Prove $$\lim\limits_{x\to 2} x^3 = 8$$ using epsilon delta definition.

I try as below.

Let $\varepsilon>0$. We choose $\delta>0$.

Consider that \begin{align} \vert x^3-8\vert &= \vert (x-2) (x^2+2x+4)\vert\\ &=\vert (x-2) \vert\vert(x^2+2x+4) \vert \\ &=\vert (x-2) \vert\vert(x-2)^2+6x \vert . \end{align}

Now I don't know how to continue this answer. I confused with $6x$.

Anyone can help me?

EDIT: I have tried as below as JC12's answer.

Let $\vert x-2\vert <1$, then $\vert x\vert -2< \vert x-2\vert <1$ then we have $$\vert x\vert -2<1 \iff \vert x\vert<3.$$

Now, $$ \vert(x-2)^2+6x \vert < \vert(3-2)^2+6\cdot 3 \vert =19. $$

Thus, \begin{align} \vert x^3-8\vert&= \vert (x-2) \vert\vert(x-2)^2+6x \vert < 19 \vert (x-2) \vert. \end{align}

Now choose $\delta=\min(1,\frac{\varepsilon}{19})$. We have \begin{align} \vert x^3-8\vert < 19 \vert (x-2) \vert< 19 \frac{\varepsilon}{19} = \varepsilon. \end{align}

So, we can conclude $\lim\limits_{x\to 2} x^3 =8$.

  • $\begingroup$ For a suitably chosen $ \delta$, can you show that $ 11 < x^2 + 2x + 4 < 13 $? $\endgroup$
    – Calvin Lin
    Commented Dec 26, 2020 at 4:05
  • $\begingroup$ I confused to show that. $\endgroup$ Commented Dec 26, 2020 at 4:12

4 Answers 4


You'll find the same approach with many other questions of a similar type, though you first notice that $|x-2|$ is the "part you want" since it is the same form as $0<|x-a|<\delta$. Say we restrict $|x-2|<1$ (this could be any other number but we just choose $1$ for sake of simplicity). Then $|x|-|2|<|x-2|<1$ and thus $|x|<3$. Then $|(x-2)^2+6x|<|(3-2)^2+6\times3|=19$. Thus you can choose your delta as $\min(1,\frac{\varepsilon}{19})$.

I leave you to finish and formalise the proof. As an added bonus, you could also try proving that $\lim_{x\rightarrow a}x^3=a^3, \forall a\in\mathbb{R}$ using a similar method.


More generally, if you want to show that $\lim_{x \to c} x^n=c^n$, note that $x^n-c^n =(x-c)\sum_{k=0}^{n-1} x^kc^{n-1-k} $.

If $c-r < x < c+r$ where $0 < r< \min(1, x)$ then $\sum_{k=0}^{n-1} x^kc^{n-1-k} \lt \sum_{k=0}^{n-1} (c+r)^kc^{n-1-k} \lt \sum_{k=0}^{n-1} (c+r)^k(c+r)^{n-1-k} =n(c+r)^n $ so

$\begin{array}\\ |x^n-c^n| &=|(x-c)\sum_{k=0}^{n-1} x^kc^{n-1-k}|\\ &=|x-c|\,|\sum_{k=0}^{n-1} x^kc^{n-1-k}|\\ &<|x-c|n(c+r)^{n-1}\\ \end{array} $

so if $|x-c| \le \dfrac{\delta}{n(c+r)^{n-1}} $ then $|x^n-c^n| \lt \delta$.


Let $\epsilon > 0$ be arbitrary and start off with $\delta_{1} := 1$ so that we have $|x-2| < \delta_{1} = 1$.

Then $x>1$ and $x < 3$ so that $(x-2)^{2} < 1$ and $6x < 18$.

Then $|(x-2)^{2} + 6x|<19$ and set $\delta _{2}$ so that $|x-2| < \delta_{2}$ where $\delta_{2} := \frac{\epsilon}{19}.$

Hence, if $\delta := \min\{\delta_{1},\delta_{2}\}$, we have the result.

The reason why it is necessary to end with a single $\delta$ is that, for this argument to work, we need $|x-2| < 1$ and $|x-2| < \frac{\epsilon}{19}$.

Then to find a fixed value that will work all the way through, choose the smaller one.


Proposition: Let $D\subseteq\mathbb{R}$. Let $f:D\to\mathbb{R}$ and $g:D\to\mathbb{R}$ be continuous. Then $fg:D\to\mathbb{R}$ defined as $(fg)(x):=f(x)g(x)$ is continuous.

Proof: Fix $x_0\in D$. Then $f$ is continuous at $x_0$. Consider a particular vaule $\varepsilon_0>0$. Then there exists a $\delta_1>0$ such that if $x\in D$ satisfies $|x-x_0|<\delta_1$, then $$||f(x)|-|f(x_0)||\leq|f(x)-f(x_0)|<\varepsilon_0$$ which implies $|f(x)|<|f(x_0)|+\varepsilon_0$ for $x\in D$ with $|x-x_0|<\delta_1$.

(This part of the proof actually showed that continuous functions are locally bounded and can also be modified to show that if $f$ is continuous, then $|f|$ is continuous.)

Now, let $\varepsilon>0$.

Since $f$ is continuous at $x_0$, then there exists a $\delta_2>0$ such that if $|x-x_0|<\delta_2$ then $$|f(x)-f(x_0)|<\frac{\varepsilon}{2(|g(x_0)|+\varepsilon_0)}$$

Since $g$ is continuous at $x_0$, then there exists a $\delta_3>0$ such that if $|x-x_0|<\delta_3$ then $$|g(x)-g(x_0)|<\frac{\varepsilon}{2(|f(x_0)|+\varepsilon_0)}$$

Take $\delta=\min\{\delta_1,\delta_2,\delta_3\}$. Then, if $x\in D$ satisfies $|x-x_0|<\delta$, we have $$\begin{align*}|f(x)g(x)-f(x_0)g(x_0)|&=|f(x)g(x)-f(x)g(x_0)+f(x)g(x_0)-f(x_0)g(x_0)|\\&\leq |f(x)||g(x)-g(x_0)|+|g(x_0)||f(x)-f(x_0)|\\&<|f(x)|\cdot \frac{\varepsilon}{2(|f(x_0)|+\varepsilon_0)}+|g(x_0)|\cdot\frac{\varepsilon}{2(|g(x_0)|+\varepsilon_0)}\\&<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\&=\varepsilon \end{align*}$$

This implies that $fg$ is continuous on $D$.

Now, in your case, we set $D=\mathbb{R}$ and $f(x)=g(x)=x$. It is trivial to show that $f$ is continuous. Then we obtain $f(x)f(x)=x^2$ is continuous so $f(x)f(x)f(x)=x^3$ is continuous, which implies that $$\lim_{x\to 2}x^3=2^3=8 $$

From here it can actually be inductively shown that for all $n\in\mathbb{N}$, $f_n:\mathbb{R}\to\mathbb{R}$ defined as $f_n(x)=x^n$ is continuous.


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