2
$\begingroup$

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

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?

$\endgroup$
4
  • $\begingroup$ $\sqrt{\frac{1}{x}+2}$ or $\sqrt{\frac{1}{x+2}}$, there's a very important difference! $\endgroup$ Commented Jul 20, 2016 at 10:52
  • $\begingroup$ We can probably give better help if you show us what you do when you get $\sqrt{2}$. $\endgroup$ Commented Jul 20, 2016 at 10:59
  • $\begingroup$ it is the first one @AlphaNumeric $\endgroup$
    – John Curry
    Commented Jul 20, 2016 at 11:10
  • $\begingroup$ @Henrik I just checked and I don't get sqrt(2). I get infinity so will edit the question. Thanks. $\endgroup$
    – John Curry
    Commented Jul 20, 2016 at 11:19

2 Answers 2

4
$\begingroup$

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)

$\endgroup$
5
  • $\begingroup$ we are looking for the limit as x tends to zero (right-hand side limit), so wouldn't the 1/x terms tend to infinity in your answer? $\endgroup$
    – John Curry
    Commented Jul 20, 2016 at 11:17
  • $\begingroup$ @JohnCurry Yes, ${1 \over x}$ tends to infinity, but it is in the denominator, so the whole fraction tends to 0. $\endgroup$
    – lisyarus
    Commented Jul 20, 2016 at 11:21
  • $\begingroup$ @lisyarus Ah, ok, I think I am rather rusty at this. So it's ok to generate the infinities on the denominator as it works out at 1/infinity which is zero. So it is a valid limit. $\endgroup$
    – John Curry
    Commented Jul 20, 2016 at 11:23
  • $\begingroup$ Thanks for clarification @lisyarus $\endgroup$
    – John Curry
    Commented Jul 20, 2016 at 11:25
  • $\begingroup$ @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. $\endgroup$
    – lisyarus
    Commented Jul 20, 2016 at 11:29
2
$\begingroup$

Use Taylor at order $1$ after some transformation: $$\sqrt{\frac{1}{x}+2} -\sqrt{\frac1x}=\frac{\sqrt{1+2x}-1}{\sqrt x}=\frac{1+x+o(x)-1}{\sqrt x}=\sqrt{x}+o(\sqrt x).$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .