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?
Best Answer
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.