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Question Number 162672 by LEKOUMA last updated on 31/Dec/21

$$Calculate$$$$\underset{x\to +\mathrm{\infty }}{\lim }(\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x})$$$$\underset{x\to 7}{\lim }\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$$

Answered by tounghoungko last updated on 31/Dec/21

$$\left(2\right)\underset{x\to 7}{\lim }\frac{\sqrt{x+2}-3+3-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$$$$L_1=\underset{x\to 7}{\lim }\frac{x-7}{(\sqrt{x+2}+3)(\sqrt[4]{x+9}-2)}$$$$L_1=\underset{x\to 7}{\lim }\frac{(x-7)4}{6(\sqrt{x+9}-4)}=\frac{2}{3}\underset{x\to 7}{\lim }\frac{8(x-7)}{(x-7)}=\frac{16}{3}$$$$$$$$L_2=\underset{x\to 7}{\lim }\frac{3-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}=\underset{x\to 7}{\lim }\frac{(7-x)}{27}.\underset{x\to 7}{\lim }\frac{4}{\sqrt{x+9}-4}$$$$L_2=\frac{4}{27}.\underset{x\to 7}{\lim }\frac{7-x}{x-7}.\left(8\right)=-\frac{32}{27}$$$$\therefore L=L_1+L_2=\frac{16}{3}-\frac{32}{27}=\frac{112}{27}$$

Commented by LEKOUMA last updated on 31/Dec/21

$$Thank$$

Answered by tounghoungko last updated on 31/Dec/21

$$\left(1\right)\underset{x\to \mathrm{\infty }}{\lim }x\sqrt[3]{1+\frac{3}{x}}-x\sqrt{1-\frac{2}{x}}$$$$[\frac{1}{x}=y]$$$$L=\underset{y\to 0}{\lim }\frac{\sqrt[3]{1+3y}-\sqrt{1-2y}}{y}$$$$L=\underset{y\to 0}{\lim }\frac{(1+\frac{3y}{3})-(1-\frac{2y}{2})}{y}=2$$

Commented by LEKOUMA last updated on 31/Dec/21

$$Thank$$

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