 Solution

No. For each $x>0$, $e^{-nx}\to 0$, whereas $e^{-n0}=1$ for every $n$. The pointwise limit on $[0,1]$ is thereforeThis limit is not 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 $n$.