Use L'Hopital's rule to show that u′(t) = −λ u(t) (question from a Probability course) I'm currently in a probability course and my professor suggested this question:
Consider a continuous and differentiable function u(t) such that u(t + s) = u(t)u(s), for any s ∈ R and t ∈ R. This function has the following properties: u(0) = 1; u is a non-increasing function (i.e. u′(t) ≤ 0, for all t); and lim u(t) = 0. 
I'm trying to show that u′ (t) = −λ u(t), for some constant λ > 0 by using L'Hopital's rule.
We're currently discussing the Poisson process and the Exponential distribution so my professor then added that we should show that the function u mentioned above has to be in the form of u(t) = e^(−λt). I'm not exactly sure what this has to do with what we've been learning, and if someone would be able to explain the significance of these two "show this" questions, I would appreciate the clarification.
 A: First of all, note that since $u(0)=1$ and $u$ is continuous, there is a positive real number $\delta$ such that $u(\delta)>0$. Then $u(n\delta)=u(\delta)^n>0$ for all $n$, and since $n\delta\to\infty$ and $u$ is non-increasing, this shows that $u(t)>0$ for all $t\geq0$.
Therefore we can define $v(t)=\log(u(t))$ for $t\geq 0$. The functional equation for $u$ implies that
$$ v(t+s)=v(t)+v(s) $$
for all $t,s\geq 0$, and since $v(t)$ is continuous this implies that $v(t)=\alpha t$ for some $\alpha\in\mathbb{R}$ (look up the Cauchy functional equation if you're not familiar with this).
Thus $u(t)=e^{v(t)}=e^{\alpha t}$, and since $u$ is non-increasing with $\lim_{t\to\infty}u(t)=0$, we can take $\alpha=-\lambda$ with $\lambda>0$.
A: $$u'(t)=lim_{\Delta x\to0}\frac{u(x+\Delta x)-u(x)}{\Delta x}$$
$$=lim_{\Delta x \to 0}u(x)\frac{u(\Delta x)-1}{\Delta x}$$
Now since both the denominator and the numerator in the fraction is in indeterminate form $\frac0 0$, we differentiate both $u(\Delta x)-1$ and $\Delta x$ with respect to $\Delta x$.
$$=u(x)lim_{\Delta x\to 0}u'(\Delta x)$$
Since $u(x)$ is differentiable, which means $u'(0)$ must exist,  so we can let $u'(0)= -\lambda$ where $\ lambda$ is a positive real number. So you have $u'(x)=-\lambda u(x)$. Since  $u'(x)$ is negative, so $u(x)$ must be positive for all $x$.
Solving the differential equation $u'=-\lambda u$ by noticing
$ln(u(x))'=\frac{u(x)}{u'(x)}$
So $$ln(u(x))=-\lambda x +C$$
$$u(x)=e^{-\lambda x}*e^C$$
Substitute $u(0)=1$ gives
$$1=e^C$$
So $$u(x)=e^{-\lambda x}$$
