Why is $\lim_{x \to 0} x = (1+\tan x)^\frac{1}{x}$ not 1?

Question

Why is it wrong to think that $$\lim_{x \to 0} (1+\tan x)^\frac{1}{x} = (1+0)^\frac{1}{x}=1^\infty=1$$ because $\lim_{x \to 0} \tan x$ and $\lim_{x \to 0} \frac{1}{x}=\infty$? I already know the correct answer but could not figure out why this way is not correct.

Answer 1

$1^\infty$ is indeterminate form, you can't say it is $1$.

Answer 2

What you are trying to do is to "split up" the limit taking procedure. In your case you to take the limit of the base which is $1$ and you then want to take the limit of the exponent, which you say is $\infty$.
You are probably familiar with this precedure from expressions like these:

\begin{equation} \lim_{x \to 0} (f(x) \cdot g(x)) = (\lim_{x \to 0} f(x)) \cdot (\lim_{x \to 0} g(x)) \end{equation}

Recall however that this only has to work when both limits of $f$ and $g$ exist as real numbers.
The very same splitting will usually work for most "nice" operations, like in your case exponentiation. (If you hit the limit $0^0$ however, it will not!)

But since $\infty$ is not a real number, it is (here) just a symbol denoting unbounded sequences, there is no reason the splitting should work.
$1^{\mathrm{bla}}$ when bla is no real number does not really make any sense at all.