Evaluate the $\displaystyle\lim_{x\to 0}\left(\frac{1}{x\sqrt{1+x}}-\frac 1x\right)$ Evaluate the limit as $x$ approaches $0$ : $$\frac{1}{x\sqrt{1+x}}-\frac 1x$$
I have so far $$\frac{x -x\sqrt{1+x}}{x(x\sqrt{1+x})}$$
Do I multiply the numerator and denominator by the conjugate of $x - x\sqrt{1+x}$? Or do I multiply the top and bottom by $x\sqrt{1+x}$?

You can get
$$\frac{1}{\color{red}{x}\sqrt{1+x}}-\frac{1}{\color{red}{x}}=\frac{1}{x\sqrt{1+x}}-\frac{\sqrt{1+x}}{x\sqrt{1+x}}=\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}.$$
Then, multiply it by $$\frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}.$$
