A strange limit I have to compute the following limit
$$\lim_{x\to \infty} \frac{x\log(1+\tan(8x))}{6^{x^2}-1}.$$
Is that a problem if $\log(1+\tan(8x))$ is not always defined? Why not?
 A: Even if we proposed to use a modified definition of the tangent function  $$T(x) = \begin{cases} \tan x, & x \in \left[\pi k, \frac{\pi}{2}(2k+1)\right), \\ 0, & {\rm otherwise,} \end{cases} \quad k \in \mathbb Z$$ the limit STILL would not exist.  To understand why, we again look at the definition of limits at infinity:  The limit of a function $f(x)$ as $x \to \infty$ equals $L$ if, for any $\epsilon > 0$, there exists some $x_0$ such that for all $x > x_0$, $|f(x) - L| < \epsilon$.  If we posit that $L = 0$ for $$ f(x) = \frac{x \log (1 + T(8x))}{6^{x^2} - 1}, $$ then we can see the definition still is not satisfied no matter how large we wish to pick $x_0$ and how small we pick $\epsilon$.  For there must exist an integer $K$ such that $\frac{\pi}{2}(2K-1) \le x_0 < \frac{\pi}{2}(2K+1)$, and in the interval $(x_0, \frac{\pi}{2}(2K+1))$, there is guaranteed to be a value of $x$ such that $|f(x)| > \epsilon$ because the denominator is at most $6^{((2K+1)\pi/2)^2} - 1$ but the numerator can be made arbitrarily large by choosing $x$ just shy of the upper bound.*  So here we have an example of a function that is defined everywhere and is equal to the original function using the tangent over their common domain, but for which the limit is still not defined.
*Explicitly, on the interval $(x_0, \frac{\pi}{2}(2K+1))$ we can write $$ f(x) = \frac{x \log(1 + T(8x))}{6^{x^2}-1} > \frac{x_0 \log(1 + T(8x))}{6^{((2K+1)\pi/2)^2} - 1} > \epsilon,$$ and solving for $x$ gives $$\tan 8x > \exp\left(\frac{\epsilon}{x_0}(6^{((2K+1)\pi/2)^2} - 1)\right) - 1.$$  Since the RHS is some fixed constant, say $C$, there is always an $x$ in the interval such that $\tan 8x > C$.
