Question about evaluating infinite limit, $\lim\limits_{x\to\infty}{\sqrt{1+4x^6}\over 2-x^3}.$

Question

I have to evaluate an positive infinite limit for $$\lim_{x\to\infty}{\sqrt{1+4x^6}\over 2-x^3}.$$ I did it my way by squaring the whole thing, which gets rid of the square root, then I just foil the denominator and divide every term by $x^6$ and my limit was $4$. However, I got the wrong answer and the solution manual did it by dividing the numerator and denominator by $x^3$ and getting a limit of 2. I don't see the reason I didn't get the same answer as the manual when I did to the bottom what I did to the top. Can someone please explain where the mistake between my answer and the manual?

Answer 1

You can't just square an expression like that, it'll give you a different value! For e.g. $\,\lim\limits_{x\to6}6x = 36\,$ but $\,\lim\limits_{x\to6}36x^2 = 1296\,.$

1. $\displaystyle\lim\limits_{x\to\infty}{\sqrt{1+4x^6}\over 2-x^3}$
2. Divide the numerator and the denominator by the highest degree term($x^3$ in this case)
3. $\displaystyle\lim\limits_{x\to\infty}{\sqrt{1/x^6 + 4}\over{(2/x^3) - 1}}$
4. Now using properties of limits:
5. $\displaystyle\lim\limits_{x\to\infty}{\sqrt{1/x^6 + 4}\over{(2/x^3) - 1}}={\sqrt{\lim\limits_{x\to\infty}1/x^6 + \lim\limits_{x\to\infty} 4}\over{\lim\limits_{x\to\infty}(2/x^3) - \lim\limits_{x\to\infty}1}}$
6. As $x$ tends towards infinity, $1/x^6$ and $2/x^3$ tend towards zero and since the limit of a constant is the constant itself (since the constant does not change as $x$ changes) $\,\lim\limits_{x\to\infty}{4} = 4\,$ and $\,\lim\limits_{x\to\infty}{1} = 1$
7. After taking these limits the rational expression becomes: $\displaystyle\dfrac{\sqrt4}{-1}=-2$

Hope this helps!

Answer 2

Divide the top and bottom by $x^3$ to get $$\lim_{x\to\infty}{\sqrt{1+4x^6}\over 2-x^3} = {\sqrt{1/x^6 + 4}\over{(2/x^3) - 1}} = -2.$$