fn is a sequence of uniformly continuous real-valued functions on R.
and fn converges pointwise to f. which is continuous but not uniformly continuous.
We can assume that $\displaystyle \lim _{x \to \infty } f(x) = L $
In this case, $\displaystyle \varepsilon > 0 \Rightarrow \left( {\exists N} \right)\left[ {x \geqslant N \Rightarrow \left| {f(x) - L} \right| < \frac{\varepsilon }{2}} \right]$.
Now I ask you is $\displaystyle f$ uniformly continuous on $\displaystyle \left[ {0,N} \right]$? (Why or Why Not?)
So what is your answer?
no, i think. for that function to be uni. continuous on [a,b], we need:
$\displaystyle \forall \epsilon>0, \ \exists \delta>0, \forall x,y\in[0,N], \ |x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon$
and f(y) takes value L at $\displaystyle y\rightarrow\infty$
and therefore we cannot have an arbitary small distance between x,y.
i appreciate your way of "teaching."