Let us consider a sequence $(p_l)_l$ of polynomials on $[0,1]$ that converge uniformely, as $l\to \infty$, to a function $f$ defined on $[0,1]$.
I denote the polynomials $p_l(t) = \sum_{k=0}^{m(l)} a_k(l) t^k$.
I have also a certain quantity $\lambda(p_l) \in \mathbb{R}$ which has the following asymptotic behaviour :
$\lambda(p_l)(x) = \sum_{k=0}^{m(l)} a_k(l)\tfrac{1}{1+k/x} \underset{x\to\infty}{\sim} \sum_{k=0}^{m(l)} a_k(l) \sum_{j=0}^{\infty}(1)^j (\tfrac{k}{x})^j$.
So:
$\lambda(p_l)(x) \underset{x\to\infty}{\sim} \sum_{k=0}^{m_l}a_k(l)  \sum_{k=0}^{m(l)} a_k(l)\tfrac{k}{x} + ... = p_l(1)  \tfrac{p_l'(1)}{x} + ...$
Naively, I want to apply the limit to obtain:
$\lambda(p_l)(x) \underset{x\to\infty}{\sim} f(1)  \tfrac{f'(1)}{x} + ...$
Questions : Do I have first check some hypothesis to write it? Do you have any reference about applying the limit in an asymptotic behavour?
Thank you a lot.

$\begingroup$ You can't expect $\lambda(p_\ell)(x) \underset{x\to\infty}{\sim} f(1)  \tfrac{f'(1)}{x} + ...$ for all $\ell$, because $p_\ell$ for fixed $\ell$ is just an arbitrary polynomial. You must have some limit as $\ell \to \infty$ in mind. $\endgroup$ – Robert Israel Jan 21 '13 at 15:23

$\begingroup$ $p_\ell(1) \to f(1)$ by your convergence assumption, but ${p_\ell}'(1)$ can be much different from $f'(1)$. In fact the uniform limit of polynomials on $[0,1]$ need not be differentiable at all. $\endgroup$ – Robert Israel Jan 21 '13 at 15:28

$\begingroup$ Of course, I had $l\to \infty$ in mind. For your second remark, what if I assume f smooth around 1? Don't the derivatives of $p_l$ converge to the derivatives of $f$ at 1, as $l\to \infty$? $\endgroup$ – pilipilax Jan 21 '13 at 15:38

$\begingroup$ No they don't. For example, if $p_j(x) = T_j(x)/j$ where $T_j$ is the $j$'th Chebyshev polynomial of the first kind, $p_j \to 0$ uniformly on $[1,1]$ but $p_j'(1) = j$. $\endgroup$ – Robert Israel Jan 21 '13 at 19:19

$\begingroup$ Here is the reasonning that maybe makes me misunderstand: Let $a\in [0,1]$ on which every $p_l$ and $f$ are differentiable. In $\tfrac{p_l(x)p_l(a)}{xa}=\tfrac{p_l(x)f(a)}{xa} + \tfrac{f(a)p_l(a)}{xa}$, the first term tends to $\tfrac{f(x)f(a)}{xa}$ and the second to $0$ as $l\to\infty$. So when $x\to a$, we have the equality of derivatives. Am I wrong? $\endgroup$ – pilipilax Jan 22 '13 at 10:23
Note that the quantity $\lambda p$ depending on the parameter $x$, as you define it for polynomials $p(t):=\sum_{k=0}^m a_k t^k$, that is
$$(\lambda p )(x)=\sum_{k=0}^m a_k \frac{1}{1+k/x}\, , $$
can be extended to a positive, bounded linear operator
$$\lambda :C^0([0,1])\to C^0_b([0,\infty))$$
taking the function $f\in C^0(I)$ to
$$(\lambda f)(x):=\int_0^1 xt^{x1}f(t)dt\, .$$ In particular, $(\lambda f)(x)=f(1)+o(1)$ as $x\to+\infty$. For $f\in C^1([0,1])$ we also have, integrating by parts $$(\lambda f)(x)=f(1)\int_0^1 t^x f'(t)dt=f(1)f'(1)/x+o(x^{1}),\qquad (\mathrm{as}\, x\to+\infty) \, .$$
$$*$$
You can also write, changing variable in the integral, ($t=e^{\tau}$)
$$(\lambda f)(x):=x \int_0^{+\infty} e^{\tau x}f(e^{\tau})d\tau\, .$$
Thus your $\lambda f$ is the Laplace transform of $f(e^{\tau})$, times $x$. Thus you can profit of the asymptotic theory for Laplace transform of functions, developed in most textbooks on the subject.