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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?

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    $\begingroup$ They multiplied by 1. You multiplied the function to itself. $\endgroup$ – randomgirl Oct 26 '15 at 1:00
  • $\begingroup$ yes, it is like you found the square of the limit $\endgroup$ – Jane Oct 26 '15 at 1:01
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You can't just sqaure an expression like that, it'll give you a different value! For e.g. $lim_{x\to6}6x = 36$ but $lim_{x\to6}36x^2 = 1296$

  1. $\lim_{x\to\infty}{\sqrt{1+4x^6}\over 2-x^3}$
  2. Divide the numerator and the denominator by the highest degree term($x^3$ in this case)
  3. $\lim_{x\to\infty}{\sqrt{1/x^6 + 4}\over{(2/x^3) - 1}}$
  4. Now using properties of limits:
  5. $\lim_{x\to\infty}{\sqrt{1/x^6 + 4}\over{(2/x^3) - 1}}$ = ${\sqrt{\lim_{x\to\infty}1/x^6 + \lim_{x\to\infty} 4}\over{\lim_{x\to\infty}(2/x^3) - \lim_{x\to\infty}1}}$
  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) $\lim_{x\to\infty}{4} = 4$ and $\lim_{x\to\infty}{1} = 1$
  7. After taking these limits the rational expression becomes: $\sqrt{4\over1} = 2$

Hope this helps!

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

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