Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$

Question

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)

Answer 1

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.

Answer 2

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}