Let $f_n(x) = x^n + \frac{e^2x}{n^2}$. As $n \to \infty$, $f_n$ converges pointwise to a function $f$ on $[0,1]$.
(i) What is $f$?
(ii) Is this convergence uniform?
(iii) Let $p$ be fixed (but arbitrary) with $0 < p < 1$. Show that $f_n \to f$ on $[0,p)$.
(iv) Is the convergence uniform on $[0,p)$?
the sequence $(f_n)$ converges pointwise to the function $f$ defined by $$f(x)=\left\{\begin{array}\\ 0\quad\text{if}\ 0\leq x<1 \\ 1\quad\text{if}\ x=1\end{array}\right.$$ and since $f$ isn't continuous whereas $f_n$ are continuous then the convergence isn't uniform.
On the interval $[0,p)$ the limit function is $f\equiv 0$ and we have $$|f_n(x)-f(x)|=x^n+\frac{e^2x}{n^2}\leq p^n+\frac{e^2p}{n^2}\to 0\quad \forall x\in[0,p)$$ so the sequence is uniformly convergent to $f$ on $[0,p)$.