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Question Number 85293 by jagoll last updated on 20/Mar/20

$$\underset{x\to 0^{+}}{\lim }\sqrt{\frac{1}{\mathrm{x}}}-\sqrt{\frac{1}{2\mathrm{x}}-2020}?$$

Commented by jagoll last updated on 20/Mar/20

$$\mathrm{what}\mathrm{is}\mathrm{result}?0\mathrm{or}\mathrm{\infty }?$$

Answered by mr W last updated on 20/Mar/20

$$=\frac{\frac{1}{2\mathrm{x}}+2020}{\sqrt{\frac{1}{\mathrm{x}}}+\sqrt{\frac{1}{2\mathrm{x}}-2020}}$$$$=\left(\frac{\frac{1}{2}+2020x}{1+\sqrt{\frac{1}{2}-2020x}}\right)\frac{1}{\sqrt{x}}$$$$=\frac{1}{1+\frac{1}{\sqrt{2}}}\times \frac{1}{0}$$$$=+\mathrm{\infty }$$

Commented by jagoll last updated on 20/Mar/20

$$\mathrm{different}\mathrm{way},\mathrm{but}\mathrm{same}\mathrm{the}\mathrm{result}$$

Answered by jagoll last updated on 20/Mar/20

Commented by prakash jain last updated on 20/Mar/20

$$\mathrm{Step}4:$$$$\mathrm{You}\mathrm{have}\mathrm{used}$$$$\lim f\left(x\right)\times g\left(x\right)=limf\left(x\right)\times limg\left(x\right)$$$$\mathrm{this}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{always}\mathrm{valid}.$$

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