How do I solve this infinite limit with the help of Taylor/Maclaurin series expansion?

Question

I'm currently stuck on solving this limit in my calculus course.

$$\lim_{x \to \infty } x(\sqrt[3]{x^3+3x^2}-\sqrt{x^2+2x-3})$$

We are currently reading about Taylor and Maclaurin series, so I'm assuming that the solution should be based on that. I've tried to use the standard expressions $(1+x)^{\alpha}$ and also tried to simplify and re-write the expression within the root-signs, but to no success. If anyone has the time to help me with this I would be most grateful!

Thanks in advance!

Answer 1

As $x \to \infty$, \begin{align} (x^3+3x^2)^{1/3} &= (x^3)^{1/3}\left(1+\frac{3}{x}\right)^{1/3} = x\left(1+\frac{1}{x}-\frac{1}{x^2} +O(x^{-3})\right) \\ &= x+1-\frac{1}{x} +O(x^{-2}) \\ \left(x^2+2x-3\right)^{1/2} &= (x^2)^{1/2}\left(1+\frac{2}{x}-\frac{3}{x^2}\right)^{1/2} = x\left(1+\frac{1}{x}-\frac{2}{x^2}+O(x^{-3})\right) \\ &=x+1-\frac{2}{x}+O(x^{-2}) \\ (x^3+3x^2)^{1/3}&-\left(x^2+2x-3\right)^{1/2} = \frac{1}{x}+O(x^{-2}) \\ x\big((x^3+3x^2)^{1/3}&-\left(x^2+2x-3\right)^{1/2}\big) = 1+O(x^{-1}) \\ \lim_{x\to\infty}x\big((x^3+3x^2)^{1/3}&-\left(x^2+2x-3\right)^{1/2}\big) =1 . \end{align}

I used the two Maclaurin series $$ (1+t)^{1/3} = 1+\frac{1}{3}t -\frac{1}{9}t^2+\dots \\ (1+2t-3t^3)^{1/2} = 1 +t-2t^2+\dots $$

Answer 2

If $x>0$, then$$x\left(\sqrt[3]{x^3+3x^2}-\sqrt{x^2+2x-3}\right)=x^2\left(\sqrt[3]{1+\frac3x}-\sqrt{1+\frac2x-\frac3{x^2}}\right).$$Now, let $f(x)=\sqrt[3]{1+3x}-\sqrt{1+2x-3x^2}$ ($x>0$). Then your limit is equal to $\lim_{x\to0^+}\frac{f(x)}{x^2}$. But $f(0)=f'(0)=0$ and $f''(0)=2$. Therefore, the Taylor polynomial of order $2$ of $f$ at $0$ is $P_{f,2,0}(x)=x^2\left(=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2\right)$. But a general property of Taylor polynomials is this: if $P_{g,n,a}$ is the Taylor polynomial of order $n$ of the function $g$ at $a$, then$$\lim_{x\to a}\frac{g(x)-P_{g,n,a}(x)}{(x-a)^n}=0.$$In particular,$$\lim_{x\to0}\frac{f(x)-x^2}{x^2}=0.$$But this is the same thing as asserting that $\lim_{x\to0}\frac{f(x)}{x^2}=1$. In other words, your limit is equal to $1$.

Answer 3

You can expand your limit around $x \to \infty$ using a laurent series. This gives you the following.

$$\lim_{x \to \infty } x(\sqrt[3]{x^3+3x^2}-\sqrt{x^2+2x-3}) = \lim_{x \to \infty } (1-\frac{1}{3x}+\frac{2}{3x^2}-\frac{2}{3x^3}+\frac{8}{9x^4}+...)$$

Of course this limit is much simpler to solve as you can decompose it into limits as

$$\lim_{x \to \infty } (1)-\lim_{x \to \infty } (\frac{1}{3x})+\lim_{x \to \infty } (\frac{2}{3x^2})-\lim_{x \to \infty } (\frac{2}{3x^3})+\lim_{x \to \infty } (\frac{8}{9x^4})+...$$

They all converge to $0$ except for the first one of course

$$\lim_{x \to \infty } (1) = 1$$