Prove that the logarithmic function is continuous on R.

[Portuga]

Homework Statement

Prove that f\left(x\right)=\log_{a}x is continuous for all \mathbb{R}.

Homework Equations

[/B]

I must find a \delta>0\in\mathbb{R} for a given \varepsilon>0
such that
<br /> \left|x-x_{0}\right|&lt;\delta\Rightarrow\left|\log_{a}x-\log_{a}x_{0}\right|&lt;\varepsilon.<br />

The Attempt at a Solution

.[/B]

I tried to use a direct proof solving \left|\log_{a}x-\log_{a}x_{0}\right|&lt;\varepsilon for x. But this gives rise to a pair of values: \delta=x_{0}\left(a^{\varepsilon}-1\right) and \delta=x_{0}\left(a^{-\varepsilon}-1\right). When I use them to build \left|x-x_{0}\right| &lt; \delta from the inequality
<br /> -\delta&lt;x-x_{0}&lt;\delta<br />
I see myself in big trouble, as there is no way to generate \left|\log_{a}x-\log_{a}x_{0}\right| &lt; \varepsilon.

Can someone give me a hint, or a new strategy? I have searched for help in other forums, but hints are very sophisticated to follow.
Thanks in advance.

 

[Orodruin]

How do you expect a function that is not defined at $x =0$ to be continuous on the entire real line?

 

[Portuga]

How do you expect a function that is not defined at $x =0$ to be continuous on the entire real line?

Oh, sorry, I forgot it! Continuous for all x&gt;0.

 

[fresh_42]

Why don't you try and see where $\log x - \log x_0 = \log \frac{x}{x_0}$ gets you? Those proofs can often be done by writing it the wrong way and start with $|f(x)-f(x_0)| < \varepsilon$ to obtain a condition for $\delta$ to get an idea and then turn the estimations, resp. conclusions around.

 

[pasmith]

You have solved <br /> -\epsilon &lt; \log_a (x/x_0) &lt; \epsilon correctly to get <br /> x_0(a^{-\epsilon} - 1) &lt; x - x_0 &lt; x_0(a^{\epsilon} - 1), but the interval is not symmetric about zero.

In this situation the closest end point to zero gives you your \delta. For suppose -b &lt; x - x_0 &lt; c for strictly positive b and c. If b &lt; c then |x - x_0| &lt; b implies -b &lt; x - x_0 &lt; b &lt; c as required; alternatively if b &gt; c then |x - x_0| &lt; c implies -b &lt; -c &lt; x - x_0 &lt; c, again as required.

 

[Portuga]

Thank you very much, pasmith! I was so close to the answer! Now I understand. Thank you very much.