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

My Try :
$$\sqrt{x^4+x+1}=\sqrt{x^4(1+\frac{1}{x^3}+\frac{1}{x^4})}=x^2\sqrt{(1+\frac{1}{x^3}+\frac{1}{x^4})}$$
Now : $$\frac{1}{x^3}+\frac{1}{x^4}=z$$
$$(1+z)^{\frac{1}{n}}= 1 + \frac1n x + \frac{1 - n}{2n^2}x^2 + \frac{2n^2 - 3n + 1}{6n^3}x^3 + O(x^4)$$
$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac12 z - \frac{1}{8}z^2 + + O(z^3)$$
$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac12 (\frac{1}{x^3}+\frac{1}{x^4}) - \frac{1}{8}(\frac{1}{x^3}+\frac{1}{x^4})^2 + O(z^3)$$
$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac{1}{2x^3}+\frac{1}{2x^4} - \frac{x^2+2x+1}{8x^8} + O(z^3)$$
And :
$$\sqrt{x^4+x+1}=\sqrt{x^4(1+\frac{1}{x^3}+\frac{5}{x^4})}=x^2\sqrt{(1+\frac{1}{x^3}+\frac{5}{x^4})}$$
So :
$$(1+(\frac{1}{x^3}+\frac{5}{x^4}))^{\frac{1}{2}}= 1 +  \frac{1}{2x^3}+\frac{5}{2x^4} -  \frac{x^2+10x+25}{8x^8}+ O(z^3)$$
$$\lim\limits_{x \to +\infty }=x^4( 1 + \frac{1}{2x^3}+\frac{1}{2x^4} - \frac{x^2+2x+1}{8x^8} + O(z^3)-( 1 +  \frac{1}{2x^3}+\frac{5}{2x^4} -  \frac{x^2+10x+25}{8x^8}+ O(z^3)))$$
$$\lim\limits_{x \to +\infty }=x^4(\frac{1}{2x^4}-\frac{5}{2x^4})=x^4(\frac{-4}{2x^4})=-2$$
is it right ?
 A: Write
\begin{align}
\sqrt{x^4+x+1}-\sqrt{x^4+x+5}
=\frac{(\sqrt{x^4+x+1}-\sqrt{x^4+x+5})(\sqrt{x^4+x+1}+\sqrt{x^4+x+5})}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}
\end{align}
and use
\begin{align}
\frac{1}{2\sqrt{x^4+x+5}}\leq\frac{1}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}\leq \frac{1}{2\sqrt{x^4+x+1}} 
\end{align}
and
\begin{align}
\sqrt{x^4+x+1}&=\sqrt{x^4(1+x^{-3}+x^{-4})}=x^2\sqrt{1+x^{-3}+x^{-4}}\\
\sqrt{x^4+x+5}&=\sqrt{x^4(1+x^{-3}+5x^{-4})}=x^2\sqrt{1+x^{-3}+5x^{-4}}
\end{align}
A: $$
\begin{aligned}
\lim _{x\to \infty }\left(x^2\:\left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)\right)
& = \lim _{x\to \infty }\left(-\frac{4x^2}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}\right)
\\& = \lim _{x\to \infty }\left(-\frac{4x^2}{x^2\sqrt{1+\frac{x}{x^4}+\frac{1}{x^4}}+x^2\sqrt{1+\frac{x}{x^4}+\frac{5}{x^4}}}\right)
\\& \approx_\infty \lim _{x\to \infty }\left(-\frac{4x^2}{2x^2}\right)
\\& = \color{red}{-2}
\end{aligned}
$$
In this case you can solve it with just an approximation without using the Taylor expansion, anyway your procedure is right.
