I hemmed and hawed over whether I should ask about this. The question seems clear enough, but it just seems far too easy. I'm worried I'm misunderstanding it.
I am confused about part b of this question in my textbook.
Part a
Use numerical and graphical evidence to guess the value of the limit $$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}$$
part b
How close to $1$ does $x$ have to be to ensure that the function in part a is within a distance $0.5$ of its limit?
So basically, as I understand it, $x$ would have to be in the range $(0.5, 1.5)$ to be within a distance of $0.5$ of $1$.
I must be misunderstanding this, right?
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$\begingroup$We know the limit is $6$.
The question is asking us to determine $\delta>0$ such that $0<|x-1|<\delta$ would ensure that
$$\left|\frac{x^3-1}{\sqrt{x}-1} -6\right| < \frac12$$
This picture might help you.
In my answer to your previous question, I did show that, at least locally,$$\frac{x^3-1}{\sqrt{x}-1}\sim 6+\frac{15}{2}(x-1)$$ So, you look for the range of$$\frac{15}{2} |x-1| \leq \frac 12\implies |x-1| \leq \frac 1{15}$$ Then, ???
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