+\frac{3}{2\mathrm{x}})}{4\mathrm{x}}=\frac{(\sqrt{5}-\sqrt{3})\sqrt{\mathrm{x}}+\frac{2}{\sqrt{5\mathrm{x}}}-\frac{3\sqrt{3\mathrm{x}}}{2\mathrm{x}}}{4\mathrm{x}}$$$$\sim \frac{\sqrt{5}-\sqrt{3}}{4\sqrt{\mathrm{x}}}\to 0(\mathrm{x}\to +\mathrm{\infty })\Rightarrow {\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=0$$

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