# Find the value of : $\lim_{x\to\infty} \sqrt{x+\sqrt{x}}-\sqrt{x}$

I tried to multiply by the conjugate:

$\displaystyle\lim_{x\to\infty} \frac{\left(\sqrt{x+\sqrt{x}}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x}}+\sqrt{x}\right)}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\displaystyle\lim_{x\to\infty} \frac{x-x+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\displaystyle\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}$

I don't even know if my rewriting has helped at all. How would I go about doing this?

• Did you try to divide by $\sqrt{x}$? – Dennis Gulko Oct 13 '13 at 7:31
• Try L'Hospital's rule – Bhargav Oct 13 '13 at 7:36
• yes, do as Dennis said. it will be 1/2 – Lei Li Oct 13 '13 at 8:57

Dividing by $\sqrt{x}$ we get: $$\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac1{\sqrt x}}+1}$$

• How can one simply divide by $\sqrt{x}$ and not multiply by it at the same time? – Hans Groeffen Oct 13 '13 at 7:34
• I multiplied both numerator and denominator by $\frac1{\sqrt x}$ – Dennis Gulko Oct 13 '13 at 7:36

Put $x=\frac1{h^2}$

$$\lim_{x\to\infty}\left(\sqrt{x+\sqrt x}-\sqrt x\right)=\lim_{h\to0}\left(\sqrt{\frac1{h^2}+\frac1h}-\frac1h\right)=\lim_{h\to0}\frac{\sqrt{h+1}-1}h$$

$$=\lim_{h\to0}\frac{{h+1}-1}{h(\sqrt{h+1}+1)}=\lim_{h\to0}\frac1{(\sqrt{h+1}+1)}$$ Cancelling out $h$ as $h\ne0$ as $h\to0$

• How did you know that it would work to put $x=\frac{1}{h^2}$? – Hans Groeffen Oct 13 '13 at 7:42
• @HansGroeffen, As there is one $\sqrt x$ inside $\sqrt$ – lab bhattacharjee Oct 13 '13 at 9:47

$\displaystyle \lim_{x\to\infty}\left(\sqrt{x+\sqrt x}-\sqrt x\right) = \lim_{x\to\infty}\sqrt{x}\left\{\left(1+\frac{1}{\sqrt{x}}\right)^{\frac{1}{2}}-1\right\}$

Now expand Using Binomialy $\displaystyle = \lim_{x\to\infty}\sqrt{x}.\left\{\left(1+\frac{1}{2.\sqrt{x}}+\frac{1}{2}.\left(\frac{1}{2}-1\right).\frac{1}{(\sqrt{x})^2}+.........\right)-1\right\} = \lim_{x\to\infty}\sqrt{x}\left\{\frac{1}{2.\sqrt{x}}+\frac{1}{2}.\left(\frac{1}{2}-1\right).\frac{1}{(\sqrt{x})^2}+.........\right\}$

$\displaystyle =\frac{1}{2}$