Square root limit problem

Question

Please help with this problem. Somewhere my algebra gets confused. I know the answer is zero but I need the steps for the problem.I will start the problem and you can tell me where my mistake is.

$$\lim_{x\to0}\left(\frac1{\sqrt{x}}-\frac1{\sqrt{x^2+x}}\right)=\lim_{x\to0}\frac{\sqrt{x^2+x}-\sqrt{x}}{\sqrt{x}\cdot\sqrt{x^2+x}}$$

Take conjugate of numerator and multiply by numerator and denominator :

$$\frac{\sqrt{x^2+x}-\sqrt{x}}{\sqrt{x}\cdot\sqrt{x^2+x}}\cdot\frac{\sqrt{x^2+x}+\sqrt{x}}{\sqrt{x^2+x}+\sqrt{x}}=\frac{x^2+x-(\sqrt{x})^2}{\sqrt{x}\cdot\sqrt{x^2+x}\left(\sqrt{x^2+x}+\sqrt{x}\right)}$$

help from here

Right now it's very difficult to figure out what your question means... you should use $\LaTeX$ markup to make your question more readable — Andrew Maurer, Jun 16 '13 at 07:58
Directions to use LaTeX in this can be found in the FAQ section — DonAntonio, Jun 16 '13 at 08:02
Yes Brian you interpreted the question correctly. I will use the latex next time. New to the site so I need to read up on it. — user56878, Jun 16 '13 at 09:26

score 1 · Answer 1 · answered Jun 16 '13 at 09:28

Notice that the limit makes sense only from the right, i.e. for $\,x>0\,$:

$$\lim_{x\to 0^+}\left(\frac1{\sqrt x}-\frac1{\sqrt{x^2+x}}\right)=\lim_{x\to 0^+}\frac{x^2+x-\left(\sqrt x\right)^2}{\sqrt x\sqrt{x^2+x}\left(\sqrt{x^2+x}+\sqrt x\right)}=$$

$$=\lim_{x\to 0^+}\frac{x^2}{\sqrt x\sqrt{x^2+x}\left(\sqrt{x^2+x}+\sqrt x\right)}\frac{\frac1{x^2}}{\frac1{x^2}}=$$

$$=\lim_{x\to 0^+}\frac{\frac{x^2}{x^2}}{\frac{\sqrt x}{\sqrt x}\cdot\frac{\sqrt{x^2+x}}{x}\left(\frac{\sqrt{x^2+x}+\sqrt x}{\sqrt x}\right)}=\lim_{x\to 0^+}\frac1{ 1\cdot\sqrt{1+\frac1x}\left(\sqrt{x+1}+1\right)}=0$$

lab bhattacharjee · Answer 2 · 2013-06-16T08:45:45.427

From where you have left,

as $x\to0,x\ne0$ divide the numerator & the denominator by $x\sqrt x$ to get

$$\lim_{x\to0}\frac{x^2+x-(\sqrt{x})^2}{\sqrt{x}\cdot\sqrt{x^2+x}\left(\sqrt{x^2+x}+\sqrt{x}\right)}$$

$$=\lim_{x\to0}\frac{x^2}{\sqrt{x}\cdot\sqrt{x^2+x}\left(\sqrt{x^2+x}+\sqrt{x}\right)}$$

$$=\lim_{x\to0}\frac{\sqrt x}{\sqrt{x+1}\left(\sqrt{x+1}+1\right)}=0$$

Alternatively,

HINT:

As far as I could read,

$$\lim_{x\to0}\frac1{\sqrt x}-\frac1{\sqrt{x^2+x}}$$

$$=\lim_{x\to0}\frac{\sqrt{x+1}-1}{\sqrt{x(x+1)}}$$

$$=\lim_{x\to0}\frac{(x+1)-1}{\sqrt{x(x+1)}}\frac1{\sqrt{x+1}+1}\text{ (rationalizing the numerator )}$$

$$=\lim_{x\to0}\frac{\sqrt x}{\sqrt{x+1}}\frac1{\sqrt{x+1}+1}\text{ as }x\to 0,\ne0$$

Square root limit problem

2 Answers2