# How do you find the Limit of lnx as x approaches 0?

Feb 6, 2017

${\lim}_{x \to 0} \ln x = - \infty$

#### Explanation:

First we prove that $\ln \left(x\right)$ is monotone increasing.

Consider:

${x}_{1} , {x}_{2} \in {\mathbb{R}}^{+}$ with ${x}_{2} > {x}_{1}$

$\ln {x}_{2} = \ln \left({x}_{2} / {x}_{1} \cdot {x}_{1}\right) = \ln \left({x}_{2} / {x}_{1}\right) + \ln {x}_{1}$

${x}_{2} > {x}_{1} \implies {x}_{2} / {x}_{1} > 1 \implies \ln \left({x}_{2} / {x}_{1}\right) > 0 \implies \ln \left({x}_{2}\right) > \ln \left({x}_{1}\right)$

which proves the point.

Since it is monotone increasing $\ln x$ has a limit for $x \to \infty$ and since the function is not bounded this limit must be $+ \infty$, so:

${\lim}_{x \to \infty} \ln x = + \infty$

Now note that:

$\ln \left(\frac{1}{x}\right) = - \ln x$

and that as the logarithm is defined only for $x > 0$

${\lim}_{x \to 0} \ln x = {\lim}_{x \to {0}^{+}} \ln x$

Substitute now $y = \frac{1}{x}$

${\lim}_{x \to {0}^{+}} \ln x = {\lim}_{y \to \infty} \ln \left(\frac{1}{y}\right) = {\lim}_{y \to \infty} - \ln \left(y\right) = - \infty$