Proving Infinite Nested Radical

Question

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.

Question:

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.

Answer 1

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 "...".