# 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?

• There is a condition on the "rule" $$\lim \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}\,.$$ Is that condition satisfied here? Aug 14 '20 at 14:29
• Think on a simpler example: $\lim_{x\to \infty}\frac{10x}{x} = \lim_{x\to \infty} 10 = 10$. However $\frac{\lim_{x\to\infty} 10x}{\lim_{x\to\infty} x} = \frac{\infty}{\infty}$ is undefined. Aug 14 '20 at 14:33
• Well, it doesn't tell us that the limit is $10$, nor that it exists at all, for that we must take a closer look at the quotients. But it tells us that separating numerator and denominator doesn't help us at all here. Aug 14 '20 at 14:36
• @Novice $\tan(\frac \pi 2 - \theta) = \cot \theta = \frac 1 {\tan\theta}$ Aug 14 '20 at 14:38
• "why is this limit not equal to 10?": hem, it is equal to $10$.
– user65203
Aug 14 '20 at 15:15

$$\lim_{x\to a}f(x)$$ you don't care about $$f(a)$$ (which could be defined or undefined), but only about $$f(x)$$ for $$x\ne a$$.
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.
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.