Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to prove the continuity of $f(x)=\log x$ using a $\epsilon-\delta$ proof

These is what I have so far but am not sure how to continue

$|\log x-\log a| < \epsilon$

$\log a- \epsilon < \log x < \log a+ \epsilon$

$\frac{a}{e^\epsilon} < x < {a}e^\epsilon$

Any help is appreciated

share|cite|improve this question
up vote 3 down vote accepted

By your inequality, the absolute value of the difference is $\lt \epsilon$ if $$\frac{a}{e^{\epsilon}}-a \lt x-a\lt ae^\epsilon -a$$ (we subtracted $a$ from each side of each of your two inequalities). Let $\delta=a\min\left(1-\frac{1}{e^{\epsilon}}, e^\epsilon -1\right)$.

Remark: Actually, $1-\frac{1}{e^{\epsilon}}$ is the smaller of the two, so in effect we are letting that be $\delta$. But we really don't need to bother finding that out: all we need to do is to show there is a $\delta$ that works.

share|cite|improve this answer
im having some trouble with the actual proof, setting delta to 1-1/e^epsilon and getting |log x-log a|<epsilon can you help? – PooperScooper Feb 28 '13 at 3:38
You proved it. To quote your post, more or less, $|\log x-\log a|\lt\epsilon$ if and only if $|\log(x/a)|\lt \epsilon$ iff $-\epsilon \lt \log(x/a)\lt \epsilon$ iff $ae^{-\epsilon}\lt x\lt ae^{\epsilon}$. All I did was to express your inequality not in terms of $x$, but in terms of $x-a$. – André Nicolas Feb 28 '13 at 3:44
I just thought that for a formal proof I need to algebraically show that |x-a|<$\delta$ implies |f(x)-f(a)|<$\epsilon$ – PooperScooper Feb 28 '13 at 3:55
You can rewrite it in that style. Suppose that $|x-a|\lt \delta$ (the $\delta$ I gave). Then reverse what I did, and get an inequality for $x$. Then reverse what you did, and get an inequality for $|\log x-\log a|$. But since almost everything was an if and only if statement (see my version of what you did, in comment above) reversal involves essentially no work. – André Nicolas Feb 28 '13 at 4:09
thank you very much sir – PooperScooper Feb 28 '13 at 14:57

Show it at $x = 1$. Then spread it around using the fact that $\log(xy) = \log(x) + \log(y)$.

share|cite|improve this answer
Indeed, the fact that the logarithm is continuous at $x=1$ suffices to show it is continuous over all the positive axis. Similarly, the fact that the exponential is continuous at $x=0$ suffices to show it is continuous over all the real line. – Pedro Tamaroff Feb 26 '13 at 3:10

Hint: from the right (i.e., $\,x>a\,$):

$$|\log x-\log a|=\log\frac{x}{a}<\epsilon\Longleftrightarrow \frac{x}{a}<e^\epsilon\Longleftrightarrow x<ae^\epsilon\;\;(\text{remember}:\;a,x>0\,\;!)\Longrightarrow$$

$$x-a<a(e^\epsilon -1)$$

and there you have your $\,\delta>0\ldots\,$

share|cite|improve this answer

What you are trying to prove is that for any fixed $x$, $\forall\, \epsilon>0\;\; \exists\, \delta>0$ such that $|\log(x+\delta)-\log x|<\epsilon$.

So, you need to find the $\delta$ in terms of $\epsilon$ and $x$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.