Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$ 
Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$ from the definition (using $\epsilon-\delta$)

Why can't I do something like this?
We want:
$|\frac{x^4-2x+1}{x-1} + \sqrt{x}-3| = |\frac{x^4-2x+1}{x-1} + \frac{x-9}{\sqrt{x}+3}| \le |\frac{x^4-2x+1}{x-1}| + |\frac{x-9}{\sqrt{x}+3}| \le |x^3+x^2+x-1| + |x-9| \le |x|^3 + |x|^2+|x-1| + |x-9| < \epsilon$
Now suppose $|x-1| < 1$, then $|x|\le 2$ and $|x-9| \le 9$, so 
$$|x|^3 + |x|^2+|x-1| + |x-9| < 2^3+2^2+9+|x-1| < \epsilon$$
Now take $\delta = \min\{1, \epsilon -21\}$. 
But this clearly is wrong because $\delta$ needs to be positive for every $\epsilon>0$, what part of my working is incorrect? (I know how to get the answer properly, just wondering why this way is incorrect, I'm quite new to $\epsilon-\delta$ proofs)

Note: This answer was posted before the condition of using the definition with $(\epsilon-\delta)$ was specified.
Factorize $x^4-2x+1$, to remove the singularity in the denominator:
\begin{aligned}
\lim_{x\to1}\frac{x^4-2x+1}{x-1} + \sqrt{x} & = \lim_{x\to1}\frac{(x-1)(x^3+x^2+x-1)}{x-1} + \sqrt{x} = \\
 & = \lim_{x\to1}(x^3+x^2+x-1)  + \sqrt{x} = \\
 & =  3
\end{aligned}
