Consider $f_n: [0,1]\longrightarrow\mathbb{R}$ is given by
$$f_n(x) = \begin{cases} \sqrt{n}, & \quad 0<x<\frac{1}{n} \\ 1, & \quad \text{otherwise.} \end{cases}$$
1.) Determine the pointwise limit $f(x)=\lim_{n\longrightarrow \infty}f_n(x)$.
Choose $x\in [0,1]$ to be non-zero. Then there exists and $N\in\mathbb{Z}$ such that $N>\frac{1}{x}$ for all $n>N$.
Then $f_n(x)=0$ at $x$, and $f(x)=\lim_{n\longrightarrow \infty} f_n(x)=1$ for all $\frac{1}{N}<x$
Is this a sufficient proof of the pointwise limit?
2.) Does $(f_n)$ converge to $f$ in the $2$-norm given by $\|g\|_2=\left(\int^1_0|g(x)|^2\,dx\right)^{1/2}$?
So, $$\|f_n\|_2=\left(\int_0^\frac{1}{n} |\sqrt{n}|^2 \, dx\right)^\frac{1}{2} + \left(\int_\frac{1}{n}^1 |1|^2 \,dx\right)^\frac{1}{2}\longrightarrow 2.$$ As $2$ doesn't equal the pointwise limit, then $f_n$ doesn't converge to $f$ in the $2$-norm.
3.) Does $f_n\longrightarrow f$ uniformly?
For $f_n\longrightarrow f$ uniformly, $\int_a^b f(x)\,dx=\int_a^b f_n(x)\,dx$.
From calculating the pointwise limit we know that $\int_a^ bf(x)\,dx=1$. Now calculating $\int_a^bf_n(x)\,dx$ gives us $$\|f_n\|_1=\int_0^\frac{1}{n} |\sqrt{n}| \,dx+ \int_\frac{1}{n}^1 |1| \,dx\longrightarrow 1.$$ Thus, $\int_a^b f(x)\,dx=\int_a^bf_n(x)\,dx$ and $f_n\longrightarrow f$ uniformly. Is this correct?
1.) is not quite correct. For fixed $x$, we have $x>1/n$ for sufficiently large $n$; and then, $f_n(x)=1$. So the pointwise limit, $f$, is the function which is identically $1$. (I think this is what you really meant to say.)
2.) is not correct. Note you need to compute $\int_0^1 | f_n(x)-1|^2\,dx$. You'll have convergence in the $2$-norm if and only if these integrals converge to $0$ as $n$ tends to $\infty$. (Note alex's comment above here.)
The reasoning in 3) is not correct. If you have uniform convergence, then you have convergence of the integrals when they exist; but the converse need not hold. (In fact, this example shows that.)
Here, you do not have uniform convergence and you can show this simply by showing that the definition for uniform convergence is not satisfied: Let $\epsilon=1$. Then given any $N>1$, you can find an $x$ such that $|f_N(x)-f(x)|>1$.
So there is no $m$ such that $\Vert f_n-f\Vert_\infty <\epsilon$ for all $n\ge m$.