# prove that lnx is continuous at 1 and at any positive real number a

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

• Related to this – Juniven Mar 14 '17 at 5:37

## 2 Answers

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 $|\ln x - \ln a| < \epsilon.$

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

(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)).

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