Prove $\ln(x)$ is continuous using $\epsilon-\delta$

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

The argument depends on the definition of $\log(x)$. Personally, I prefer to define the logarithm by $$ \log(x) = \int_{1}^{x} \frac{1}{t}\,\mathrm{d}t, $$ where $x > 0$.

Fix some $a > 0$. To show that $\log$ is continuous at $a$, let $\varepsilon > 0$, choose $\delta < \frac{a}{2} \min\{\varepsilon, 1\}$, and suppose that $$0 < |x-a| < \delta.$$ Then $$ |\log(a) - \log(x)| = \left| \int_{1}^{a} \frac{1}{t}\,\mathrm{d}t - \int_{1}^{x} \frac{1}{t}\,\mathrm{d}t \right| = \left| \int_{x}^{a} \frac{1}{t}\,\mathrm{d}t \right| \le \left| \int_{x}^{a} \frac{1}{\min\{x,a\}} \,\mathrm{d}t \right|, $$ where the inequality follows from the observation that $1/t$ is a decreasing function, hence the value of the integrand is maximized at one of the endpoints of integration. From the assumption that $|x-a| < a/2$, it follows that $x > a/2$. Thus $$\min\{x,a\} \ge \min\{a/2,a\} = \frac{a}{2} \implies \frac{1}{\min\{x,a\}} \le \frac{2}{a}.$$ Therefore $$ |\log(a)-\log(x)| \le \left| \int_{x}^{a} \frac{1}{\min\{x,a\}} \,\mathrm{d}t \right| \le \left| \int_{x}^{a} \frac{2}{a} \,\mathrm{d}t\right| = \frac{2}{a} |a-x| < \frac{2}{a} \delta < \frac{2}{a} \cdot \frac{a}{2} \varepsilon = \varepsilon. $$

Answer 5

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 $x$.