# Thread: Definition of Limit proof: x -> infinity of sqrt(x) - x = -infinity

1. ## Definition of Limit proof: x -> infinity of sqrt(x) - x = -infinity

First post! yay!

Okay so yesterday I had a problem set due and it had two definition of limit at infinity proofs on it. I ended up leaving them... well not blank but all I did was write the definition.

the question was:
prove the following statements using the appropriate definition of limit:
(i) Limit as x-> infinity sqrt(x) - x = -infinity

I got as far as:

Given N<0, there exists M>0 such that
x>M implies sqrt(x) - x < N

which I believe is the appropriate definition.
I always draw a blank on these proofs. The first thing I did when I made my account here was print the sticky on definition of limit proofs. I am going through it now.

Anyways any help on this would be amazing, Thanks!

2. Take into account that is...

$\displaystyle \displaystyle \sqrt{x}-x = - \frac{x\ \sqrt{x} - x}{\sqrt{x}} = - x\ \frac{\sqrt{x}-1}{\sqrt{x}}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. If $\displaystyle N\ge 3$ then $\displaystyle N^2>2N$ or $\displaystyle N^2-N>N$.

If $\displaystyle J\in \mathbb{Z}^-~\&~J<-2$ then if $\displaystyle x\ge J^2$ can you conclude that $\displaystyle \sqrt{x}-x^2<J~?$

4. Hey thanks for the help guys, sorry I haven't been able to follow up on my question, I had a bunch of assignments due on Monday. So I was able to prove this limit as follows:
I started playing with the inequality sqrt(x) - x < N. I multiplied by -1 to get x-sqrt(x)>-N --> sqrt(x)(sqrt(x) - 1) > -N. then since N<0 by assumption, --> -N>0.

Now if sqrt(x) - 1 > sqrt(-N) and sqrt(x) > sqrt(x) - 1 > sqrt(-N) then we get sqrt(x)(sqrt(x)-1) > sqrt(-N)sqrt(-N) = -N.

so I worked with sqrt(x) - 1 > sqrt(-N) ---> x > (1 + sqrt(-N))^2. So I concluded by choosing M = (1+sqrt(-N))^2 and showed that it works.

Can someone just confirm with me that this proof works? and also how do i post the math symbols like you guys did above me?