<s>[INST] 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)
 [/INST] You went off the track when you introduced the term
$$\left|\frac{x-9}{\sqrt x+3}\right|\ .$$
As $x\to1$, this term will tend to $2$ and therefore it can't be made arbitrarily small.  Something like
$$\eqalign{\left|\frac{x^4-2x+1}{x-1}+\sqrt{x}-3\right|
  &\le\left|\frac{x^4-2x+1}{x-1}-2\right|+\left|\sqrt{x}-1\right|\cr
  &=\left|x-1\right|\left|x^2+2x+3\right|+\left|\frac{x-1}{\sqrt x+1}\right|\cr}$$
would have a better chance of working.
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