The only limit that gets to $1$? $$\lim_{x\to 0} \frac{\sin(\pi x)}{e^{ax}-1}=1$$
has only one real solution, when $a=\pi$
I find this absolutely fascinating. Is there a proof or explanation of this phenomenon?
 A: It is a consequence of 


*

*$\displaystyle \lim_{x\to 0} \dfrac{\sin(k x)}{x}=k$

*$\displaystyle \lim_{x\to 0} \dfrac{e^{ax}-1}{x}=a$


which should be fairly obvious from the early terms of the series expansions


*

*$\sin(k x) = kx -\dfrac{(kx)^3}{3!}+\cdots$

*$e^{ax} = 1 + ax +\dfrac{(ax)^2}{2!}+\dfrac{(ax)^3}{3!}+\cdots$

A: Hint. By L'Hospital rule, you just have
$$
\lim_{x\to 0} \frac{\sin(\pi x)}{e^{ax}-1}=\lim_{x\to 0} \frac{\pi\cos(\pi x)}{a\:e^{ax}}=\frac{\pi}a.
$$
A: Hint: $$\lim_{x\to0}\dfrac{\sin(\pi x)}{e^{ax}-1}=\lim_{x\to0}\dfrac{\sin(\pi x)}{x}\dfrac{x-0}{e^{ax}-e^{a\cdot 0}}=\lim_{x\to0}\pi\dfrac{\sin(\pi x)}{\pi x}\left(\dfrac{e^{ax}-e^{a\cdot 0}}{x-0}\right)^{-1},$$
$\displaystyle\lim_{x\to0}\dfrac{\sin(\pi x)}{\pi x}=1$, and $\displaystyle\lim_{x\to0}\dfrac{e^{ax}-e^{a\cdot 0}}{x-0}=\left.\dfrac{\rm d}{{\rm d}x}e^{ax}\right|_{x=0}=ae^{a\cdot 0}=a$.
A: for a isnot 0,sin($\pi x$)~$\pi$x,$e^ax-1$~$ax$
hence the limit is $\frac{\pi}{a}$
(because sinx=x+o(x),$e^x$=1+x+o(x))
A: HINT: $$\dfrac{\sin\pi x}{e^{ax}-1}=\dfrac\pi a\cdot\dfrac{\sin\pi x}{\pi x}\cdot\dfrac1{\dfrac{e^{ax}-1}{ax}}$$
Can you take it from here?
