Prove by epsilon-delta that $\lim \limits_{x \to 1} x^3-4=-3$

Question

I need to prove that $\lim \limits_{x \to 1} x^3-4=-3$ with epsilon-delta.

My work

$\forall \varepsilon > 0 ,\exists \space \delta > 0: 0<|x-1|< \delta \implies |x^3-4+3| < \varepsilon$

Working with the consequent:

$|x^3-4+3| < \varepsilon \iff |x^3-1| < \varepsilon \iff |x-1||x^2+x+1| < \varepsilon$

Multiplying the antecedent by $|x^2+x+1|:$

$|x-1|< \delta$ $/\cdot |x^2+x+1| \iff |x-1||x^2+x+1|< \delta |x^2+x+1|$

Here i found a relation but i don't know how to proceed, and i don't know if this way is the best way to prove this. Any hints?

Answer 1

Complete the square: $x^2+x+1=(x+\frac12)^2+\frac34$. Then if $\delta \lt1$, we get $\vert x^2+x+1\vert\lt (\frac52)^2+\frac34=7$.

So set $\delta =\operatorname {min}(\frac{\epsilon}7,1)$.

Answer 2

Assume that $\delta < 1/2$, then $1/2<x<3/2$.

$$|x^2+x+1|=x^2+x+1 < 5$$

$$ |x-1||x^2+x+1|< \delta |x^2+x+1|<5\delta =\epsilon$$

You can take it from here.

Answer 3

Hint: For $\lim_{x\to a}f(x)=L$, we essentially want to show that $|f(x) - L|<\epsilon$ for every $|x-a|<\delta$; that is, we want to show that $|x^3-1|<\epsilon$ for every $|x-1|<\delta$. Let $\delta=1$. Then $|x-1|<\delta$ or simply $|x-1|<1\implies x\in(0,2)$; hence, we also see that $|x^2+x+1|<|2^2+2+1|=7$. This implies that we should choose $\delta=\min\{1,\frac{\epsilon}{7}\}$. Why? Because, if so, then we have $$ \begin{align*} |(x^3-4)-(-3)| &= |x^3-1|\\[0.5em] &= |x-1||x^2+x+1|\\[0.5em] &< \frac{\epsilon}{7}\cdot 7\\[0.5em] &= \epsilon. \end{align*} $$ Hopefully that supplies the missing detail you needed.