# 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 $$\frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}.$$

• @user137452: Is there anything unclear? – mathlove Aug 6 '14 at 11:09
• What happened to the x's? – user137452 Aug 6 '14 at 11:10
• @user137452 I built upon his answer. Please take a look at this for a more detailed explanation. – Varun Iyer Aug 6 '14 at 11:12
• @user137452: Look at the parts colored in red. – mathlove Aug 6 '14 at 11:13
• @mathlove You're welcome - thanks for all your great questions and answers. – user84413 Aug 6 '14 at 18:34

For $x→0^+$
Here we can use the substitution $x=\tan^2⁡θ$
$$\lim_{x→0)}\dfrac{1}{x√(1+x)}-\dfrac{1}{x}$$ $$=\lim_{θ→0} cot^2⁡θ (cos⁡θ-1)$$ $$=\lim_{θ→0} \dfrac{cos^2⁡θ(cos⁡θ-1)}{sin^2⁡θ}$$ $$=\lim_{θ→0}-\dfrac{cos^2⁡θ (cos⁡θ-1))}{(cos⁡θ-1)(cos⁡θ+1)}$$ $$=-\dfrac{1}{2}$$

• The final answer is probably right, but the method you give only gives you the limit as $x \to 0^+$ (because $\tan^2 \theta \ge 0$). – Najib Idrissi Aug 7 '14 at 8:32
• Thanks. I didn't notice that. – Bumblebee Aug 7 '14 at 12:01
• I add that on my answer. – Bumblebee Aug 7 '14 at 12:13
• If I use the substitution $x=-\sin^2\theta$ then the I can find the left limit also. – Bumblebee Aug 13 '14 at 9:33