1
$\begingroup$

Definition :. We say that a function f is continuous at a provided that for any ε > 0, there exists a δ > 0 such that if |x−a| < δ then |f(x)−f(a)| < ε.

(a) Use the definition of continuity to prove that lnx is continuous at 1. [Hint: You may want to use the fact |lnx| < ε ⇔−ε < lnx < ε to find a δ.]

(b) Use part (a) to prove that lnx is continuous at any positive real number a. [Hint: ln(x) = ln(x/a) + ln(a). This is a combination of functions which are continuous at a. Be sure to explain how you know that ln(x/a) is continuous at a.]

for part a how can I find δ that works for if |x−a| < δ then |f(x)−f(a)| < ε. and for part b how can I show that ln(x/a) is continuous at a ?

please help me with part a and b

$\endgroup$
  • 1
    $\begingroup$ Related to this $\endgroup$ – Juniven Mar 14 '17 at 5:37
2
$\begingroup$

Every case here can be proved by showing that if $|x-a| < \delta = \min(a/2,a \epsilon/2)$, then $|\ln(x/a)| =|\ln x - \ln a| < \epsilon$.

You need the fact that if $y > 0$, then $\ln(1+y) < y$ which follows from $e^y > 1 +y$.

(1) If $x = a$, then for any $\epsilon >0$ we always have $|\ln x - \ln a| = 0 < \epsilon$

(2) If $x > a$ then

$$|\ln x - \ln a| = \ln \left(\frac{x}{a}\right) = \ln \left(1 + \frac{x-a}{a} \right) < \frac{x-a}{a} = \frac{|x-a|}{a},$$

and with $|x-a| < a \epsilon$ we have $| \ln x - \ln a| < \epsilon.$

(3) If $x < a$ then

$$|\ln x - \ln a| = \ln \left(\frac{a}{x}\right) = \ln \left(1 + \frac{a-x}{x} \right) < \frac{a-x}{x} = \frac{|x-a|}{x},$$

and with $|x-a| < \min (a/2, a\epsilon/2) $ we have $x > a/2$ and $|x-a|/x < 2|x-a|/a < \epsilon$.

Thus, if $|x- a| < \delta = \min(a/2,a\epsilon/2, a \epsilon) = \min(a/2,a\epsilon/2)$ then $|\ln x - \ln a| < \epsilon.$

$\endgroup$
  • $\begingroup$ how about part a , δ=e^ε or δ=(e^ε )-1? thanks $\endgroup$ – rian asd Mar 14 '17 at 15:58
  • $\begingroup$ I presented another approach. If you want to follow that route then see the link in the comment above. In particular, $$| \ln x| < \epsilon \iff -(1 - e^{-\epsilon}) < x - 1 < e^{\epsilon} -1 \iff |x-1| < \min(e^{\epsilon} -1, 1 - e^{-\epsilon})$$ $\endgroup$ – RRL Mar 14 '17 at 16:10
1
$\begingroup$

(a) Fix $\epsilon>0$. Then $|\ln(x)-\ln(1)|=|\ln(x)|$. By the hint, $|\ln(x)| < \epsilon$ if and only if $e^{-\epsilon}< x < e^\epsilon$. Does this suggest a $\delta$ that would work in the definition of continuity?

(b) $\ln(x/a)$ is continuous at $a$ it is the composition of $x \mapsto x/a$ (which is continuous), $y \mapsto \log y$ (which is continuous at $y=1$ by part (a)).

$\endgroup$
  • $\begingroup$ do you mean for part a δ=e^ε ? thanks $\endgroup$ – rian asd Mar 14 '17 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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