If $0.9999\ldots=1$, then why is $\lim_{n\to\infty}\frac{\tan(89.[n\,\text{"$9$"s}]^\circ)}{\tan(89.[(n-1)\;\text{"$9$"s}]^\circ)}$ not equal to $10$? 
If $0.9999\ldots=1$, then why is this limit not equal to $10$?
$$L = \lim_{n \to \infty} \frac{\tan(89.\overbrace{9999...}^{\text{n times}} \space ^\circ)}{\tan(89.\underbrace{999...}_{\text{n-1 times}} \space ^\circ)}$$

We can rewrite this limit as
$$\begin{gather}
L = \lim_{n \to \infty} \frac{\tan\left( \frac \pi 2 - \frac{\pi}{180 \times 10^n}\right)}{\tan\left( \frac \pi 2 - \frac{\pi}{180 \times 10^{n-1}} \right)} \\ 
L = \lim_{n \to \infty} \frac{\tan\left( \frac{\pi}{180 \times 10^{n-1}}\right)}{\tan\left( \frac{\pi}{180 \times 10^{n}}\right)}
\end{gather}$$
let $t = \frac{\pi}{180 \times 10^n}, t \to 0$
$$L = \lim_{t \to 0}\frac{\tan10t}{\tan t} \\
\boxed{L = 10}$$
However, according to the well-known proof 0.9999 = 1, shouldn't the limit be $\frac{\tan(90^\circ)}{\tan(90^\circ)}$, which is undefined? Where am I going wrong here?
 A: I am afraid that you are missing the essence of a limit. When you compute
$$\lim_{x\to a}f(x)$$ you don't care about $f(a)$ (which could be defined or undefined), but only about $f(x)$ for $x\ne a$.
A: We cannot simply assume that both of the limits $\lim_{n \rightarrow \infty}f(x)$ and $\lim_{n \rightarrow \infty}g(x)$ exist, so the formula
$$ \lim_{n \rightarrow \infty} \dfrac{f(x)}{g(x)}= \dfrac{\lim_{n \rightarrow \infty}f(x)}{\lim_{n \rightarrow \infty}g(x)}$$
doesn't have to work. 
As a result, you can't think that if $89,999\ldots\approx90$ then in the formula which you analyse is the same in numerator and denominator and the result is simply $1$. You ought to go deeper and that is what you have done after.
Remember that intuition is very helpful in mathematics, but we cannot always blindly believe in all which it says to our mind.
A: As pointed out in the comments, $$\lim \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}$$ only when $\lim f(x)$ and $\lim g(x)$ exist. In this case, both of these limits do not exist, which is why
$$\lim_{n \to \infty} \frac{\tan(89.\overbrace{9999...}^{\text{n times}} \space ^\circ)}{\tan(89.\underbrace{999...}_{\text{n-1 times}} \space ^\circ)} \ne \frac{\tan 90^\circ}{\tan 90^\circ}$$
The limit must be evaluated as shown in the question to obtain the answer, which is 10.
