Two questions about the initial value theorem for pdfs

Question

Let $f$ be a probability density function (pdf) of a non-negative random variable. Let the Laplace transform of $f$ be defined as \begin{align} \phi(t)=\int_0^\infty e^{-tx} f(x)\, dx. \end{align}

The initial value theorem states that \begin{align} \lim_{t \to \infty} t \phi(t)=\lim_{x \to 0^{+}} f(x). \end{align}

I have the following questions about this result

1. What is the minimal condition on $f$ such that this identity holds? For example, will it hold if we only impose $\lim_{x \to 0^{+}} f(x)<\infty$? Note that one sufficient condition is that $f$ is bounded, which follows from the dominated convergence theorem.
2. What happens if $\lim_{x \to 0^{+}} f(x)=\infty$? Does $\lim_{t \to \infty} t \phi(t)=\lim_{x \to 0^{+}} f(x)=\infty$? Or, do we have an example where $\lim_{t \to \infty} t \phi(t)<\infty$?

Here is the proof of IVT

We use the assumption that $f$ is bounded.

\begin{align} \lim_{t \to \infty} t \phi(t) &=\lim_{t \to \infty} t \int_0^\infty e^{-tx} f(x) \,dx \\ &=\lim_{t \to \infty} \int_0^\infty e^{-x} f\left(\frac{x}{t}\right) dx \\ &= \int_0^\infty e^{-x} \lim_{t \to \infty} f\left(\frac{x}{t}\right) dx \quad \text{ using DCT and assumption $f$ is bounded}\\ &= f(0^{+}) \int_0^\infty e^{-x} \lim_{t \to \infty} dx \\ &= f(0^{+}) \end{align}