$$\lim_\limits {x \to \pi} \frac{(e^{\sin x} -1)}{(x-\pi)}$$
I found $-1$ as the answer and what I did was:
$\lim_\limits {x \to \pi} \frac{(e^{\sin x} -1)}{(x-\pi)}$ $\Rightarrow$ $\lim \frac{(f(x) - f(a))}{(x-a)}$ $\Rightarrow$ $f(x)=(e^{\sin x})$
$f(a)=1$
$x=x$
and $a=\pi$
So I concluded that the limit of the first function would be the same as the derivative of $f(x)$ so I did:
$\frac{d}{dx} (e^{\sin x})$ $\Rightarrow$ $-1$
But isn't it the same as using the L'HÔPITAL rule??
If you want more than just the limit, Taylor series are simple and convenient.
First, let us set $x=\pi+y$ and we shall consider what happens around $y=0$. $$\frac{(e^{\sin (x)} -1)}{(x-\pi)}=\frac{e^{-\sin (y)}-1}{y}$$ Now, let us use $$e^z=1+z+\frac{z^2}{2}+O\left(z^3\right)$$ So, by Taylor, $$e^{-\sin (y)}=1-\sin(y)+\frac{\sin^2(y)}{2}+\cdots$$ $$\sin(y)=y-\frac{y^3}{6}+O\left(y^4\right)$$ Combining $$e^{-\sin (y)}=1-y+\frac{y^2}{2}+O\left(y^3\right)$$ $$\frac{e^{-\sin (y)}-1}{y}=-1+\frac{y}{2}+O\left(y^3\right)$$ which shows the limit and also how it is approached.
The given limit $$\lim_{x \to \pi} \frac{e^{\sin x} - 1}{x - \pi}$$ is by definition the derivative of $f(x) = e^{\sin x}$ evaluated at $x = \pi$, since $$f'(a) \equiv \lim_{x \to a} \frac{f(x) - f(a)}{x-a}.$$ Thus $$f'(x) = \cos x e^{\sin x},$$ and $$f'(\pi) = (\cos \pi) e^{\sin \pi} = -1.$$ This is completely distinct from L'Hopital's rule, which requires the consideration of the derivatives of the numerator and denominator of particular indeterminate forms.
