# Limit Problem

Homework Helper
I just read the following fact and it somewhat defies my usual logic.

$$\lim_{x\to\infty} \sqrt{x^2+x} - x =\frac{1}{2}$$.

Now my usual logic tells me that as x becomes large, the leading term in a polynomial sequence is the major one and therefore the $x^2+x$ becomes $x^2$ so $\sqrt{x^2+x}$ becomes just x.

The same logic seems to work for this limit here:https://www.physicsforums.com/showthread.php?t=166906

However the limit is obviously not equal to zero, where is the error in this logic?

divide top and bottom by x after you get rid of the radical on the top

Homework Helper
I know perfectly well how to evaluate the limit that way, Multiply by its conjugate, use difference of two squares and simplify, so on so forth. What I wanted to know was why the other method does not work.

I just read the following fact and it somewhat defies my usual logic.

$$\lim_{x\to\infty} \sqrt{x^2+x} - x =\frac{1}{2}$$.

Now my usual logic tells me that as x becomes large, the leading term in a polynomial sequence is the major one and therefore the $x^2+x$ becomes $x^2$ so $\sqrt{x^2+x}$ becomes just x.

Im sure your logic will tell you that
$$\sqrt{x^2+x}$$
is the same as
$$\sqrt{x^2+x + \frac{1}{4} }$$
when x becomes large.
And this is the same as
$$x + \frac{1}{2}$$ .

mjsd
Homework Helper
the logic you have just mentioned doesn't work either
$$\lim_{x\rightarrow\infty} \sqrt{x^2+x+1} - \sqrt{x^2-3x}$$ gives you zero too if you blindly replaces all polynomials with just the term of highest degree. what makes the eventual "substitution" to work in that other example was not that you replaces everything using that logic of yours ... but to look at "limit" as x goes large, namely:

$$\lim_{x\rightarrow\infty} \sqrt{x^2+x+1} - \sqrt{x^2-3x} = \lim_{x\rightarrow\infty} \frac{4x+1}{\sqrt{x^2+x+1} + \sqrt{x^2-3x}}$$
then divide top and bottom by x to get

$$\lim_{x\rightarrow\infty} \frac{4+1/x}{\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}} + \sqrt{\frac{x^2}{x^2}-\frac{3x}{x^2}}}$$

then if you now take the limit, you get what you want.

the lesson here is that $$\lim_{x\rightarrow \infty}\sqrt{x^2+x}\neq \sqrt{x^2}$$

It does work, you can solve this by inspection. get the conjugate and and the denominator the highest order variable is x^2, ignore the others. Walla, same at the other thread.

Homework Helper
Im sure your logic will tell you that
$$\sqrt{x^2+x}$$
is the same as
$$\sqrt{x^2+x + \frac{1}{4} }$$
when x becomes large.
And this is the same as
$$x + \frac{1}{2}$$ .

Yes it does. In fact, I usually complete the square when questions come up like that.

mjsd - But look at Hurkyl's comment in that thread...Sure he was not replacing the polynomials with the leading term right from the start, but he did on an equivalent expression! The fact the he couldn't do it at the start but can on an equivalent expression confuses me even more!

O and completeing the square works for that limit as well.

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Homework Helper
It does work, you can solve this by inspection. get the conjugate and and the denominator the highest order variable is x^2, ignore the others. Walla, same at the other thread.

Yes exactly, so it does work once we have changed the form. How come it works like that, and what form does it need to be in for it to work?

Yes exactly, so it does work once we have changed the form. How come it works like that, and what form does it need to be in for it to work?

remember when you were first learning limits? They say you will get a fraction answer if you only focused on highest order and if the order are some on top and bottom? Are the orders same on top and bottom? you sould say yes. because limit approaches zero on all others if you rationalize the top and bottom. Also, you cannot directly evauate that limit, so you have to change its idenitity to legally find it limit that is in disguise.

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Homework Helper
Ahh I see. So for this method to work, the degree of the numerator must be the same as the denominator. Ok thanks guys!

Hurkyl
Staff Emeritus
Gold Member
As a zeroth-order approximation, $\sqrt{x^2 + x}$ is equal to x, plus some terms less important than x.

Therefore, $\sqrt{x+2} - x$ is equal to 0, plus some terms less important than x.

Alas, this doesn't tell us the limit! So, we try a first-order approximation.

$\sqrt{x^2 + x} \approx x + 1/2$, plus some terms less important than 1. (using the fact $(A + \epsilon)^n \approx A^n + nA^{n-1} \epsilon$, plus terms less important than $A^{n-1}\epsilon$)

Therefore, $\sqrt{x+2} - x \approx 1/2$, plus some terms less important than 1.

Happily, terms less important than 1 don't matter here, so the answer is 1/2. (It would matter if, say, we had a fraction where both the numerator and denominator were zero + terms less important than 1)

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Homework Helper
Where are you getting these nth order approximations from? What would the 2nd order approximation be?

Hurkyl
Staff Emeritus
Gold Member
Where are you getting these nth order approximations from? What would the 2nd order approximation be?
Here, I'm using the binomial theorem:
$$(A + \epsilon)^n = A^n + n A^{n-1} \epsilon + \frac{n(n-1)}{2} A^{n-2} \epsilon^2 + \cdots$$
including the fact that if A dominates $\epsilon$, then each term in this series dominates the tail.

In general, you would use a differential approximation, or a Taylor series if you needed even more terms.

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hey Hurkle what class did you get that theorem from? I only have limited knowlege of only Calc, so I use the obvious method

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Homework Helper
Umm ok but how did we know the the first order approximation didn't give us the limit and that its zero. We did you know it wasnt zero and you had to get a better approximation?

Hurkyl
Staff Emeritus
Gold Member
Umm ok but how did we know the the first order approximation didn't give us the limit and that its zero. We did you know it wasnt zero and you had to get a better approximation?

The error in the approximation $\sqrt{x^2 + x} \approx x$ is asymptotically less than x. Actually, because it's differentiable, we can do better: we know it's asymptotically on the order of 1 or smaller.

So, we know the error in the approximation $\sqrt{x^2 + x} - x \approx 0$ is on the order of 1. So, all this tells us is when x gets big, $\sqrt{x^2 + x} - x$ remains bounded.

The error in the approximation $\sqrt{x^2 + x} \approx x + 1/2$ is asymptotically less than 1. Actually, because it's differentiable, we can do better: we know it's asymptotically on the order of 1/x or smaller.

So, we know the error in the approximation $\sqrt{x^2 + x} - x \approx 1/2$ is on the order of 1/x. So we know the limit is 1/2 as x gets big.

Note that if we were instead looking at

$$\lim_{x \rightarrow +\infty} \sqrt{4x^2 + x} - x$$

we only need the zeroth-order approximation: $\sqrt{4x^2 + x} - x \approx 2x - x = x[/tex]. The error is still asymptotically on the order of 1, so x dominates, and we know the limit is [itex]+\infty$.

Hurkyl
Staff Emeritus
Gold Member
The error in the approximation $\sqrt{x^2 + x} \approx x$ is asymptotically less than x. Actually, because it's differentiable, we can do better: we know it's asymptotically on the order of 1 or smaller.
Formally, this means
$$\exists c: \forall x:|\sqrt{x^2 + x} - x| < c \cdot 1$$
Or, if you know big-Oh notation,
$$\sqrt{x^2 + x} = x + O(1)$$

The less strict condition is simpler:
$$\lim_{x \rightarrow +\infty} \frac{\sqrt{x^2 + x} - x}{x} = 0$$
or
$$\sqrt{x^2 + x} = x + o(x)$$

The error in the approximation $\sqrt{x^2 + x} \approx x + 1/2$ is asymptotically less than 1. Actually, because it's differentiable, we can do better: we know it's asymptotically on the order of 1/x or smaller.
And this means
$$\exists c: \forall x:|\sqrt{x^2 + x} - (x + 1/2)| < c \cdot (1/x)$$
or
$$\sqrt{x^2 + x} = x + 1/2 + O(1/x)$$

The less strict one is
$$\lim_{x \rightarrow +\infty} \frac{\sqrt{x^2 + x} - (x + 1/2)}{1} = 0$$
or
$$\sqrt{x^2 + x} = x + 1/2 + o(1)$$

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You seem like you got the proofs down packed pretty good Hurkle. Are you a professor? Where did you learn to master proofs? Advanced Calc?

Homework Helper
I'm sorry If I seem slow, but what do you mean by asymptotically on the order of?

Hurkyl
Staff Emeritus
Gold Member
The intuition from asymptotic analysis (also given by nonstandard analysis) is that by asking for the limit, you want the value of the function "rounded" to the nearest constant. (or to the nearest standard number, in the nonstandard picture)

The error in $\sqrt{x^2 + x} \approx x$ is on the order of a constant, which is not enough precision for the answer we desire.

The error in $\sqrt{x^2 + x} \approx x + 1/2$ is on the order of 1/x, which is enough precision to get the desired answer.

Homework Helper
Why is the error in $\sqrt{x^2 + x} \approx x + 1/2$ on the order of 1/x, how do we know that? And why is that enough precision for the desired answer but the order of a constant is not?

Hurkyl
Staff Emeritus
Gold Member
And why is that enough precision for the desired answer but the order of a constant is not?
That one's easy: if I told you

"the value of that function is zero, plus some function of x that's bounded by c"

would you be able to figure out the limit? No. The best you could say is that, if the limit exists, it's somewhere between -c and c.

But if I told you

"the value of that function is 1/2, plus some function of x that's bounded by c/x"

would you be able to find the limit? Yes, because the limit of that error term is zero.

Homework Helper
Thank you soo soo very much, I understand now. Ill read over the posts again tomorrow to see if I get it, otherwise Ill post here again. Thanks!

Homework Helper
Ok I re read it and I need help again :(

As To Post 21, you answered the 2nd bit but not the 1st, which im still completely lost on. How do we know what order the error is on for a certain approximation, and how does it being differentiable help us do better..

Hurkyl
Staff Emeritus
Gold Member
The case where x goes to 0 is easier to describe, I think.

Suppose we make the approximation f(x) ~ f(0). The error term E(x) is given by
f(x) = f(0) + E(x)

Exercise 1: Prove that if f is continuous, then $\lim_{x \rightarrow 0} E(x) = 0$.

Exercise 2: Prove that if f is differentiable, and L > 0, then there exists a C such that whenever $|x| < L$, we have $|E(x)| < C|x|$.

Suppose f is differentiable, and we make the approximation f(x) ~ f(0) + x f'(0). The error term E(x) is given by
f(x) = f(0) + x f'(0) + E(x)

Exericse 3: Prove that $\lim_{x \rightarrow 0} E(x) / x = 0$.

Exercise 4: Figure out what the next exercise should be.

You can keep iterating to higher derivatives, and you're essentially led to [URL [Broken] theorem[/url] to cover the general case.

Exercise 5a: Suppose f is continuous and strictly positive, g is continuous, and $\lim_{x \rightarrow +\infty} g(x) / f(x) = 0$. Consider the approximation $\left( f(x) + g(x) \right)^n \approx f(x)^n$. Using exercise 1, show that $\lim_{x \rightarrow +\infty} E(x) / f(x)^n = 0$.

Exercise 5b: If the f and g are also differentiable, use exercise 2 to show there is an L and a C such that for all x > L, $E(x) < C |f(x)^{n-1} g(x)|$.

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