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?

\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 $\sqrt{1-\frac{3}{z^2}}<1$.
