Evaluting the limit $\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}$ I'm attempting to evaluate the limit
$\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}$
I got it reduced to the following
$\lim_{x\rightarrow\infty}\frac{\sqrt{\frac{1}{\left(x-2\right)^{2}}-\frac{3}{\left(x-2\right)^{4}}}+1}{1-\frac{3}{\left(x-2\right)^{2}}-1}$
But putting in $\infty$ I get $\frac{1}{0}$ and, what's worse, Mathematica tells me the limit is equal to $-\infty$. Where am I going wrong?
 A: Hint: multiply top and bottom by $\sqrt{x^2-4x+1}+x-2$.
Additional hint upon request: $\lim_{x\rightarrow \infty}\sqrt{x^2-4x+1}=\infty$ and $\lim_{x\rightarrow \infty} x-2=\infty$.
A: \begin{align*}
&\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}\\
=&\lim_{x\rightarrow\infty}\frac{1}{\sqrt{(x-2)^2-3}-(x-2)}\\
=&\lim_{x-2\rightarrow\infty}\frac{1}{\sqrt{(x-2)^2-3}-(x-2)}\\
=&\lim_{z\rightarrow\infty}\frac{1}{\sqrt{z^2-3}-z}\\
=&\lim_{z\rightarrow\infty}\frac{\frac{1}{z}}{\sqrt{1-\frac{3}{z^2}}-1}
\end{align*}
Note that the absolute value of the last expression is $\infty$ and the denominator is negative as $\sqrt{1-\frac{3}{z^2}}<1$.
A: $$F=\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}$$
$$=\lim_{x\rightarrow\infty}\frac{1}{\sqrt{(x-2)^2-3}-x+2}$$
Let us put $x-2=\sqrt3\csc2\theta$
$x\to\infty\implies \theta\to0$
So, $$F=\lim_{\theta\to0}\frac1{\sqrt3\cot2\theta-(\sqrt3\csc2\theta+2)+2}$$
$$=-\frac1{\sqrt3}\lim_{\theta\to0}\frac{\sin2\theta}{1-\cos2\theta}$$
$$=-\frac1{\sqrt3}\lim_{\theta\to0}\frac{2\sin\theta\cos\theta}{2\sin^2\theta}$$ (using $\sin2\theta=2\sin\theta\cos\theta,\cos2\theta=1-2\sin^2\theta$)
$$=-\frac1{\sqrt3}\lim_{\theta\to0}\cot\theta$$ as $\theta\to0\implies \sin\theta\to0\implies \sin\theta\ne0$
What is $\cot0?$
A: Hint: multiply numerator and denominator by the denominator's conjugate:
$$\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}=\frac{\sqrt{x^2-4x+1}+x-2}{x^2-4x+1-x^2+4x-4}=\;\ldots$$
Added under request : 
$$\left(\sqrt{x^2-4x+1}+x-2\right)\frac{\frac1x}{\frac1x}=\frac{\sqrt{1-\frac4x+\frac1{x^2}}+1-\frac2x}{\frac1x}\xrightarrow[x\to\infty]{}\ldots$$
