# Limit of square root problem

Can you please help me out with this limit problem. Actually, I tried to solve it by the conjugate method but it didn't work with me.

Thank you.

$$\lim_{x \to 0}\; \bigg( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x²+x}} \bigg)$$

• I edited your question. Please check and make sure that I did not alter what you are asking. – Thomas Jan 24 '13 at 15:13
• what was the result from your conjugate method? – still_learning Jan 24 '13 at 15:15
• The limit is not defined as the function is not defined in a deleted neighbourhood of $0$. The function is defined on $(0, \infty)$ so you could instead consider the corresponding one-sided limit. – Michael Albanese Jan 24 '13 at 15:41

$$\lim_{x \to 0} \frac{\sqrt{x^2+x} - \sqrt{x}}{\sqrt{x(x^2+x)}}= \lim_{x \to 0} \frac{\sqrt{x}(\sqrt{x+1} - 1)}{x\sqrt{x+1}}= \lim_{x \to 0} \frac{x^{\frac{3}{2}}}{x\sqrt{x+1}(\sqrt{x+1}+1)}=0$$ The third step is due to mupltiplying both numerator and denominator by $\sqrt{x+1}+1$
Take $\frac{1}{\sqrt{x}}$ common and then try the conjugate method.
Also, $$\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x^2+x}}=\dfrac{1}{\sqrt{x}}\left(1-(1+x)^{-1/2}\right)=\dfrac{1}{\sqrt{x}}\left(\dfrac{1}{2}x+o(x)\right)\to 0 \mbox{ as } x\to 0^+$$