# Question #36d91

Jan 8, 2017

We can demonstrate that for every for every ${x}_{0} \in \left(0 , + \infty\right)$ we have:

${\lim}_{x \to {x}_{0}} \log x = \log {x}_{0}$

using the $\delta - \epsilon$ definition of limit.

#### Explanation:

We have to prove that for every ${x}_{0} \in \left(0 , + \infty\right)$ we have:

${\lim}_{x \to {x}_{0}} \log x = \log {x}_{0}$

that is for every $\epsilon > 0$ we can find ${\delta}_{\epsilon} > 0$ such that for every $x \in \left({x}_{0} - {\delta}_{\epsilon} , {x}_{0} + {\delta}_{\epsilon}\right)$ we have:

$\left\mid \log x - \log {x}_{0} \right\mid < \epsilon$.

Now we can restrict ourselves to natural logarithms since:

${\log}_{a} x = \ln \frac{x}{\ln} a$

and a continuous function multiplied by a constant is still continuous.
Then we note the following properties of logarithms:

$\left(i\right) \ln \left(x\right) - \ln \left({x}_{0}\right) = \ln \left(\frac{x}{x} _ 0\right)$
$\left(i i\right) {x}_{1} < {x}_{2} \implies \ln {x}_{1} < \ln {x}_{2} , \text{ since } e > 1$

Now, given ${x}_{0} \in \left(0 , + \infty\right)$ and $\epsilon > 0$ we choose:

${\delta}_{\epsilon} = \min \left({x}_{0} \left({e}^{\epsilon} - 1\right) , {x}_{0} \left(- {e}^{-} \epsilon + 1\right)\right)$

Since $\epsilon > 0$, ${e}^{\epsilon} > 1$ and ${e}^{- \epsilon} < 1$ so both number are positive.

Now we consider $x \in \left({x}_{0} - {\delta}_{\epsilon} , {x}_{0} + {\delta}_{\epsilon}\right)$ and we distinguish the two cases:

(1) $x > {x}_{0}$

$\left\mid \ln x - \ln {x}_{0} \right\mid = \ln x - \ln {x}_{0} = \ln \left(\frac{x}{x} _ 0\right) < \ln \left(\frac{{x}_{0} + {x}_{0} \left({e}^{\epsilon} - 1\right)}{x} _ 0\right) = \ln {e}^{\epsilon} = \epsilon$

(2) $x < {x}_{0}$

$\left\mid \ln x - \ln {x}_{0} \right\mid = \ln {x}_{0} - \ln x = \ln \left({x}_{0} / x\right) < \ln \left({x}_{0} / \left({x}_{0} - {x}_{0} \left(- {e}^{-} \epsilon + 1\right)\right)\right) = \ln \left(\frac{1}{e} ^ - \epsilon\right) = \ln \left({e}^{\epsilon}\right) = \epsilon$

In either case we have:

$\left\mid \ln x - \ln {x}_{0} \right\mid < \epsilon$

and the continuity is proved.