How to prove these two functions are equal on the interval [-1,0] I created a function by first considering some well known limits:
$$\lim_{n\to\infty } \frac{n}{2}\sin\left ( \frac{2\pi}{n} \right )=\pi $$
$$\lim_{n\to\infty } \left ( 1+\frac{1}{n} \right )^{n}=e$$
Now, recalling Euler's identity:
$$e^{i \pi }=-1$$
The following limit can be created, which is a combination of Euler's identity and the above two limits;
$$\lim_{n\to\infty } {\left ( 1+\frac{1}{n} \right )^{i \frac{n^{2}}{2}\sin\left ( \frac{2\pi}{n} \right ) }}=-1$$
This led me to the idea of creating the following function:
$$h(x)=\left | \left ( 1+\frac{1}{x} \right )^{\frac{\ i}{2}x^{2}\sin \left (\frac{2\pi }{x}  \right )} \right |$$
With        $ x \in \mathbb{R}$
This function oscillates with a decaying amplitude and I find it very interesting.
Now it was a while ago that I was messing around with this function, and I wanted to find a way to express it such that it wasn't the Absolute value of a complex function... I can't remember quite how I found it but I think the following function is equivalent on the interval $\left [ -1,0 \right ]$
$$g(x)=e^{-\frac{\pi }{2}x^{2}\sin \left (\frac{2\pi }{x}  \right )}$$
Alas, to my question, are these functions equivalent on this interval (they appear to be when they are plotted), and how might I go about proving this? I suspect I could take advantage of the series expansion for the natural log but not sure if that's the best way about it... Could someone point me in the right direction? 
Thanks for reading. 
 A: Let's take
$$h(x)=\left|
\left(1+\frac1x\right)^{i\tfrac12 x^2 \sin\left(\tfrac{2\pi}{x}\right)}
\right|$$
Using the equation $a^b=e^{b\ln(a)}$ we get
$$h(x)=\left|
\exp\left(
i\underbrace{\tfrac12 x^2 \sin(\tfrac{2\pi}{x})}_{K}\ln\left(1+\frac1x\right)
\right)
\right|$$
The logarithm may return complex numbers, so we use the identity $\ln(x)=\operatorname{Ln}|x|+i\arg (x)$, where $\operatorname{Ln}$ means the principal value of complex-valued natural logarithm:
$$h(x)=\left|
\exp\left(
iK\left(\operatorname{Ln}\left(1+\frac1x\right)+i\arg\left(1+\frac1x\right)\right)
\right)
\right|$$
$$=\left|
\exp\left(
iK\operatorname{Ln}\left(1+\frac1x\right)-K\arg\left(1+\frac1x\right)
\right)
\right|=\left|
\exp\left(
iK\operatorname{Ln}\left(1+\frac1x\right)\right)\cdot
\exp\left(-K\arg\left(1+\frac1x\right)
\right)
\right|$$
Here, we use the Euler's formula $e^{ix}=\cos x+i\sin x$ for $x\in\mathbb R$. It implies that $$\left|e^{ix}\right|=\sqrt{\cos(x)^2+\sin(x)^2}=1$$
So our function is equal to the following
$$h(x)=\left|
\exp\left(-K\arg\left(1+\frac1x\right)
\right)
\right|$$
At this point $\exp$ function is real-valued so $|e^x|=e^x$, and taking the principal value of $\arg$,
$$\operatorname{Arg}(1+\tfrac1x)=\begin{cases}
0,&\text{if }1+\tfrac1x>0\implies x\in \mathbb R \setminus [-1;0]\\
\pi,&\text{if }1+\tfrac1x<0\implies x\in(-1;0)
\end{cases}$$
Thus $h(x)=1$ for all $x\in(-\infty;-1)\cup(0;+\infty)$, and for $x\in(-1;0)$ we have:
$$h(x)=
\exp\left(-\tfrac{\pi}{2} x^2 \sin(\tfrac{2\pi}{x})\right)=g(x)$$
For $x=0$ both $f(x)$ and $g(x)$ doesn't exist, and for $x=-1$, $h(x)$ doesn't exist, $g(x)=1$.
