Statement (a):
Let's say $\lim_{x\to \infty} = a$. if $a\neq 0$, then by definition of limit, there is a number $X$ such that for any $x \geq X$, we have $|f(x)| \geq a/2$. This means that from that point on, the integral $\int_0^xf(x)\,dx$ will increase (or decrease, depending on whether $a$ is positive or negative) by at least $a/2$ for each unit we increase $x$, and thus the improper integral doesn't exist. So if $a$ exists and the integral is finite, then $a = 0$ by this contradiction.
Statement (b) and (c):
I do these together because they can be disproven with the same counter-example. Start out with the function that is constantly $0$. Now, take the interval $[0.5, 1.5]$, and raise the graph as a triangle with a top in the point $(1, 2)$. Now the total integral is $1$. Then for the interval $[1.75, 2.25]$, raise a triangle so the top is at $(2, 2)$. The total integral is now $1.5$. Continue doing this around every integer point along the $x$ axis, always halving the interval length. You will end up with a function that integrates to $2$, but has bumps all over with height $2$. So no limit. It is also non-negative, so it fulfils (c).
Statement (d):
Taking the function above, you can "smooth out" the corners so that the function is differentiable, and the change to the total integral is finite. Now you have a differentiable function which has derivatives of whatever size you want, if you just go far enough out, so the derivative has no limit.
Hint Can you find a function $f: [n,n+1) \to [0,\infty)$ such that $\int_n^{n+1} f(x) dx \leq \frac{1}{n^2}$, $f(n)=f(n+1)=0$, but $f(n+\frac{1}{2}) > n$?
Try to draw the graph of such a function....
Best Answer
Define
$$f_k(x)=\begin{cases} k &\text{if } k\leq x \leq k+\frac{2^{-k}}{k} \\ 0\text{ otherwise} \end{cases}$$
Then define
$$f(x)=\sum_{i=1}^\infty f_i(x)$$
Then
$$\int_0^\infty f(x)dx=\int_0^{\infty}\sum_{i=1}^\infty f_i(x)=\sum_{i=1}^\infty \int_k^{k+\frac{2^{-k}}{k}}k=\sum_{i=1}^\infty 2^{-i}=1<\infty$$
EDIT Added the continuity condition.
Define
$$g_k(x)=\begin{cases} \frac{2^{-k-1} (x-k)}{k^2} &\text{if } k\leq x \leq k+\frac{2^{-k-1}}{k} \\ \frac{k^4+2^{-k-1} k (k-x)+2^{-2 (k+1)}}{k^3} &\text{if } k+\frac{2^{-k-1}}{k}\leq x \leq k+\frac{2^{-k}}{k} \\ 0\text{ otherwise} \end{cases}$$
Basically, these are just triangles that are under $f_k(x)$ that start at the beginning of the step function around $x=k$, hit the middle of the top of the step, and go back to the end of the step. This implies
$$g_k(x)\leq f_k(x)$$
Then for
$$g(x)=\sum_{k=1}^\infty g_k(x)$$
we have
$$g(x)=\sum_{i=1}^\infty g_i(x)\leq \sum_{i=1}^\infty f_i(x)<\infty$$
Since $g(x)$ is continuous, we are done.