$\ln{\left(\frac{1}{0}\right)} = -\infty$?

Question

I have shown it using a theorem that I made, but I am not sure, as $\lim_{\alpha \to 0^{-}}{\left(\frac{1}{\alpha}\right)} = -\infty$, and $\lim_{\alpha \to 0^{+}}{\left(\frac{1}{\alpha}\right)} = \infty$, and, therefore, no limit exists for $\lim_{\alpha \to 0}{\left(\frac{1}{\alpha}\right)}$, and, thus, no limit exists for $\lim_{\alpha \to 0}{\left(\ln{\left(\frac{1}{\alpha}\right)}\right)}$.

Edit

Under these circumstances, I suppose that it would not make sense to evaluate $\lim_{\alpha \to 0^{-}}{\left(\ln{\left(\frac{1}{0}\right)}\right)}$, as the logarithm of negative numbers is undefined, which means that one would only have to evaluate the limit $\lim_{\alpha \to 0^{+}}{\left(\ln{\left(\frac{1}{0}\right)}\right)}$, which I believe to be equal to $-\infty$.

Answer 1

The left hand limit will not be defined for $\ln$, be careful. Then note that \begin{align*} \lim_{x \rightarrow 0^{+}} \ln\bigg(\frac{1}{x}\bigg) &= \lim_{x \rightarrow 0^{+}} \ln(1) - \ln(x) \\ &= \lim_{x \rightarrow 0^{+}} - \ln(x) \\ &= \infty \end{align*}

Answer 2

The domain of the (real) logarithm function is the set $(0,\infty)$. In particular, $\displaystyle \lim_{\alpha \to 0^-} \ln \left( \frac 1 \alpha \right)$ is undefined. On the other hand, $\displaystyle \lim_{\alpha \to 0^+} \ln \left( \frac 1 \alpha \right) = \infty$ via the limit laws.