Does $\int_0^\infty |f'(x)| dx < \infty$ conclude $\lim_{x\to \infty} f(x)<\infty $

Question

$f:[0,\infty) \to \mathbb R $ is $C^1$ and $$\int_0^\infty |f'(x)| dx < \infty$$ then can we prove that $\lim_{x\to \infty} f(x)$ exists and $$\lim_{x\to \infty} f(x)<\infty $$

My attempt: If $f'(x)$ be non-negetive (non-positive) for large amounts of $x$ then the easily we conclude that$$\lim_{x\to \infty} f(x)<\infty $$

Answer 1

Consider the auxiliary function $$F(x):=\int_0^x|f'(t)|\>dt\qquad(x\geq0)\ .$$ By assumption, the $\lim_{x\to\infty} F(x)$ exists. By Cauchy's criterion it follows that for each $\epsilon>0$ there is an $M\geq0$ with $$|f(y)-f(x)|\leq\int_x^y|f'(t)|\>dt=F(y)-F(x)<\epsilon\qquad(M<x\leq y)\ .$$ The "essential part" of Cauchy's criterion then allows to conclude that $\lim_{x\to\infty} f(x)$ exists (and is finite).

Answer 2

Hint: prove first that $f$ is bounded around $\infty$. Then assume that $x_n, y_n\to \infty$ with $x_n< y_n< x_{n+1}$ and that $f(x_n)\to X$, $f(y_n)\to Y$ and $X\neq Y$ and try to find a contradiction.