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

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 $-\infty$.

• The left hand limit (i.e. $\lim_{x \to 0^-} \ln(1/x)$) is not defined, since you are evaluating $\ln$ outside of its domain. $\lim_{x \to 0^+} \ln(1/x)$ is $+\infty$, since it is the same as $\lim_{y \to +\infty} \ln(y)$ – Ian Jul 15 '15 at 14:03
• @Ian Yes, I have edited my question in order to state that, and, now, I believe that I am correct. – Taylor Jul 15 '15 at 14:04
• $\lim_{\alpha \to 0^{+}}{\left(\ln{\left(\frac{1}{\color{red}{\alpha}}\right)}\right)} = \color{red}{\infty}$ – Zain Patel Jul 15 '15 at 14:06
• @ZainPatel Oops... :') – Taylor Jul 15 '15 at 14:07
• What I'm trying to say is that as you make $\alpha$ closer to $0$, $\frac{1}{\alpha}$ gets closer and closer to infinity and $\ln$ of something getting close to infinity gets closer to infinity itself. $i.e \ln \infty = \infty$. Not $-\infy$ as your question suggests. – Zain Patel Jul 15 '15 at 14:18

## 2 Answers

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*}

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.