<s>[INST] Epsilon Delta proof for rational function containing radicals I am having trouble constructing the delta-epsilon proof for this limit. I am currently trying to find the delta in terms of epsilon:
$$\lim_{x \to 1} \frac{x^3-1}{\sqrt{x}-1}$$
So far I have gotten to
$$6-\epsilon < (x-1)*\frac{x^2+x+1}{\sqrt{x}-1} < 6+\epsilon$$
I am having trouble bounding the fraction $\frac{x^2+x+1}{\sqrt{x}-1}$ so that I can isolate $x-1$ and find the delta. It has a vertical asymptote at x=1, which makes it difficult to find an upper/lower bound.
Am I on the right track to this problem? If not, how would I go about finding the delta?
 [/INST] We have that for $x\neq 1$
$$\frac{x^3-1}{\sqrt{x}-1}=\frac{(x-1)(x^2+x+1)}{\sqrt{x}-1}=$$
$$=\frac{(\sqrt{x}-1)(\sqrt{x}+1)(x^2+x+1)}{\sqrt{x}-1}=(\sqrt{x}+1)(x^2+x+1)$$
and since for $|x-1|<1 \iff 0<x<2$, using triangle inequality and that $\forall a,b \:a\ge b\ge 0 \implies |x^a-1|\le |x^b-1|$, we obtain
$$\left|\frac{x^3-1}{\sqrt{x}-1}-6\right|=\left|(\sqrt{x}+1)(x^2+x+1)-6\right|\le\left|x^2\sqrt{x}+x^2+x\sqrt x+x+\sqrt{x}-5\right|\le$$
$$\leq |x^2\sqrt{x} -1|+\ldots +|\sqrt{x}-1| \leq 5|x^3 - 1| \le 5|x-1||x^2+x+1|\le 35|x-1|<\varepsilon$$
and it suffices to assume $\delta =\min\left(1,\frac{\varepsilon}{35}\right)$.
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