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I have a limit:

$$\lim_{x \to \infty} \left(\sqrt{x^4+5x^2+1} - x^2\right)$$

I know the answer is $\frac{5}{2}$, and you can get it by multiplying and dividing both sides by the conjugate ($\sqrt{x^4+5x^2+1} + x^2$) to get

$$ \lim_{x \to \infty} \left(\frac{x^4+5x^2+1-x^4}{\sqrt{x^4+5x^2+1} + x^2}\right) = \lim_{x \to \infty} \left(\frac{5x^2+1}{x^2 \sqrt{1+\frac{5}{x^2}+\frac{1}{x^4}} + x^2}\right) $$

It is then pretty easy to see that the stuff under the square root in the denominator will go to 1, and we will get $\frac{5x^2}{2x^2}$ = $\frac{5}{2}$. This seems to be the correct answer according to WolframAlpha as well. However, I got to thinking that there seems to be another completely legitimate way to do this limit, as follows:

$$ \lim_{x \to \infty} \left(\sqrt{x^4+5x^2+1} - x^2\right) = \lim_{x \to \infty} \left(x^2 \left(\sqrt{1+\frac{5}{x^2}+\frac{1}{x^4}}\right) - x^2\right) $$

Again, the stuff inside the square root will go to 1, so we will get

$$ = \lim_{x \to \infty} \left( x^2 - x^2\right ) = 0$$

Obviously this second way is wrong, but can someone explain why?

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"the stuff inside the square root will go to 1"? Really? – Byron Schmuland Feb 1 '14 at 4:05
As x goes to infinity, 5/(x^2) goes to 0, and 1/(x^4) goes to 0, so my thought was that the limit of the whole square root would be 1 – Chad Russell Feb 1 '14 at 4:07
"x goes to infinity"?? – Byron Schmuland Feb 1 '14 at 4:07
whoops, I meant all the limits to go to infinity, not 0! I'll edit it, sorry – Chad Russell Feb 1 '14 at 4:08
up vote 6 down vote accepted

You made a step which does not follow from your previous equations. For example, $$1=\lim_{x\to \infty} (x^2+1-x^2)=\lim_{x\to \infty} (x^2(1+x^{-2})-x^2)\neq \lim_{x\to \infty}(x^2-x^2)=0$$

This shows straightforwardly that such a manipulation isn't valid without extra justification.

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No, it is $\infty-\infty$ that is (relevantly) indeterminate here. A limit that gives rise to $\infty(1-\frac1\infty)$ goes (determinately) to$~\infty\times1=\infty$, but that is only what happens to the first additive operand. – Marc van Leeuwen Feb 1 '14 at 15:14
@MarcvanLeeuwen You're completely right. I wrote way too quickly. Don't know how that slipped by me. – NeuroFuzzy Feb 1 '14 at 16:29

Writing $$ \lim_{x \to \infty} \left(\sqrt{x^4+5x^2+1} - x^2\right) = \lim_{x \to \infty} \left(x^2 \left(\sqrt{1+\frac{5}{x^2}+\frac{1}{x^4}}\right) - x^2\right) $$ is fine, but you cannot ignore what is inside the radical.Remember that, for small value of $y$, $\sqrt{1+y}$ has a Taylor expansion which is $1+\frac{y}{2}+O\left(y^2\right)$. If we ignore the third term is the radical (wich is much smaller that the second), we then have $$\sqrt{\frac{1}{x^4}+\frac{5}{x^2}+1}=1+\frac{5}{2 x^2}+O\left(\left(\frac{1}{x}\right)^4\right)$$
Expanding now, we find again the correct limit.

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The second method is wrong as it results in an indeterminate form. If I write the last term as \begin{equation} x^2\left(\sqrt{1 + \frac{5}{x^2} + \frac{1}{x^4}} - 1\right). \end{equation} After putting the limiting value of $x$ I get $x^2$((a term that tends to $1$ but is greater than $1)-1)$ or $ \infty $*(a number tending to $0)$. Hence an indeterminate form.

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