$f: [a. \infty ) \rightarrow \mathbb R$ is differentiable and $\int_a^\infty f(t)dt $ and $\int_a^\infty f^{\prime}(t)dt $ are convergent then $f(t) \rightarrow 0$ as $t \rightarrow \infty$
my answer :
$\int_a^x f^{\prime}(t)dt =f(x) - f(a)$
$\int_a^\infty f^{\prime}(t)dt $ is convergent $\implies$ $\lim_{x \rightarrow \infty} f(x) = l $(where $l$ is finite ).to show $l=0$.
if $l \neq 0 \implies \exists <x_n> \rightarrow \infty $ such that $f(x_n) \geq \epsilon$ or $f(x_n) \leq \epsilon$ if i suppose $f(x_n) \geq \epsilon$ then as $f(x)$ is continuous then $f(x) \geq \frac{\epsilon}{2} \forall x \in(x_n-\delta,x_n+\delta)$ $\implies $ $\int_{x_n-\delta}^{x_n +\delta} f(t)dt \geq \epsilon\delta$ which will contradict the convergence of $\int_a^\infty f(t)dt $