Show that $e^{x}=\left | P(x) \right |$ has at least one real solution We have $P$ a polynomial that is not identically $0$. Show that $e^{x}=\left | P(x) \right |$ has at least one real solution.
I got $f(x) = e^{x}- \left | P(x) \right |$, which is continuous. I need to get an $x$ where $f$ is negative, and one more where $f$ is positive. In that case there will be an $x_{0}$ and $f(x_{0})=0$. 
The limit when $ x\rightarrow -\infty $ is $\displaystyle{ \lim_{x\rightarrow -\infty}f(x)=\lim_{x\rightarrow -\infty}e^{x}-\left | P(x) \right |= 0 - (+\infty) = - \infty . }$ The limit when $ x\rightarrow +\infty $ is $\displaystyle{ \lim_{x\rightarrow +\infty}f(x)=\lim_{x\rightarrow +\infty}e^{x}-\left | P(x) \right |= \lim_{x\rightarrow +\infty}e^{x}(1 - \frac{\left | P(x) \right |}{e^{x}}) }$ και το $ \lim_{x\rightarrow +\infty} \frac{\left | P(x) \right |}{e^{x}} $ είναι $\frac{+\infty}{+\infty}$ , so we use del'Hopital $n$-times (depending on $P$) and we will find that $ \lim_{x\rightarrow +\infty} \frac{\left | P(x) \right |}{e^{x}}=0 $ so $\lim_{x\rightarrow +\infty}e^{x}(1 - \frac{\left | P(x) \right |}{e^{x}})= +\infty$. We have that $f$ is continuous,so there is necessarily an $x_{0}$ where $f(x_{0})=0$. So there is necessarily at least one real solution. 
The problem is that $\left | P(x) \right |$ is not always differentiable. How can I fix it? Is there another solution? Thanks

$P$ being a polynomial then at infinity $|P(x)|\sim |a_n\,x^n|$ with $n=\deg(P)$.
Assuming the equation has no real solution, then $\begin{cases}\lim\limits_{x\to-\infty} |P(x)|=+\infty\\\lim\limits_{x\to-\infty} e^x=0\end{cases}$
 implies that $\forall x\in\mathbb R,\ |P(x)|>e^x$.
Now at $+\infty$ we get $|P(x)|\sim |a_n|x^n=o(e^x)$ so we can never realize $\dfrac{|P(x)|}{e^x}>1$ since this quantity tends to zero.
So there must exists at least one real solution to our equation.
