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.