Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a bounded function with the properties $f(0)=0$ $\lim_{x\to\infty} f(x)=1$. For each $n\in\mathbb{N}$ define $f_n(x)=f(x+e^n)$ with $(x\in\mathbb{R})$
Show that $(f_n)_{n\in\mathbb{N}}$ is pointwise but not uniformly convergent
So to prove pointwise, I would do the following: $f_n(x)=f(x+e^n)$ so as $n\rightarrow\infty$ you would have $e^n \rightarrow\infty$ So I am unsure how to prove this? What would the limit function be?
For each $x \in \Bbb R$, $\lim_{n \to \infty} x + e^n = \infty$, and $\lim_{t\to\infty} f(t)=1$, therefore: $$ \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} f(x + e^n) = 1 \, .$$ So $(f_n)$ converges pointwise to the function $F(x) = 1$.
To prove that the convergence is not uniform, compute $f_n(x) - F(x)$ for $x = -e^n$.