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


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.

  • $\begingroup$ 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? $\endgroup$ – PooperScooper Feb 28 '13 at 3:38
  • $\begingroup$ 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 $x-a$. $\endgroup$ – André Nicolas Feb 28 '13 at 3:44
  • $\begingroup$ I just thought that for a formal proof I need to algebraically show that |x-a|<$\delta$ implies |f(x)-f(a)|<$\epsilon$ $\endgroup$ – PooperScooper Feb 28 '13 at 3:55
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Feb 28 '13 at 4:09

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 $\,\delta>0\ldots\,$


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

  • 1
    $\begingroup$ 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. $\endgroup$ – Pedro Tamaroff Feb 26 '13 at 3:10

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 $x$.


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