Chapter 4 #3

Use the definition to find the derivative of $f(x)=\sqrt {x}$, for $x>0$.  Is $f$ differentiable at zero?  Explain.

$\lim_{t\rightarrow 0} \dfrac {\sqrt {x+t}-\sqrt {x}}{t}$

     $=\lim_{t\rightarrow 0} \dfrac {\sqrt {x+t}-\sqrt {x}}{t} * \dfrac {\sqrt {x+t}+\sqrt {x}}{\sqrt {x+t}+\sqrt {x}}$

     $=\lim_{t\rightarrow 0} \dfrac {x+t-x}{t[\sqrt {x+t}+\sqrt {x}]}$

     $=\lim_{t\rightarrow 0} \dfrac {1}{\sqrt {x+t}+\sqrt {x}}$

     $=\dfrac {1}{2\sqrt {x}}$

$f(x)=\sqrt {x}$ is not differentiable at $x=0$ because the limit of $T(x)=\dfrac {f(x)-f(x_0)}{x-x_0}$ as $x$ approaches zero does not exist.

$\lim_{x\rightarrow 0} \dfrac {\sqrt {x}-\sqrt {0}}{x-0}$

     $=\lim_{x\rightarrow 0} \dfrac {1}{\sqrt {x}}$

     $=\dfrac {1}{0}$ and is undefined.

If continuous, then Riemann integrable

Theorem 5.4 
If $f:[a,b]\rightarrow R$ is continuous, then $f\in R(x)$ on $[a,b]$.

*The domain of the function is closed and bounded, making it compact.  Having a compact domain implies the function is uniformly continuous, and with this we can say that $f$ is bounded (by previous theorems).


By the definition of uniform continuity, 

          choose an $\epsilon >0$ then there is a $\delta >0$ 

          such that $\left| x-y\right| <\delta$ with $x,y\in [a,b]$ 

          implies $|f(x)-f(y)|<\dfrac {\epsilon}{b-a}$.


Let $P=\{x_0, ..., x_n\}$ be a partition 

          such that $x_i-x_{i-1}<\delta$ for $i=1, ...,n$.

          Then there are $t_i,s_i\in [x_{i-1}, x_i]$ 

          such that $M_i(f)=f(t_i)$ and $m_i(f)=f(s_i)$.

Since $x_i-x_{i-1}<\delta$ then certainly $|t_i-s_i|<\delta$ 

          which implies$|f(t_i)-f(s_i)|<\dfrac {\epsilon}{b-a}$.  

          By a previous theorem, we know $f(t_i)\geq f(s_i)$, so we have a stronger inequality:

          $0\leq f(t_i)-f(s_i)<\dfrac {\epsilon}{b-a}$.

We now have the following:

$U(P,f)-L(P,f)=\sum _{i=1}^{n}[f(t_i)-f(s_i)](x_i-x_{i-1})$

                                $<\sum _{i=1}^{n}\dfrac {\epsilon}{b-a}(x_i-x_{i-1})$

                                $=\dfrac {\epsilon}{b-a}(b-a)=\epsilon$

Therefore, $f\in R(x)$ on $[a,b]$.

Notes

4.1 The Derivative of a Function


"The basic idea in calculating the slope of the tangent line at a point $P_0$ on the curve is to find the slope of the line through $P_0$ and another point $P$ on the curve, and then to take the limit of that slope as $P$ approaches $P_0$.  If that limit exists, it is defined to be the slope of the curve at $P_0$."

If the coordinates of point $P_0$ are $(x_0, y_0)$ and the coordinates of point $P$ are $(x, y)$, then the slope of the line passing through those points is

          $\dfrac {y-y_0}{x-x_0}$    or    $\dfrac {f(x)-f(x_0)}{x-x_0}$.

Let $f:D \rightarrow R$ with $x_0 \in D$ an accumulation point of $D$ and $x_0 \in D$ with $x \neq x_0$, define $T(x)= \dfrac {f(x)-f(x_0)}{x-x_0}$.

Let $f:D \rightarrow R$ with $x_0$ an accumulation point of $D$ and $x_0 \in D$.  For each $t \in R$ such that $x_0 + t \in D$ and $t \neq 0$, define $Q(t)= \dfrac {f(x_0 + t)-f(x_0)}{t}$. 

Prove that the two definitions above are equivalent.

Limit of a Function

 How would one prove that the limit of this function, 
$f(x)=x-2$ where $x\neq-2$,
is $L=-4$ at the excluded point?







Lets define $f:(-2,0) \rightarrow R$ by $f(x) = \dfrac {x^2 - 4} {x+2}$. 
    
By definition of a limit, if $x_0$ is an accumulation point of the domain, then $f$ has a limit $L$ at $x_0$ iff for each $\epsilon >0$, there is a $\delta >0$ such that if $0<|x-x_0|<\delta$ then $|f(x) - L|<\epsilon$.


Choose $\epsilon = \delta$
     If $0<|x+2|<\delta$ 
          Then $|f(x) - L| = |\dfrac {x^2 - 4} {x+2} +4| = |(x-2)+4| = |x+2|<\epsilon = \delta$ 

Therefore, $f(x)$ has a limit at $x=-2$, and that limit is $L=-4$.

Uniformly continuous, continued

Aha!  Class on Friday revealed some important information in how this proof works out.   

Since $D$ is a bounded set, there are $x_1, x_2, ... , x_n \in D$ such that
$D\subset \cup^n _{i=1} (x_i - \delta, x_i + \delta)$

Because D is bounded, we can find an open cover: a collection of open sets that contains the original set.

For all $x \in D$, then $x_i - \delta \leq x \leq x_i + \delta$
Then $|x - x_i| < \delta$
Which implies $|f(x) - f(x_i)| < 1$
     But that just means $f(x_i) - 1 \leq f(x) \leq f(x_i) + 1$

Therefore, $f(D)$ is bounded.

Uniformly continuous

I will begin with a definition of uniformly continuous directly from the book

A function $f: D \to R$ is uniformly continuous on $E\subset D$ iff for every $\epsilon >0$, there is \delta >0 such that if $x,y\in D$ with $|x-y|<\delta$, then $|f(x) - f(y)|<\epsilon$. If $f$ is uniformly continuous on $D$, we say $f$ is uniformly continuous.

That definition is helpful in proving this theorem

3.9 If $f: D\to R$ is uniformly continuous and $D$ is a bounded set, then $f(D)$ is a bounded set.

Choose \epsilon =1. 
     Then \exists \delta >0 such that
          if $|x-y| <\delta$
          then $|f(x) - f(y)| <1$

Since $D$ is a bounded set, there are $x_1, x_2, ..., x_n \in D$ such that
     $D\subset \cup^n _{i=1} (x_i-\delta, x_i+\delta)$
          Then $f(D)\subset \cup^n _{i=1} (f(x_i)-1, f(x_i)+1)$

Now we have $|x-x_i| <\delta$ where $x \in D$
     Then $|f(x) - f(x_i)| <1$

Thus, $f(D)$ is bounded!

Well, I find this proof easy to follow, except for the union of the neighborhoods.  How does that come from D being bounded?