Epsilon Delta proof for rational function containing radicals

Question

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?

Answer 1

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 $\delta =\min\left(1,\frac{\varepsilon}{35}\right)$.

Answer 2

Since $x\to 1$ set $x=(1+u)^2$ where $u\to 0$

Then $f(x)=\dfrac{(1+u)^6-1}{|1+u|-1}$, since $u$ is small $|1+u|=1+u$ and this simplifies to

$$f(x)= 6+15u+20u^2+15u^3+6u^4+u^5$$

But $u$ being small means we can take $|u|<1$ which implies $|u|^n<|u|$ for any $n$.

Thus $$|f(x)-6|\le (15+20+15+6+1)|u|=57|u|$$

For $|u|<1$ we have $|u+2|>1$ so $|u|<|u(u+2)|=|u^2+2u|=|(1+u)^2-1|=|x-1|$

Just take $$\delta=\min(1,\frac {\varepsilon}{57})\quad\text{we get}\quad|u|<|x-1|<\delta\implies |f(x)-6|\le 57|u|<\varepsilon$$