Proving limits using precise definition of limits.

Question

$$ \lim_{x \to 1} \frac{x^3-1}{x-1}=3 $$ How to prove it using precise definition of limits? While solving it, I get stuck at |(x-1)(x+2)|<ε. I don't know how to take out the inequality for x only as this inequality contains quadratic form.

Answer 1

Note that\begin{align}\frac{x^3-1}{x-1}-3&=x^2+x+1-3\\&=x^2+x-2\\&=x^2-1+x-1\\&=(x-1)(x+1)+x-1.\end{align}Now, take $\varepsilon>0$. If $\lvert x-1\rvert<1$, then $\lvert x+1\rvert<3$ and therefore$$\bigl\lvert(x-1)(x+1)+x-1\bigr\rvert<4\lvert x-1\rvert.$$So, take $\delta=\min\left\{1,\frac\varepsilon4\right\}$ and then$$\lvert x-1\rvert<\delta\implies\bigl\lvert(x-1)(x+1)+x-1\bigr\rvert<4\times\frac\varepsilon4=\varepsilon.$$

Answer 2

Let $0 <\epsilon<1$. If $|x-1| <\epsilon /4$ then $|\frac {x^{3}-1} {x-1} -3|=|x^{2}+x+1-3|=|(x-1)(x+1)+(x-1)|<|x-1|(|x-1|+2)+|x-1|<(\epsilon/4) (3)+\epsilon/4 <\epsilon$.

Answer 3

As you've probably noticed, when $x \ne 0$, \frac{x^3-1}{x-1} = x^2 + x + 1$, so you just need to show that |x^2 + x + 1 - 3| = |x^2 + x -2| gets arbitrarily close to 0 (in epsilon delta terms).

Answer 4

The standard trick to deal with this issue is to declare that you will never pick an $x$ outside of $(0,2)$. In other words, no matter what $\varepsilon$ is, your $\delta$ will never be greater than $1$.

With this declaration, we get $|x+2|\leq 4$, and this gives $$ |(x-1)(x+2)\leq 4|x-1| <4\delta $$ So as long as $\delta$ is less than or equal to $\frac\varepsilon4$, everything will work fine.

So with the assumption that $\delta\leq 1$, everything works as long as we also have $\delta\leq \frac\varepsilon 4$. Thus one possible choice for $\delta$ will be $$ \delta = \min\left(1, \frac\varepsilon4\right) $$ With this choice of $\delta$, and assuming $0<|x-1|<\delta$, we get $$ \left|\frac{x^3-1}{x-1} - 3\right|= |(x-1)(x+2)|\leq 4|x-1|<4\delta\leq \varepsilon $$ and we're done.