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

Question

When I had the calculus class about the limit, one of my classmate felt confused about this limit:

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

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

This result gets verified by wolframalpha.

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.

Answer 1

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 $p-1$.

Answer 2

@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...)

Answer 3

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: $\displaystyle\lim_{x \to \infty} \sqrt[n]{p(x)} - \sqrt[n]{q(x)} = a-b$.