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

(a) Use the definition of continuity to prove that $\ln(x)$ is continuous at $1$. [Hint: You may want to use the fact $|\ln(x)| < \epsilon \iff − \epsilon < \ln(x) < \epsilon$ to find a $\delta$.]

(b) Use part (a) to prove that $\ln(x)$ is continuous at any positive real number $a$. [Hint: $\ln(x) = \ln(x/a) + \ln(a)$. This is a combination of functions that 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 $\delta$ that works for if $|x−a| < \delta$ then $|f(x)−f(a)| < \epsilon$, and for part (b) how can I show that $\ln(x/a)$ is continuous at $a$?

Please help me with parts (a) and (b).

amWhy
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rian asd
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2 Answers2

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

RRL
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  • 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
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(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)).

angryavian
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