<s>[INST] Find limit of the function $$\lim_{x\to 0^+} \frac { \sqrt {x+bx^2}-\sqrt x}{bx^{3/2}}$$
I tried taking x common and all but it doesn't work.Are there are any shortcuts for finding the answer?Any help is greatly appreciated
Thanks in advance
 [/INST] First of all $b$  cannot be zero.
By multiplying both numerator and denominator by $\sqrt{x+bx^2}+\sqrt{x}$
the function becomes:

$$\frac{(\sqrt{x+bx^2}+\sqrt{x})(\sqrt{x+bx^2}-\sqrt{x})}{(\sqrt{x+bx^2}+\sqrt{x})bx^{3/2}}=\frac{bx^2}{(\sqrt{x+bx^2}+\sqrt{x})bx^{3/2}}$$ $$=\frac{\sqrt{x}}{\sqrt{x+bx^2}+\sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x(bx+1)}+\sqrt{x}}=\frac{\sqrt{x}}{\sqrt{x}(\sqrt{1+bx}+1)}=\frac{1}{\sqrt{1+bx}+1} \to \frac{1}{2}$$ as $x \to 0^+$

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