# Clarification: Breaking up a limit going towards infinity

1. Sep 20, 2015

### Zaculus

1. The problem statement, all variables and given/known data
Find $$\lim_{x\to \infty} \sqrt{x^2+1}-x$$

2. Relevant equations

3. The attempt at a solution
This is mostly for a refresher. I know it's zero because when we multiply the top and bottom by the conjugate to obtain $$\frac{1}{\sqrt{x^2+1}+x}$$ and the denominator increases without bound, thus the limit must be 0. But it is not alright to do the following, correct?
$$\lim_{x\to \infty} \frac{1}{\sqrt{x^2+1}+x} = \lim_{x\to \infty} \frac{1}{x(\sqrt{1+\frac{1}{x^2}} + 1)} =\lim_{x\to \infty} \frac{1}{2x} = 0,$$
knowing that $$\frac{1}{x^2}$$ tends to 0. The reason I don't think it's technically correct is that even though I'm not explicitly writing it, I'm technically using
$$\lim_{x\to \infty} x(\sqrt{1+\frac{1}{x^2}}+1) = \lim_{x\to \infty}x \lim_{x\to \infty}(\sqrt{1+\frac{1}{x^2}}+1) = (\lim_{x\to \infty}x)2 = \lim_{x\to \infty} 2x,$$
but I am not allowed to do the first equality above because $$\lim_{x\to \infty}x$$ does not exist, right? Thanks for your input.

2. Sep 20, 2015

### andrewkirk

I think this is easiest to solve by just using the definition of a limit. You need to show that, for any $\epsilon>0$, you can find $N>0$ such that $x>N\Rightarrow |f(x)-0|<\epsilon$ where $f(x)$ is the function of $x$ in your first line.

It is simple algebra with inequalities to work out how big $N$ needs to be for the inequality to be satisfied.

3. Sep 20, 2015

### andrewkirk

Actually, you can do that line you were worried about:

$$\lim_{x\to \infty} \frac{1}{\sqrt{x^2+1}+x} = \lim_{x\to \infty} \frac{1}{x(\sqrt{1+\frac{1}{x^2}} + 1)} =\lim_{x\to \infty} \frac{1}{2x} = 0,$$

The second equality is not justifiable, but you can avoid that if you use the fact that the limit of a product is equal to the product of limts, if both limits exist. Work with the fraction $\frac{1}{x(\sqrt{1+\frac{1}{x^2}} + 1)}$ by factorising it into two factors, each of which has an easy limit, rather than with the denominator as you tried above, which led you towards illegality.

4. Sep 20, 2015

### Zaculus

Ah yes, I can break it up into $\frac{1}{x}$ and $\frac{1}{(\sqrt{1+\frac{1}{x^2}} + 1)}$ the former having a limit of 0 and the latter having a limit of 1/2. In this way, I can "leave the former limit alone," obtain 1/2 from the second one, and then recombine it back into $\frac{1}{2x}$.

My main motive for asking this question was that I am a TA for a calc one class, and in the process of grading the quizzes, students did something like this in their solutions, but it is hard to ascertain whether they did it the illegal way in my original post or the more justifiable way in yours. Perhaps I give them the benefit of the doubt? I know there's not a right answer to this, but I'm trying to be consistent since there are other similar illegal things a few of done to obtain 0.

5. Sep 21, 2015

### HallsofIvy

Staff Emeritus
I am puzzled by this. You say you are a TA for a Calculus class but you also appear to be saying that you do not know the basic laws for limits?
They are:
1) If $lim_{x\to a} f(x)= F$ and $\lim_{x\to a} g(x)= G$ then $\lim_{x\to a} (f+ g)(x)= F+ G.$
2) If $/lim_{x\to a} f(x)= F$ and $\lim_{x\to a} g(x)= G$ then $\lim_{x\to a} (fg)(x)= FG.$
3) If $/lim_{x\to a} f(x)= F$ and $\lim_{x\to a} g(x)= G$ then $\lim_{x\to a} (f/g)(x)= F/G$ provided $G\ne 0$.
4) If $/lim_{x\to a} f(x)= F$ and there exist some $\epsilon>0$ such that f(x)= g(x) for $0< |x- a|< \epsilon$ then $\lim_{x\to a} g(x)= F.$

You are using #2.

Last edited by a moderator: Sep 21, 2015
6. Sep 21, 2015

### Zaculus

Yes I do know them. For whatever reason, I decided to try and ask the question as if I were a student to somehow gain some hints about to explain things to them. I'm still new to the process, but perhaps I should be more up front in my motives from the start.