A sequence of functions evaluated at a sequence

[AxiomOfChoice]

What are the rules if you have a sequence f_n of real-valued functions on \mathbb R and consider the sequence f_n(x_n), where x_n is some sequence of real numbers that converges: x_n \to x. All I have found is an exercise in Baby Rudin that says that if f_n \to f uniformly on E, then f_n(x_n) \to f(x) if x_n \to x is in E. But the exercise seems to indicate that it is possible to have f_n(x_n) \to f(x) for every sequence x_n\to x without having f_n \to f uniformly. (I believe the canonical example f_n(x) = x^n on E = [0,1] works here.)

I ask because I recently had a colleague who claimed that if x_n \to x, then \left( 1 + \frac{x}{n} \right)^n \to e^x. She asked what the rule that made this possible was, and I replied that I wasn't sure if it was, in fact, true, since f_n(x) = \left( 1 + \frac xn \right)^n obviously does not converge uniformly to e^x on \mathbb R (even though, obviously, f_n \to e^x pointwise).

 

[PeroK]

For $n$ large enough:

$f_n(x_n)$ is close to $e^{x_n}$; and $x_n$ is close to $x$ so $e^{x_n}$ is close to $e^x$

 

[AxiomOfChoice]

For $n$ large enough:

$f_n(x_n)$ is close to $e^{x_n}$; and $x_n$ is close to $x$ so $e^{x_n}$ is close to $e^x$

So is it always true that if f_n \to f pointwise and x_n \to x, then f_n(x_n) \to f(x)?

 

[PeroK]

So is it always true that if f_n \to f pointwise and x_n \to x, then f_n(x_n) \to f(x)?

Why don't you try to prove it?

Hint: Does $f_n(x)$ converge uniformly to $e^x$ on any bounded interval?

 

[AxiomOfChoice]

Why don't you try to prove it?

Hint: Does $f_n(x)$ converge uniformly to $e^x$ on any bounded interval?

Ok, I think I see the strategy you are suggesting. Since \{ x_n \} is bounded (since it converges), we have -M \leq x_n \leq M for all n and for some M. I can restrict my attention to [-M,M]. On that interval, \left( 1 + \frac xn \right)^n converges uniformly to e^x, and I can apply the result of the Baby Rudin exercise.

 

[AxiomOfChoice]

Here is what I have so far. I have decided to confine my attention to $[0,\infty)$. Suppose $f_n \to f$ uniformly on a set $E$, where each $f_n$ is continuous. (The latter is a hypothesis I inadvertently omitted from my previous posts.) Let $\{ x_n \}$ be a sequence of points in $E$ with $x_n \to x$. Then, given $\epsilon > 0$, there exists $N \in \mathbb N$ such that $n > N$ implies $|f_n(x) - f(x)| < \epsilon/2$ for all $x\in E$ and $|f(x_n) - f(x)| < \epsilon/2$, since the limit function $f$ is continuous. Then for $n > N$,

$|f_n(x_n) - f(x)| < |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$.

The terms of $\{ x_n \}$ are bounded; suppose they are contained in an interval $[-M,M]$. Then

$M_n = \sup_{x\in [-M,M]} \left| \left(1 + \frac{x}{n} \right)^n - e^x \right|$

is a decreasing sequence bounded from below by $0$; hence it converges to $0$; hence $\left( 1 + \frac{x}{n} \right)^n \to e^x$ uniformly on $[-M,M]$, which is what we wanted.