Prove using epsilon delta definition that $\lim_{x\rightarrow1}f(x) = 3$ when $f(x) = \frac{x^3 - 1}{x-1}$.

Question

I get to the point where I get (x+2)(x-1) is less than epsilon and do not know how to proceed just tell me a hint since I want to do it myself

Answer 1

HINT

\begin{align*} x^{3} - 1 = (x-1)(x^{2} + x + 1) \end{align*}

Answer 2

Given $\epsilon$, you want a $\delta$ such that $|x-1|<\delta \implies |x^2+x+1-3|<\epsilon.$

As you pointed out we need $|x^2+x-2|=|x-1||x+2|<\epsilon$.

So: We know $|x-1|<\frac{\epsilon}{|x+2|}$

$|x^2+x+1 -3|$ has a maximum value of 4 if $|x-1|<1$.

We need to divide the maximum value of epsilon in that neighborhood of 1 to guarantee being in that neighborhood. The largest value of x in that range is 2 which maximizes the denominator.

So it is expected that $|x-1|<\epsilon/4$.

Going Forward:

$$1-\epsilon/4<x<1+\epsilon/4$$ $$1-\epsilon/2+\epsilon^2/4<x^2<1+\epsilon^2/4+\epsilon/2$$ $$2-3\epsilon/4+\epsilon^2/4<x^2+x<2+\epsilon^2/4+3\epsilon/4$$ $$-3\epsilon/4+\epsilon^2/4<x^2+x-2<\epsilon^2/4+3\epsilon/4$$

Now the squared epsilon term guarantees $-3\epsilon/4$ is smaller than $x^2+x-2$. Continuing to assume by hypothesis $\epsilon$ is less than 1, then so must its square, so we know $\epsilon^2/4+3\epsilon/4<4\epsilon/4=\epsilon$.

Taking then bound with largest absolute value, we are then guaranteed that

$|x^2+x-2|<\epsilon$.

Answer 3

Remember that $\lim_{x\to\ a} f(x) = L$ means that for all $\epsilon > 0$ there exists a $\delta > 0$ such that $$\text{if } \mid x- a \mid < \delta \text{ then } \mid f(x) - L\mid < \epsilon$$

Read the absolute values as distance. So in order to prove that $\lim_{x\to\ 1} \frac{x^3 - 1}{x - 1} = 3$ we have to fix an arbitrary $\epsilon > 0$ and find a $\delta >0$ that guarantees the condition above. So we have the following

$$\text{if } \mid x - 1 \mid < \delta \text{ then } \mid \frac{x^3-1}{x-1} - 3\mid < \epsilon$$ $$ \mid \frac{x^3-1}{x-1} - 3\mid = \mid x-1\mid \mid x+2\mid < \epsilon$$

Notice that $\mid x-1 \mid < \delta$, in other words, $\delta$ is an upper bound. Multiplying by $\mid x+2\mid$ we get we have $$\mid x-1\mid \mid x+2\mid < \mid x+2\mid \delta$$ If we choose $\delta = 1$(1 is an arbitriary number), then $\mid x-1\mid < 1$ is also bounded by $1$.

So here's the trick, we have found an upper bound for the original expression, namely $\mid x+2\mid \delta$, if $\mid x+2\mid \delta < \epsilon$ the $\delta$ that guarantees the inequality is $\delta < \frac{\epsilon}{\mid x+2 \mid}$

So if we choose $\delta = min\{1, \frac{\epsilon}{\mid x+2 \mid}\}$, then $\mid x-1\mid \mid x+2\mid <\epsilon$

Answer 4

I'll present an alternate way to get rid of the $x+2$ factor of other proofs, I'll use this property instead: $$0\le u\le 1\implies 0\le u^2\le u$$

Note that this works also for any power of $u$ if needed.

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

Since $|x-1|\ge 0$ and close to zero, it can also be made $<1$ if needed, so apply for $u=|x-1|$

$$|f(x)-3|\le |x-1|^2+3|x-1|\le 4|x-1|$$

and epsilon-delta proof is now straightforward, choose the minimum between $\delta<\frac{\epsilon}4$ (for the limit) and $\delta< 1$ (for the condition $u<1$)