What delta do you specify in $\epsilon$-$\delta$ formula

Question

I'm not sure what delta to choose to satisfy this question:

$$\lim_{x\rightarrow1}\frac{x^{3}-1}{x-1}=L$$ For this limit $L$, given $\epsilon>0$ , find $\delta>0$ in terms of $\epsilon$ such that: $$\lvert \frac{x^{3}-1}{x-1}-L\rvert<\epsilon$$ For all $x$ satisfying $0<\lvert x-1\rvert<\delta$.

The work I have so far is:

Guess the limit $L=3$. Then $$\lvert \frac{x^{3}-1}{x-1}-3\rvert<\epsilon$$ $$=\lvert \frac{(x-1)(x^{2}+x+1)}{x-1}-3\rvert<\epsilon$$ $$=\lvert {(x^{2}+x+1)}-3\rvert<\epsilon$$ $$=\lvert x^{2}+x-2\rvert<\epsilon$$ $$=\lvert (x+2)(x-1)\rvert<\epsilon$$ $$=\lvert x+2\rvert \lvert x-1\rvert<\epsilon$$ Then using $\lvert x-1\rvert<\delta$ I'm not sure which delta to choose. I assume using $\delta=1$ would work because it is still in the neighborhood for the limit $L$, and I know I am looking for some $\lvert x+2 \rvert<$ some number, I am just unclear how I should go about choosing my delta.

Any help on how to choose a $\delta$ is appreciated as I feel confident in solving the rest of the problem.

Answer 1

Given an $\epsilon>0$, you would find $ \delta>0$ such that

$$|x-1|<\delta \;\; \implies $$ $$\;\;|(x+2)(x-1)|<\epsilon$$

As $ x $ is near $ 1$, you can assume that, for example, that $ |x-1|<\color{red}{1 }$ or which is equivalent $$-1 < x-1 <1$$ which gives

$$2 < x+2 < 4$$

thus, we look for $ \delta>0 $ such that

$$|x-1|<\delta \text{ and } |x-1|<1 \; \implies$$ $$ \; |x-1||x+2|<4|x-1|<\epsilon$$ it is sufficient to take $$\delta = \min(\color{red}{1},\frac{\epsilon}{4})$$

Answer 2

Rephrasing :

Let $|x-1|<1$;

$|x+2||x-1| =$

$|(x-1)+3||x-1| \le$

$(|x-1|+|3|)|x-1|\le$

$ |x-1|^2+3|x-1| <$

$|x-1|+3|x-1|=4|x-1|$;

Choose $\delta =\min (1,\epsilon/4)$.

Answer 3

If you stipulate that $\delta=1$ would limit your $|x-1|$ term so that you have $|x-1|<1$, then think about what this does to the other term.

There are two ways I do them - (A) by triangle inequality, or (B) intervals

(A) With triangle inequality: $|x+2|=|(x-1)+3|\leq|x-1|+|3|$

And since $|x-1|<1$ then I can say $|x+2|\leq|x-1|+|3|<|1|+|3|=4$

So $|x+2|<4$

(B) With intervals: $|x-1|<1$ is equivalent to $-1<x-1<1$

And since $-1<(x+2)-3<1$ then I could say $-1+3<x+2<1+3=4$. And from that right side, I'd get $|x+2|<4$

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Either method will get the same thing: $|x+2|<4$

From here you return to the original function in factored form: $|x+2||x-1|<\epsilon$

Since $|x+2|<4$ then you have $4|x-1|<\epsilon$ which means $|x-1|<\frac{\epsilon}{4}$

Now you have two options to limit $|x-1|$, either $\delta=1$ (from stipulation at top) or $\delta=\frac{\epsilon}{4}$ (from results of that stipulation). And since you want both choices to hold, you make $\delta=\min\{1,\frac{\epsilon}{4}\}$

Generally, I like triangle inequality more than intervals since intervals can get very cumbersome if equations get more complex than quadratics

Answer 4

Working with limits other than in $0$ is too much confusing, just set $x=1+u$ with $u\to 0$ it is so much easier because it triggers familiar things.

$f(x)=\dfrac{x^3-1}{x-1}=x^2+x+1=3+3u+u^2$

So $|f(x)-3|=|3u+u^2|\le 3|u|+|u|^2\le 4|u|\to 0$

With $\epsilon,\delta$ just take $\delta=\min(\epsilon,1)$ thus $\begin{cases}|u|<1\implies |u|^2<|u| \\ |u|<\epsilon\implies |f(x)-3|<4\epsilon\end{cases}$

Answer 5

We don't care what happens when $x\le 0$ or $x\ge 2.$

If $0<|x-1|<1$ then $0<|x+2|<4,$ implying $|x+2|\cdot |x-1|<4|x-1|.$

So if

(i) $0<|x-1|<1$

and (ii) $|x-1|<\epsilon/4$

then $|x+2|\cdot |x-1|<4\cdot |x-1|<4\cdot \epsilon/4=\epsilon.$

Conditions (i), (ii) can be stated together as $0<|x-1|<\min(1,\epsilon/4).$