Let $f:\mathbb R\to \mathbb R$ be a continuous function such that $\int_0^\infty f(x) dx$ exists. Then I claim that only a) is correct out of the following four options-
a)if $\lim_{n\to \infty} f(x)$ exists then it is must be equal to $0$.
b) the limit in a) must exist and equal to $0$,
c) if f is non negative then limit in a) must exist.
d) if f is differentiable, then $\lim_{n\to \infty} f'(x)$ must exist and equal to $0$.
Proof:
Choice $(a)$:
Let $g$ be a function defined as $g(t):=\int_0^tf(x)\,dx $ for all $t\in \mathbb R$. By fundamental theorem of Calculus, $g$ is differentiable for every $t\in \mathbb R$ and $g'(t)=f(t)$.
$\int_0^\infty f(x) \,dx$ exists $\implies \lim_{t\to \infty} g(t)$ exists $\implies \text{the sequence } (g(n))$ converges $\implies \lim_{n\to \infty} g(n+1)-g(n)=0$.
By LMVT on $[n, n+1]$, where $n\in \mathbb N$, there $\exists c_n \in (n,n+1)$ such that $$\frac{g(n+1)-g(n)}{n+1-n}=g'(c_n)=f(c_n)$$
Taking limit on both sides, it follows that $f(c_n)\to 0$ which implies that "if $\lim_{x\to \infty} f(x)$ exists then it must be zero."
Choice $(b)$: Counterexample can be constructed as follows: Let's consider a triangle OAB of height $1$ and base of length $1$ with $O$ as origin, let's call this triangle T1. Then, inductively define $Tn$ as the triangle which has base length equal to half of base length of $T_{n-1}$. Also, suppose all Tn's have the same height that is 1. These Tn's define a function g with domain $[0,\infty)$ and range =$[0,1/2]$. This function can be extended to $f:R\to R$ by setting $f(x)=0$ on $x<0$ and $f(x):=g(x)$ on $x\ge 0$. Note that $f$ is continuous on $\mathbb R$ and that $\int_0^\infty f(x)\,dx=\frac 12 +(\frac 12)^2+...=1$ but $\lim_{x\to \infty} f(x)$ does not exist.
By "choice $(b)$" section above, choice $(c)$ is also not correct.
Choice d): while writing this post, this post came up and I'm satisfied with the very nice explanation for option d) by Mr. Arthur.
If someone reviews my proof and lets me know of any mistake therein, I'll greatly appreciate it. Thanks in advance.
