I'm trying to solve this limit but I can't figure out how to get rid of the undefined form. $$\eqalign{ & \mathop {\lim }\limits_{x \to {0^ + }} \left( {{1 \over {\sqrt x }} - {1 \over {\sqrt {{x^2} + x} }}} \right) \cr & = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{{\sqrt {{x^2} + x} - \sqrt x } \over {\sqrt {{x^3} + {x^2}} }}} \right) \cr} $$ $$\eqalign{ & = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{{{x^2}} \over {\left( {\sqrt {{x^3} + {x^2}} } \right)\left( {\sqrt {{x^2} + x} + \sqrt x } \right)}}} \right) \cr & = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{{{x^2}} \over {\left( {x\sqrt {x + 1} } \right)\left( {x\sqrt {x + 1} + x\sqrt {{1 \over x}} } \right)}}} \right) \cr} $$ $$ = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{x \over {\left( {(x + 1) + \sqrt {1 + {1 \over x}} } \right)}}} \right)$$
Now what?
Multiply the conjugate of the denominator in problems like this always helps: \begin{align} & \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x^2 + x}} \\ = & \frac{\sqrt{x^2 + x} - \sqrt{x}}{\sqrt{x}\sqrt{x^2 + x}} \\ = & \frac{(\sqrt{x^2 + x} - \sqrt{x})\color{red}{(\sqrt{x^2 + x} + \sqrt{x})}}{\sqrt{x}\sqrt{x^2 + x}\color{red}{(\sqrt{x^2 + x} + \sqrt{x})}} \\ = & \frac{x^2}{\sqrt{x}\sqrt{x^2 + x}(\sqrt{x^2 + x} + \sqrt{x})} \\ = & \frac{x^{1/2}}{\sqrt{x + 1}(\sqrt{x + 1} + 1)} \to 0 \end{align} as $x \to 0+$.