Problem: Show that $\lim_{x\to 4}x^3=64$.

My steps:

$|f(x)-L|<\epsilon$ so $|x^3-64|<\epsilon$
$|x^3|<64+\epsilon \rightarrow 64-\epsilon < x^3 < 64+\epsilon$
So $\delta<\sqrt[3]{64+\epsilon}-4$
Let $\epsilon>0$. Pick $\delta$ so that $\delta<\sqrt[3]{64+\epsilon}-4$.
Suppose $0<|x-4|<\delta$. Then $|x-4|<\sqrt[3]{64+\epsilon}-4$.
Thus, $|x|<\sqrt[3]{64+\epsilon}$ and cube to get $|x^3|<64+\epsilon$.
Then $|x^3-64|<\epsilon$ which completes the proof.

But this is what the answer said:

Let $\epsilon>0$. Pick $\delta$ so that $\delta<1$ and $\delta<\frac{\epsilon}{61}$.
Suppose $0<|x−4|<\delta$. Then $4−\delta<x<4+\delta$.
Cube to get $(4−\delta)^3 <x^3 <(4+\delta)^3$.
Expanding the right-side inequality, we get
The other inequality is similar.

So my questions are:

  1. How can I get $\delta<\frac{\epsilon}{61}$?
  2. Why $x^3<\delta^3+12*\delta^2+48*\delta+64<\delta+12\delta+48\delta+64=64+\epsilon$?
  3. When we need to pick $\delta=\min(1,\frac{\epsilon}{61}$) in this case?
  4. Does my proof has problems ?

I am just a newbie to calculus... Thanks in advance.

  • $\begingroup$ I think your proof is equivalent, but it just doesn't follow the typical order of $\epsilon$-$\delta$ -proof. In $\epsilon$-$\delta$ proof one discovers what $\delta$ one needs, by "looking in the future" in the proof. So you do things starting from the "Suppose $0 < |x-4| < \delta$" and then decide on the $\delta+12\delta+48\delta+64$ part, what $\delta$ you need in order to make the conclusion that you wish, which is $x^3 < 64 + \epsilon$. $\endgroup$
    – mavavilj
    Mar 22, 2018 at 14:40
  • $\begingroup$ Notice what the line just before $64+\epsilon$ reads if you plug in $\epsilon/61$. Thus this becomes the constraint that you want to use for $\delta$. $\endgroup$
    – mavavilj
    Mar 22, 2018 at 14:46

4 Answers 4


Well, one is not supposed to use inverse functions like cube roots in expression for $\delta$ otherwise this leads to a circular argument.

The typical approach is to analyze your goal namely $$|x^3-64|<\epsilon$$ which has to be achieved using a specific kind of means namely $0<|x-4|<\delta$. So one first tries to simplify our goal by rewriting it as $$|x-4||x^2+4x+16|<\epsilon$$ The first factor in LHS can be controlled using our means but the second factor needs more work. One just needs to bound this factor suitably (again using the same means). We note that if $0<|x-4|<1$ then $3<x<5, x\neq 4$ and hence $|x^2+4x+16|<61$. Further we can also start with the bound $0<|x-4|<2$ and get $|x^2+4x+16|<76$. The important thing to observe here is that this factor $|x^2+4x+16|$ is bounded. Let's choose the simpler bound $61$ which is possible if $0<|x-4|<1$.

Now we have the obvious inequality $$|x^3-64|=|x-4||x^2+4x+16|<61|x-4|$$ provided that $0<|x-4|<1$. If the RHS of the above inequality is less than $\epsilon$ then the LHS will also be less than $\epsilon$ and hence our goal is achieved if we have $61|x-4|<\epsilon $ or $|x-4|<\epsilon/61 $. Combining the above we see that if $0<|x-4|<\min(1,\epsilon /61)$ then $|x^3-64|<\epsilon $. And hence $\delta$ can be taken as $\min(1,\epsilon /61)$. In case you have read the last paragraph carefully you can immediately see that the $\delta$ can also be taken as $\min(2,\epsilon /76)$.

Such problems do not have a unique answer and most importantly they are not supposed to be solved via algebraic manipulation of inequalities.

  • $\begingroup$ Thank you for the explanation ! $\endgroup$ Mar 24, 2018 at 9:04
  1. for any given (strictly) positive $\epsilon$ just pick a positive number that is smaller than $\epsilon/64$ such a number exist, because this quotient is (strictly) positive.
    1. Because of 3 ( if $\delta < 1$ then $\delta^3 <\delta^2 < \delta$
    2. I suppose that should be "why" : for 2 to be true
    3. I'll leave that one to another

Note that in you proof is ok from here


you should set


and what you find is the optimal value for $\delta$.

Note that by binomial expansion


then for $\epsilon$ small also $\frac{\epsilon}{61}<\frac{\epsilon}{48}$ fulfills the condition.

The other limit is obtained by

$$|x^3-64|<\epsilon\iff |x-4||x^2+4x+16|<\epsilon\iff |x-4|<\frac{\epsilon}{ |x^2+4x+16|}=\delta$$

thus for $0<|x-4|<1 \implies3< x<5 \quad x\ne 4$ the minimum value for $\delta$ is attained for $x=5$ that is

$$ \delta_{min}=\frac{\epsilon}{ |5^2+4\cdot5+16|}=\frac{\epsilon}{61} $$

then the condition is satisfy for $\delta=\min\left(1,\frac{\epsilon}{61}\right)$.

  • $\begingroup$ Using cube roots involves the fact that cube roots exist. And existence of cube roots is a direct consequence of continuity of $x^3$ in which case the limit in question is already available so this is circular. In general one should not use inverse functions to find $\delta$ in terms of $\epsilon$. $\endgroup$
    – Paramanand Singh
    Mar 22, 2018 at 15:18
  • $\begingroup$ @ParamanandSingh so the first approach is not correct? I will look to your answer, Thanks! $\endgroup$
    – user
    Mar 22, 2018 at 15:21
  • $\begingroup$ Also see math.stackexchange.com/a/2104039/72031 (especially last paragraph) and math.stackexchange.com/a/2487862/72031 $\endgroup$
    – Paramanand Singh
    Mar 22, 2018 at 15:25
  • $\begingroup$ @ParamanandSingh Thanks, I've tried to make the second part more clear, I'll take a look. Your help is always valuable. $\endgroup$
    – user
    Mar 22, 2018 at 15:45

In regards to finding a value for $\delta$ that satisfies the proof, we only really care about values of $x$ that are close to $4$, so setting $\delta < 1$ and $\delta < \frac{\epsilon}{61}$ is like saying the distance between $x$ and $4$ must be no more than $1$ or $\frac{\epsilon}{61}$.

By setting these restrictions on $\delta$, we can then make our lives easier in terms of solving this proof because the values acceptable for $x$ also become restricted, i.e if $\delta \lt 1 \rightarrow\lvert x - 4 \rvert \lt 1$ then its safe to make the claim that $3 \lt x \lt 5; x \not=4$ (because the difference between $x$ and $4$ cannot be greater than 1).

So, to answer question 1, "How can I get $\delta \lt \frac{\epsilon}{61}$ ?", it's the result we get when we restrict $\delta \lt 1$.

$$\begin{align} \lvert ~f(x) - 64~\rvert &\lt \epsilon\\ \lvert ~x^3 - 64~\rvert &\lt \epsilon && \text{factor $x^3 - 64$} \\ \lvert ~(x - 4)(x^2 + 4x + 16)~| &\lt \epsilon && \text{isolate $(x - 4)$}\\ \lvert ~x - 4~ \rvert &\lt \frac{\epsilon}{\lvert~x^2 + 4x + 16~\rvert} \end{align}$$

Knowing that when $\delta \lt 1$, then $3 \lt x \lt 5; x \not=4$, so our goal is to apply this inequality to $x^2 + 4x + 16$ and see what we get.

$$\begin{align} 3 \lt x &\lt 5 && \text{let's find the min and max of $x^2$}\\ 9 \lt x^2 &\lt 25 \\\\ 3 \lt x &\lt 5 && \text{let's find the min and max of $4x$}\\ 12 \lt 4x &\lt 20 \\\\ 9 \lt x^2 &\lt 25 && \text{let's bring this all together now}\\ 9 + 12 \lt x^2 + 4x &\lt 25 + 20 \\ 21 \lt x^2 + 4x &\lt 45 && \text{all that's left to add is $16$}\\ 21 + 16 \lt x^2 + 4x + 16 &\lt 45 + 16\\\\ 37 \lt x^2 + 4x + 16 &\lt 61 \end{align}$$

What we just showed is that when $\delta \lt 1$, then $37 \lt x^2 + 4x + 16 \lt 61$ which in turn means that, $$\begin{align} \lvert ~x - 4~ \rvert &\lt \frac{\epsilon}{\lvert~x^2 + 4x + 16~\rvert} \lt \frac{\epsilon}{61} \\\\ \lvert ~x - 4~ \rvert &\lt \frac{\epsilon}{61} && \text{$\square$} \end{align}$$

For your second question: "Why $x^3<\delta^3+12\cdot \delta^2 + 48\cdot \delta + 64 < \delta + 12\delta + 48\delta + 64 = 64 + \epsilon$ ?", if think about the value for $\delta$, it's always less than $1$ so when we when multiply two or more values that are less than $1$, we get a smaller number. This explains why, $$\delta^3+12\delta^2 + 48\delta + 64 < \delta + 12\delta + 48\delta + 64$$

In regards to how $\delta + 12\delta + 48\delta + 64 = 64 + \epsilon$, we just simplify the right side of the equation to get $$\begin{align} \delta + 12\delta + 48\delta + 64 &= 61\delta + 64 && \text{sub. in $\frac{\epsilon}{61}$ for $\delta$}\\ 61 \cdot \frac{\epsilon}{61} + 64 &= \epsilon + 64 && \text{$\square$} \end{align}$$

Unfortunately, for your third question, I don't really know what your asking. However, what I can say about knowing when to set $\delta$ to either $1$ or $\frac{\epsilon}{61}$ is that it depends on the value for $\epsilon$; if $\epsilon \lt 61$ then set $\delta = \frac{\epsilon}{61}$, otherwise set $\delta = 1$.

Lastly, in terms of your proof having problems, these questions are not easy to begin with, so making a misstep in the beginning of your proof will very easily lead you astray. The good thing is that you tried to get the $\epsilon$ inequality to resemble the $\delta$ inequality which is the right thing to do, the problem is how you chose to do it.

Trying to isolate for $x$ and then manipulating the inequality to resemble the $\delta$ inequality isn't really an effective approach because things can get complicated very quickly (especially in this case). I would try and factor first and see where that takes me. That being said, when it comes to $\epsilon-\delta$ proofs, there is no silver bullet approach to finding the solution. You will need to practice them and with experience, you will start to have an idea of what to do.

As a side note, here's a reference to a tutorial on Limits I wrote a couple months ago for Mathbook.

  • $\begingroup$ Thank you! Your answer is really clear and helped me a lot =) $\endgroup$ Mar 24, 2018 at 9:05
  • $\begingroup$ You very welcome. I'm glad I could be of help. $\endgroup$
    – JetJet13
    Mar 24, 2018 at 17:40

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