This question asks to prove the limit of the infinitely nested radical. Now, I only have vague idea of what rigor means in proving something, but seeing my "answer" being radically different from those provided from others, I guess there are some critical errors in my reasoning, but again, I'm too noob to see it.


Prove $$\lim_{x\to 0^+} \sqrt{x+\sqrt[3]{x+\sqrt[4]{\cdots}}}=1$$

My "Answer":

We can see the nested radical is always positive, so taking the square

$$\lim_{x\to 0^+} (x +\sqrt[3]{x+\sqrt[4]{x + \sqrt[5]\cdots}})=1$$

$$\lim_{x\to 0^+} x + \lim_{x\to 0^+}\sqrt[3]{x+\sqrt[4]{x + \sqrt[5]\cdots}}=1$$

Where $$\lim_{x\to 0^+} x = 0 $$

So we are left with

$$\lim_{x\to 0^+}\sqrt[3]{x+\sqrt[4]{x + \sqrt[5]\cdots}}=1$$

This can obviously continue indefinitely until we are left with

$$\lim_{x\to 0^+, n \to \infty}x^{\frac{1}{n}}=1$$

And we know

$$\lim_{n \to \infty}\frac{1}{n}= 0$$

So the question can be re-written as

$$\lim_{x\to 0^+}x^x=1$$

Where it is known numerically that the limit of the above does indeed equal to one. (Where I just end here, pseudo-complete)

I think the error lies where I just take the square without considering the RHS, where it should be more like

$$\lim_{x\to 0^+} \sqrt{x+\sqrt[3]{x+\sqrt[4]{\cdots}}}= a , a > 0$$

And taking the square

$$ \lim_{x\to 0^+} (x +\sqrt[3]{x+\sqrt[4]{x + \sqrt[5]\cdots}})= a^2 $$

$$ \lim_{x\to 0^+}\sqrt[3]{x+\sqrt[4]{x + \sqrt[5]\cdots}} = a^2 $$

$$ \lim_{x\to 0^+} x = 0 $$

again, and continuing but this time

$$ \lim_{x\to 0^+, n \to \infty}x^{\frac{1}{n}}= ((a^{2})^3)^4... $$

But here, I'm stuck. I do not know how to argue any further than this.

  • 3
    $\begingroup$ Where you begin with your "answer": you already assume what has to be proven, namely that the limit exists and equals one. Why? $\endgroup$ – DonAntonio Feb 29 '16 at 11:08
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    $\begingroup$ How do you define this infinitely nested radical as a sequence? $\endgroup$ – lhf Feb 29 '16 at 11:11
  • $\begingroup$ @Joanpemo then how could I start the formulation? like where do I begin? $\endgroup$ – VladeKR Feb 29 '16 at 11:27
  • $\begingroup$ @VladeKR I really don't know. Hopefully someone will give some ideas later. $\endgroup$ – DonAntonio Feb 29 '16 at 11:44
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    $\begingroup$ @Joanpemo: Haha yes this limit is not a nice one. But it's still not as nasty as nasty ones go. You see, there are a number of standard techniques that one should try; first is asymptotic expansions, next is bounding limit supremum and infimum and then using the general squeeze theorem. Usually these kinds of techniques together are sufficient, such as for this type of limit question. But if both fail, then one is quite doomed. For example, does $\lim_{n\in\mathbb{N}\to\infty} \frac{\tan(n)}{n^2}$ exist? $\endgroup$ – user21820 Mar 1 '16 at 6:15

You made multiple logical errors in your 'proof' One of them boils down to the following:

We shall 'prove' that $\lim_{n\to\infty} \underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{n\text{ times}} = 0$.

As you can see, $\lim_{n\to\infty} \frac{1}{n} = 0$, so all we need to prove is $\lim_{n\to\infty} \underbrace{\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}}_{n-1\text{ times}} = 0$.

This can 'obviously' continue indefinitely until we are left with proving $\lim_{n\to\infty} \frac{1}{n} = 0$.

This is obviously true, hence we are 'done'.

This error is marked in bold. Look very carefully at the theorem that says that the limit of a sum of two expressions is the sum of their limits if they exist. You can use induction to show that this theorem extends to any sum of finitely many expressions, but it does not extend to an infinite sum!

Your other errors are with randomly using limit notation, and I cannot really pinpoint for you what is wrong. Basically, in mathematics you must be able to write down precise definitions of everything without using any "...". That is the most prolific source of errors. Consider for example:

Let $x = 1 + 1 + \cdots$.

Then $1 + x = 1 + ( 1 + 1 + \cdots ) = 1 + 1 + \cdots$.

Thus $x = 1 + x$ and hence $0 = 1$. TADA!

The error is in the first line in the very second that I wrote "...".

  • 1
    $\begingroup$ @VladeKR: If you want to learn to do proper proofs of limits, one good place to start is a good textbook on real analysis such as Spivak's Calculus book. But before that, you might want to learn how to perform valid logical deductions, and one excellent book is "How to Prove It" by Daniel Velleman. (I'm not sure whether the PDF available online is legal, but you'll not regret if you buy the book.) $\endgroup$ – user21820 Feb 29 '16 at 11:42

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