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Question Number 118966 by bramlexs22 last updated on 21/Oct/20

$$\underset{x\to 0}{\lim }\frac{\sqrt{x+{bx}^2}-\sqrt{x}}{bx\sqrt{x}}=?$$

Answered by benjo_mathlover last updated on 21/Oct/20

$$\underset{x\to 0}{\lim }\frac{\sqrt{x}(\sqrt{1+bx}-1)}{bx\sqrt{x}}=$$$$\underset{x\to 0}{\lim }\frac{\sqrt{1+bx}-1}{bx}=\underset{x\to 0}{\lim }\frac{(1+bx)-1}{bx(\sqrt{1+bx}+1)}$$$$=\underset{x\to 0}{\lim }\frac{bx}{bx(\sqrt{1+bx}+1)}=\underset{x\to 0}{\lim }\frac{1}{\sqrt{1+bx}+1}=\frac{1}{2}$$

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