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?
Best Answer
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 $x\ne a$.