Show that if $a\sin x + b\cos x + ce^x$ is the zero function for constants $a, b, c$ then $a = b = c = 0$ 
Suppose that $a\sin x + b\cos x + ce^x$ is the zero function. Prove that $a=b=c=0$.

Does the zero function simply mean that $a\sin x + b\cos x + ce^x = 0$? I am going under the assumption that it is.

Geometrically speaking, I think it is pretty clear that $a=b=c=0$ must be true if $a\sin x + b\cos x + ce^x = 0$ as there are no x-values that make the equation true otherwise.
Is there a purely algebraic way of saying this that does not utilize showing a graph?

Given
$a\sin x + b\cos x + ce^x = 0, \tag 1$
we multiply through by $e^{-x}$, which yields
$ae^{-x}\sin x + be^{-x}\cos x + c = 0; \tag 2$
letting $x \to \infty$, we find that
$c = 0; \tag 3$
we are left with
$a\sin x + b\cos x = 0; \tag 4$
now set
$x = \dfrac{\pi}{2}, \tag 5$
and obtain
$a = a\sin \dfrac{\pi}{2} + b\cos \dfrac{\pi}{2} = 0; \tag 6$
thus
$a = 0; \tag 7$
in a similar manner, choosing
$x = 0 \tag 8$
we find
$b = a \sin 0 + b \cos 0 = 0 \tag 9$
as well.  
We have in fact established that $\sin x$, $\cos x$, and $e^x$ are linearly independent.
