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\centerline{\bf Math 313 , Introduction to Analysis }
\centerline{uniform continuity - worked example 2}

\bigskip
 
   {\bf Example:} Show that the function $f(x) = \ln(x) $ on the   interval $[1, \infty)$ is uniformly continuous.
   
   {\bf Solution:} Given $\epsilon >0$ we want to find $\delta > 0$ such that for all $1 \leq c$ and any $| h | < \delta$, then $|f(c+h) - f(c) | < \epsilon$. 
      By the definition of $f(x)$ this means we need 
  $$|f(c+h) - f(c) | = |\ln(c+h) - \ln(c)| < \epsilon $$
The first problem is now, we need a definition of the   function $\ln(x)$! Recall from Calculus I that $\ln(x)$ is the unique differentiable  function defined for $x > 0$ satisfying the two conditions
$$\ln(1) = 0 ~ {\rm and }~ \ln'(x) = \frac{1}{x}$$
Since $\ln(x)$ is differentiable at every point $x >0$, it is also continuous at every point $x > 0$. 

We also need the rules of $\ln(x)$, that $\ln(a \cdot b ) = \ln(a) + \ln(b)$. 

Then since $\ln(1) = 0$, this implies $0 = \ln(b/b) = \ln(b) - \ln(1/b)$, and so $\ln(1/b) =   - \ln(b)$.

Combining these two properties we get $\ln(a/b) = \ln(a) - \ln(b)$.

Now we can rewrite the estimate we need to be
  $$|f(c+h) - f(c) | = |\ln(c+h) - \ln(c)|  = |\ln \Big( \frac{c+h}{c} \Big)| = |\ln \Big(1 + \frac{h}{c} \Big)| < \epsilon $$
When $h \to 0$ the argument  $1 + \frac{h}{c} \to 1$, so we need continuity of $\ln(x)$ at $x = 1$, or continuity of the composition $y \mapsto \ln(1 + y)$ at $y = 0$. This is true since $\ln(1+y)$ has a derivative at $y =0$ so it is continuous at $y=0$. We write out what this means:

For $\epsilon > 0$ given, there is some $\lambda > 0$ so that 
$$|y| = | y - 0| < \lambda ~ \Longrightarrow ~ | \ln(1 + y) - \ln(1)| = |\ln(1+y)| < \epsilon$$
This means that if $|\frac{h}{c} |  < \lambda$ then $ |\ln(1+\frac{h}{c})| < \epsilon$. 
Rewrite this as $|h| < c \cdot \lambda$ then  $ |\ln(1+\frac{h}{c})| < \epsilon$.
 
 Since $c \geq 1$, if we assume $ |h| < \lambda $ then $ |h| < c \cdot \lambda$, and then $ |\ln(1+\frac{h}{c})| < \epsilon$.
    
   We need to find $\delta > 0$ so the conclusion $ |\ln(1+\frac{h}{c})| < \epsilon$ is true, so take $\delta = \lambda$, the same $\lambda$ chosen  using continuity of $\ln(1+y)$ at $y = 0$. Then 
   $$|h| < \delta \Longrightarrow   |\ln(1+\frac{h}{c})| < \epsilon \Longleftrightarrow  |\ln(c+h) - \ln(c)| < \epsilon $$
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