One-sided infinite limit of csc(x) as x approaches 0 from the left

Write $\csc(x) = \dfrac{1}{\sin(x)}$. As $x \to 0^-$, meaning $x$ is a small negative number close to $0$, the sine function is also small and negative, since $\sin(x) \approx x$ for $x$ near $0$ and $\sin(x) < 0$ for small negative $x$. So $\sin(x) \to 0^-$ as $x \to 0^-$.

Since the denominator $\sin(x)$ approaches $0$ through negative values while the numerator stays fixed at $1$, the fraction $\dfrac{1}{\sin(x)}$ becomes a large negative number, growing without bound in magnitude. Therefore $\lim_{x \to 0^-} \csc(x) = -\infty$, confirming that $x=0$ is a vertical asymptote of $f(x)=\csc(x)$.