# Cannot find limit using epsilon delta definition

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 $$\lim\limits_{x\to 2} x^3 =8$$.

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

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.