Applying Euler's formula for limit of $\frac{\sin(x)}x$ as x approaches $0$ in exponential form

Question

I'm currently studying Euler's formula and the representation of

$\frac{\sin(x)}x$ as $\frac{1}{2ix}(e^{ix}-e^{-ix})$

and my understanding would be taking the limit as $x$ approaches $0$ would be

$$\lim_{x\rightarrow0} \frac{1}{2ix}(e^0-e^{-0})$$

which I believe is not right

I think I'm asking two questions:

1. Clarity of Euler's formula (I sort of understanding it from the series derivation but in practice I don't understand this at all)
2. Clarity on taking the limit for the real numbers using this complex representation

Edit: Just wanted to say thank you to everyone who commented, all the answers were very helpful!

Answer 1

Euler's formula states that $e^{ix}=\cos(x)+ i\sin(x)$ for any real number $x$, where $ i=\sqrt{-1}$ is the imaginary unit (see 1 and 2). It can obtained from the Maclaurin series expansions of $\cos(x)$, $\sin(x)$ and $e^x$. Then we have

$$\frac{e^{ix}-e^{-ix}}{2xi}=\frac{\cos(x)+ i\sin(x)-(\cos(-x)+i\sin(-x))}{2xi}=\frac{2i\sin(x)}{2xi}=\frac{\sin(x)}{x}$$

since $\cos(x)$ is an even function and $\sin(x)$ is an odd function.

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It is well know that $\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=$ $1$. If you want to start from the other definition we have

$$\lim_{x\rightarrow 0}\frac{e^{ix}-e^{-ix}}{2ix}=\lim_{x\rightarrow 0}\frac{\left(1+ix-\frac{x^2}{2}+...\right)-\left(1-ix-\frac{x^2}{2}+...\right)}{2ix}=\lim_{x\rightarrow 0}\frac{2ix+...}{2ix}=1$$

by power series expansion. You can also use L'Hopital's rule.

Answer 2

We have that by standard limit $\frac{e^x-1}x \to 0$ as $x \to 0$ we obtain

$$\frac{\sin x}x=\frac{e^{ix}-e^{-ix}}{2xi}=\frac1{e^{ix}}\frac{e^{2ix}-1}{2ix}\to 1\cdot 1=0$$