Why is $\frac{1}{\frac{1}{0}}$ undefined?

Is the fraction

$$\frac{1}{\frac{1}{0}}$$

undefined?

I know that division by zero is usually prohibited, but since dividing a number by a fraction yields the same result as multiplying the number by the fraction's reciprocal, you could argue that

$$\frac{1}{\frac{1}{0}} = (1)\left(\frac{0}{1}\right) = 0$$

Is that manipulation permissible in this case? Why or why not?

Any expression having an undefined term somewhere inside is undefined as a whole. The rule $\frac1{\frac1x}=x$ holds only for $x\ne0$.

Another way to think about this is order of operations:

$$ \frac{1}{\frac{1}{0}}=1/(1/0) $$

I always compute what's inside the parenthesis first, which gives me undefined, and I have to stop there.

The expression is undefined because the value you are trying to divide $1$ by is undefined. Therefore the operation cannot take place. If you had something like $\frac{1}{\frac{1}{\ln 0}}$ then that would also be undefined because you cannot evaluate $\ln 0$. I think your misunderstanding comes from the fact that you treat $\frac{1}{0}$ as wholly separate term and ignore it's value (not even undefined).

Yeah, this is undefined. When we divide some number with some N.D. number it results in N.D. number. I think you are making a mistake by considering $1/0$ a fraction. Its not a fraction because denominator is zero, its a number which is not defined. So, here "fraction's reciprocal" doesn't make any sense.

But taking limit of $1/{(1/x)}$ as $x\to 0$ makes sense.

A fraction $\frac{a}{b}$ is defined as a solution of the equation $bx=a$. Of course, the equation $0x=1$ has no solutions in $\mathbb{R}$. If you very want to solve this equation you could do as follows.

Consider $\mathbb{R}$ as a (multiplicative) semigroup and try to embed it in such a semigroup $S$ that $\exists x\in S: 0x=1$. Of course, you get nothing, since $0=0\cdot 1=0\cdot 0x=0x=1$. Then you can consider more general situation: take as $S$ some magma (see in Wikipedia "Magma (algebra)"). Since the multiplication will be non-associative, the contradiction disappears: you get $0(0x)=0$, i.e. $0\cdot 1=0$. I believe that such a magma exists, but I didn't try to build it.