Suppose $f$ is continuous and $F(x) = \int_{a}^{x} f(t) \ \mathrm{d}t$ bounded on $[a,b)$.
Does $\lim_{x \to b^{-}} F(x)$ exist?
Since $F$ is bounded on $[a,b)$, there exists $L > 0$ such that
$$-L \leq F(x) \leq L$$
for all $x \in [a,b)$. taking the limits
$$\lim_{x \to b^{-}} - L \leq \lim_{x \to b^{-}} F(x) \leq \lim_{x \to b^{-}} L$$ $$- L \leq \lim_{x \to b^{-}} F(x) \leq L$$
The squeeze theorem cannot be used since $-L \neq L$. Must this be true?
Take any continuously differentiable and bounded function $g:[a, b)\to \Bbb R$ for which $\lim_{x \to b^{-}} g(x)$ does not exist, and set $f = g'$.
Then $f$ is continuous on $[a, b)$ and $F(x) = \int_{a}^{x} f(t) dt = g(x)-g(a)$ is bounded, but $\lim_{x \to b^{-}} F(x)$ does not exist.
A simple example is $f(x) = \frac{d}{dx} \sin(1/x) = -\cos(1/x)/x^2$ on $[-1, 0)$.
Note that the situation is different if $f$ is continuous and bounded. Then $F$ is uniformly continuous, and that implies the existence of $\lim_{x \to b^{-}} F(x)$.