Prove that the set of functions $\{x, e^x, \sin(x)\}$ is linearly independent. I am supposed to prove that ${x,e^x,\sin(x)}$ is a linearly independent set.
I know that if $\{x,e^x,\sin(x)\}$ is linearly independent, then we would have $ax+be^x+c\sin(x)=0$, for all $x \in \mathbb{R}$, and $a,b,c \in \mathbb{R}$, where a=b=c=0.
However, I'm skeptical how to proceed.
My gut tells me to begin taking derivatives, since I'm given a set of real function, and I may be able to to show that $a$, $b$, and $c$ must be equal to zero.
\begin{align*}
(ax+be^x+c\sin(x))'&= a1+be^x+c\cdot cos(x)=0\\
(ax+be^x+c\sin(x))''&=a0+be^x-c\cdot sin(x)=0\\
(ax+be^x+c\sin(x))'''&=a0+be^x-c\cdot cos(x)=0\\
(ax+be^x+c\sin(x))''''&=a0+be^x+c\cdot sin(x)=0\\
\end{align*}
Now subtracting $c\cdot \sin(x)$, we have
\begin{equation*}
be^x=-c(\sin x)
\end{equation*}
From this I can't determine much, I know that $e^x >0$ for all $x\in\mathbb{R}$.
Can someone give me a push? 
 A: $ax +be^x +c\sin(x) =0$, $x=0$ implies $b=0$, $ax+c\sin(x) = 0$, $x=\pi$, $a=0$ thus $c=0$
A: Since $x$, $e^x$, and $\sin x$ are analytic, then their non-vanishing Wronskian is sufficient to conclude that the set is linearly independent.  
The Wronskian is equal to the determinant $W$ of the matrix given by 
$$\begin{bmatrix}
x & e^x & \sin x\\
1 & e^x & \cos x\\
0 & e^x &  -\sin x\\
\end{bmatrix}$$
Therefore, we have 
$$W=
\begin{vmatrix}
x & e^x & \sin x\\
1 & e^x & \cos x\\
0 & e^x &  -\sin x\\
\end{vmatrix}
=-xe^x(\sin x+\cos x)+2e^x \sin x$$
Inasmuch as $W\ne 0$ for all $x\in \mathscr{R}$ (e.g., take $x=\pi/2$), then $x$, $e^x$ and $\sin x$ are linearly independent.
A: Suppose
$ax +be^x +c\sin(x) =0
$.
If
$b \ne 0$,
then,
for large enough $x$,
$b e^x$
is much larger than
the other two terms,
which is a contradiction.
Therefore
$b = 0$,
so we have
$ax+c\sin(x)
= 0
$.
Again,
if $a \ne 0$,
we can choose $x$
large enough so  that
$|ax| > |c|$.
Since
$|\sin(x)| \le 1$,
this can not hold.
Therefore
$a = 0$,
so that
$c \sin(x) = 0$.
This obviously implies that
$c=0$.
And we are done.
