<s>[INST] Two answers for the same limit 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? 
 [/INST] 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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