MHB Limits of Complex Expressions: Infinity or 0?

[Dethrone]

1.
$$=\lim_{{x}\to{\infty}} (x^3+\sqrt{x^6+x^3+1}$$
Multiplying top and bottom by $$x^3-\sqrt{x^6+x^3+1}$$, we get:
$$=\lim_{{x}\to{\infty}} \frac{x^3+1}{x^3-\sqrt{x^6+x^3+1}}$$
Dividing by the highest power:
$$=\lim_{{x}\to{\infty}} \frac{1+\frac{1}{x^3}}{1-\sqrt{1+\frac{1}{x^3}}}$$
$$=\lim_{{x}\to{\infty}} \frac{1+\frac{1}{x^3}}{1-\sqrt{1+\frac{1}{x^3}+\frac{1}{x^6}}}$$

The answer is still undefined, how can I proceed?

2.
$$=\lim_{{x}\to{\infty}}\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}$$
Factoring out an $x^3$
$$=\lim_{{x}\to{\infty}}x(\sqrt[3]{1+\frac{1}{x}}-\sqrt[3]{1-\frac{1}{x}}) $$

What do I do now? Wolfram Alpha says the answer is infinity, but a student-written answer key says it is 0, because "the x outside the brackets goes to infinity, but the brackets will put it back to zero, eventually winning". That sounds like a very bad reason, but any ideas?

 

[Opalg]

1.
$$=\lim_{{x}\to{\infty}} \bigl(x^3+\sqrt{x^6+x^3+1}\bigr)$$

You don't need to multiply top and bottom by anything here. Each term is at least as large as $x^3$, so their sum gets very large as $x$ increases ... .

2.
$$=\lim_{{x}\to{\infty}}\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}$$
Factoring out an $x^3$
$$=\lim_{{x}\to{\infty}}x\Bigl(\sqrt[3]{1+\frac{1}{x}}-\sqrt[3]{1-\frac{1}{x}}\Bigr) $$

Good start. Now use the binomial expansion $(1+s)^{1/ 3} = 1 + \frac13s +$ (higher powers of $s$), taking $s = \frac1x$ and then $-\frac1x$. You should find that the answer is neither $0$ nor infinity.

 

[Dethrone]

I've never formally learned binomial expansion. By what I've researched, does the expansion have an infinite number of terms for non-integer exponent values? If so, I will only need to do the first few terms, right? What is the formula that will give me the terms? And is this in any way related to taylor series expansion, or is it the taylor series expansion?

Anyways, in the meantime, I'll be using the series expansion as given by Wolfram Alpha, but I do want to learn how to get them after.

$$\displaystyle =\lim_{{x}\to{\infty}}x\Bigl([1+\frac{1}{3x}+\frac{1}{9x^2}...]-[1-\frac{1}{3x}-\frac{1}{9x^2}...]\Bigr)$$
$$\displaystyle =\lim_{{x}\to{\infty}}[x+\frac{1}{3}+\frac{1}{9x}...]-[x-\frac{1}{3}-\frac{1}{9x}...]$$
$$=\frac{2}{3}$$

Is that correct?

Spoiler

This question is from a Calc 1 exam... If it involves series expansion, I guess this is a really mean question. If anyone's curious as to where I got this question from: http://skule.ca/courses/exams/MAT194H1_20139_6313949059622013_final.pdf

 

[Opalg]

The answer $2 /3$ is correct. Another way to do it, without using Newton's binomial theorem, is to rationalise the expression. Just as you can sometimes simplify $\sqrt a - \sqrt b $ by writing it as $\dfrac{(\sqrt a - \sqrt b)(\sqrt a + \sqrt b)}{\sqrt a + \sqrt b} = \dfrac{a-b}{\sqrt a + \sqrt b}$, so you can do the same sort of thing with cube roots. In fact, $$ a^{1 /3} - b^{1 /3} = \frac{(a^{1 /3} - b^{1 /3})(a^{2 /3} + a^{1 /3}b^{1 /3} + b^{2 /3})}{a^{2 /3} + a^{1 /3}b^{1 /3} + b^{2 /3}} = \frac{a-b}{a^{2 /3} + a^{1 /3}b^{1 /3} + b^{2 /3}}.$$ That trick will get you the answer $2 /3$ here, and it's probably the method that was intended in that exam.

 

[Dethrone]

Great! Was the binomial expansion that you were referring to a taylor series expansion?

 

[Opalg]

Was the binomial expansion that you were referring to a taylor series expansion?

Yes, the binomial expansion of $(1+x)^r$ is the same as its Taylor series.