# Limits of a function and its derivative

Let $u:[0,\infty) \to \mathbb{R}$ be a continuously differentiable function in $t$, and let $$t^{n-1} u'(t) + \frac{1}{2} t^n u(t) = C$$ for some constant $C$ and positive integer $n$.

Suppose that $\displaystyle\lim_{t\to +\infty} u(t) = 0$ and $\displaystyle\lim_{t\to +\infty} u'(t) = 0$.

How can one show that then $C=0$?

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What do we assume on $n$ and $t$ in the equation? – Davide Giraudo Sep 5 '11 at 20:15
Good question. $t$ is in the domain of $u$ where $u$ is differentiable, so $t \in (0,\infty)$. $n$ is some integer. – user1736 Sep 5 '11 at 20:58

It's a linear differential equation. Can't you just solve it and then see what the limit conditions imply about $C$ and the constant of integration?

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@Gerry: That's possible, but in the book that I am reading, the author first shows that the constant is equal to zero, then simplifies the equation to $u'(t) = \frac{1}{2} t u(t)$, and finally proceeds to solve this differential equation for $u$ from there. I'm guessing that this means that it is not necessary, or at the very least, that he did not intend the reader to solve the equation in the initial form. – user1736 Sep 5 '11 at 6:10