# A question about substitute equivalent form into limit: $\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$

$$\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$$

What he thought that since $$x^2 > x$$ and $$x^2 > 3x$$ when $$x \to \infty$$ so the first square root must be $$x$$ and same for the second. Hence, the limit must be $$0$$.

It is obviously problematic.

And what I thought is that make prefect square under the limit, though I know the right solution is to rationalize the numerator.

After perfect-squaring, $$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}$$ I assert that since there is a perfect square and a square root. As $$x \to \infty$$, the constant does not matter. So

$$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}= \lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2}-\sqrt{\left(x+\frac{3}{2}\right)^2}\\ = \frac{1}{2}-\frac{3}{2} = -1$$

But when I conquer this limit: this post.

Here is my argument:

$$\sqrt{ax^2+bx+c} =O(\sqrt a(x+\frac{b}{2a}))$$ when $$x \to \infty$$

$$\lim_{x\to\infty}\frac{\sqrt{x^2-2x+3}+\sqrt{4x^2+5x-6}}{x+\sqrt{x^2-1}} =\frac{x-1+2(x+5/4)}{x+x} = \frac{3}{2}$$

Very concise get this answer but it gets downvoted.

I do not know what is wrong with my strategy. But in general case: for instance this problem about cubic root my strategy seems to work really efficient:

$$\lim_{x\to \infty} \sqrt[3]{x^3+6x^2+9x+1}-\sqrt[3]{x^3+5x^2+x+1}$$

My solution is:

$$\lim_{x\to \infty} \sqrt[3]{x^3+bx^2+cx+d} = \lim_{x\to \infty} \sqrt[3]{\left(x+\frac{b}{3}\right)^3}$$

So the limit becomes: $$\lim_{x\to \infty} \sqrt[3]{(x+2)^3+O(x)}-\sqrt[3]{\left(x+\frac{5}{3}\right)^3 + O(x)} =\lim_{x\to \infty} (x+2) -\left(x+\frac{5}{3}\right) = \frac{1}{3}$$

To put all into a nutshell, what is wrong with my solution to these three problem. Is there any counterexample to this substitution. Any help, you will be appreciated.

• I don't think anything is wrong, it's only that the down voting person has wanted more rigor in the argument. You cannot write $O(n)$ just like that, it will need justification. Furthermore, to show that the limit actually exists, you should show that there is an upper bound to the sequence. I would not expect this in general, which means this solution is fine for me, but somebody else may be put off by the argument's slightly loose nature. Aug 11 '16 at 0:50
• @астонвіллаолофмэллбэрг How to generalize my statement and prove it is fine to the mathematics axioms or theorem about real anlysis. This is the key. By the way, I cannot generalize it. Could you help me? Write it in the full solution. Aug 11 '16 at 0:53
• I will do that for the general case. Aug 11 '16 at 0:57
• The problem with your classmate's approach to the first limit is that each individual square root tends to infinity, and he is in effect saying that $\infty-\infty=0$. But you can never do arithmetic with $\infty$ in this way. Aug 11 '16 at 1:11
• @David yes but what problem is in my solution? Aug 11 '16 at 1:15

Consider the general case $$A=\sqrt{x^2+ax+b}-\sqrt{x^2+cx+d}=x\left(\sqrt{1+\frac a x+\frac b {x^2}}-\sqrt{1+\frac c x+\frac d {x^2}}\right)$$ and use the fact that, for small $y$, using the generalized binomial theorem or Taylor series, $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ For the first radical, replace $y$ by $\frac a x+\frac b {x^2}$ and by $\frac c x+\frac d {x^2}$ for the second radical. You then obtain $$A=x\left(1+\frac{a}{2 x}+\frac{\frac{b}{2}-\frac{a^2}{8}}{x^2}+O\left(\frac{1}{x^3}\right)-\left(1+\frac{c}{2 x}+\frac{\frac{d}{2}-\frac{c^2}{8}}{x^2}+O\left(\frac{1}{x^3}\right)\right)\right)$$ $$A=\frac{a-c}{2}+\frac{-a^2+4 b+c^2-4 d}{8 x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.

Doing the same with $$B=\sqrt[3]{x^3+ax^2+bx+c}-\sqrt[3]{x^3+dx^2+ex+f}$$ and using $$\sqrt[3]{1+y}=1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)$$ you should arrive to $$B=\frac{a-d}{3}+\frac{-a^2+3 b+d^2-3 e}{9 x}+O\left(\frac{1}{x^2}\right)$$ Doing the same with $$C=\sqrt[p]{x^p+a_1x^{p-1}+a_2x^{p-2}+\cdots}-\sqrt[p]{x^p+b_1x^{p-1}+b_2x^{p-2}+\cdots}$$ the limit will just be $$\frac{a_1-b_1} p$$

Your strategy works well for the limit because you just ignore the terms of degrees lower than $p-1$.

@ZackNi As David indicated, you are assuming "infinity minus infinity = 0,which works out well because both radicals are squareroots and the headcoefficients of the polynomials are both 1.You can also get"infinity minus infinity = 0 from a squareroot and cuberoot, but using your approach will then yield to a wrong answer to the limit. Generally, $\infty-\infty$ and $\infty*0$ situations should be converted into situations like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ on which other techniques can be applied to get the desired limit. In your problem the conjugate approach is also a good way to go. (My comment post got messed up and I can't delete it...)

• Yes I know that rationalize the numerator is the conjugate approach. But my approach always drive me to right answer so I really feel confused about what is wrong in my answer indeed. Aug 11 '16 at 1:54
• @ZackNi That is because the nature of the problems are as such. Here is food for thought: To what line does $y=x\sqrt{1+4/x}$ approach when $x$ goes to infinity?. One can argue that the square root goes to $1$ for large $x$ and so the asymptote would be $y=x$. Alas, that is incorrect. Think about it... Aug 11 '16 at 1:59
• do you know how to read critically. Here is my approach: $y = \sqrt{x^2+4x} = \sqrt{x^2+4x+4 -4} = \sqrt{(x+2)^2 - 4}$ So when $x \to \infty$ ,$y = x+2$. This is my solution! Aug 11 '16 at 2:07
• Yes, the solution is correct, but the step after "so when ....." is algebraically not that elegant.... Aug 11 '16 at 2:51
• The answer by our Indian friend (which you accepted) is really a good answer. I hope you will post some more limit problems in the future of the nature I mentioned. The comment section is too small to go into all sorts of examples. Good luck! Aug 11 '16 at 2:55

Let our problem be of the form $\displaystyle\lim_{x \to \infty} \sqrt[n]{p(x)} - \sqrt[n]{q(x)}$, where $p$ and $q$ are $n$th degree monic(coefficient of $x^n$ is $1$) polynomials. First of all, without loss of generality, we will rewrite $p$ and $q$ in the forms: $$p = (x + a)^n + \sum_{i=0}^{n-1} a_ix^i, q = (x + b)^n + \sum_{i=0}^{n-1} b_ix^i$$

where $a_i$ and $b_i$ are real constants which may be zero.(This we did in the cubic and quadratic case).

Now, note that $x+a \leq \sqrt[n]{(x + a)^n + \sum_{i=0}^{n-1} a_ix^i} \leq \sqrt[n]{(x + a)^n} + \sqrt[n]{\sum_{i=0}^{n-1} a_ix^i}$.Note that $$\sqrt[n]{\sum_{i=0}^{n-1} a_ix^i} \leq \sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}$$ However, the term on the left goes to $0$ as $x \to \infty$, because $i < n,$ so $x^i \to 0$. A similar logic follows for $q$. To complete the argument,

However, this is better expressed by: $x+a - ((x+b) + \sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}) \leq \sqrt[n]{p(x)} - \sqrt[n]{q(x)} \leq (x+a +\sum_{i=0}^{n-1} \sqrt[n]{a_ix^i}) - (x+b))$.

On taking limits on both sides, we get $a-b$ on both sides, and using squeeze theorem, we get the result: $\displaystyle\lim_{x \to \infty} \sqrt[n]{p(x)} - \sqrt[n]{q(x)} = a-b$.

• Thanks a lot really excellent solution. Aug 11 '16 at 2:01
• You are welcome, @ZackNi. Aug 11 '16 at 2:41
• Good answer indeed +1 Aug 11 '16 at 2:56
• How can you apply the squeeze theorem when the limit is for $x\to\infty$? The left side converges to $-\infty$ while the right side converges to $+\infty$, there is no squeeze. -- More precisely, you are actually claiming that $\lim_{x\to\infty}x^i=0$. Aug 11 '16 at 10:46
• How does the left and right side converge to $-\infty$ and $+\infty$? Because $\frac{i}{n} < 1$, it follows that $x^{\frac{i}{n}} \to 0$ as $x \to \infty$. Aug 13 '16 at 9:10