# Why is the limit of this function $0$?

$$\lim_{x\to 0} \left(\sqrt{\dfrac{1}{x}+2} - \sqrt{\dfrac{1}{x}}\right)$$

I keep working this limit out as infinity as each term separately works out at infinity or non-existent but online calculators and the book solution tell me the limit is $0$.

Can anyone help me understand the steps to generate the right hand limit?

• $\sqrt{\frac{1}{x}+2}$ or $\sqrt{\frac{1}{x+2}}$, there's a very important difference! Commented Jul 20, 2016 at 10:52
• We can probably give better help if you show us what you do when you get $\sqrt{2}$. Commented Jul 20, 2016 at 10:59
• it is the first one @AlphaNumeric Commented Jul 20, 2016 at 11:10
• @Henrik I just checked and I don't get sqrt(2). I get infinity so will edit the question. Thanks. Commented Jul 20, 2016 at 11:19

Multiply by $$\frac{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}.$$ You get $$\frac{2}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$ which tends to 0. (expanding in sight of binomial theorem is a pretty common trick)
• @JohnCurry Yes, ${1 \over x}$ tends to infinity, but it is in the denominator, so the whole fraction tends to 0. Commented Jul 20, 2016 at 11:21
• @JohnCurry Problems can arise with limits like $\frac{1}{a-b}$, where both $a$ and $b$ tend to positive infinity. In our case we have $\frac{1}{a+b}$, and $a+b$ will always tend to positive infinity. Commented Jul 20, 2016 at 11:29
Use Taylor at order $1$ after some transformation: $$\sqrt{\frac{1}{x}+2} -\sqrt{\frac1x}=\frac{\sqrt{1+2x}-1}{\sqrt x}=\frac{1+x+o(x)-1}{\sqrt x}=\sqrt{x}+o(\sqrt x).$$