Solution
= Solution
\b[No.] For each $x>0$, $e^{-nx}\to0$, whereas $e^{-n0}=1$ for every $n$. The <pointwise convergence>[pointwise limit] on $[0,1]$ is therefore
$$
f(x)=\begin{cases}1,&x=0,\\0,&x>0.\end{cases}
$$
This limit is not <continuous function>[continuous] at $0$. Since every $f_n(x)=e^{-nx}$ is continuous, the <uniform limit theorem> rules out uniform convergence. Directly, $|f_n(x)-f(x)|=e^{-nx}$ for $x>0$, whose <supremum> over $(0,1]$ is $1$ for every $n$.