I'm basically learning analysis on my own, since I have a hard time going to class due to personal circumstances. At any rate, my understanding is subpar and I was hoping someone could tell me whether my proof is correct, as well as clarify some points for me. The definition I'm working with is the following:
Let $f:D\rightarrow\mathbb R$ with $x_{0}\in D$. Then, the function is said to be differentiable at $x_{0}$ if and only if the limit as $x$ approaches $x_{0}$ of $\ \dfrac{f(x)-f(x_{0})}{x-x_{0}}$ exists for each $x\not=x_{0}$.
Why must $x_{0}$ be an accumulation point of $D$? Why doesn't it suffice to say that $x_{0}$ is in $D$ with $x\not=x_{0}$? Also, does the fact that we take the limit as $x$ approaches $x_{0}$ imply that the derivative may not be defined at $x_{0}$? If so, why do we end up plugging in $x_{0}$ for $x$ nonetheless?
I can still do the problem because I can just mechanically apply the definition using the hand-waved notions from calculus, but I would like to gain a deeper understanding of the concept. Thanks in advance.
Problem. Use the definition to find the derivative of $f(x)=\sqrt(x)$, for $x>0$. Is $f$ differentiable at zero? Explain.
Solution. Let $f:(0,\infty)\rightarrow \mathbb R;f(x)=\sqrt{x}$, and $x_{0}\in (0,\infty)$ be an accumulation point of $(0,\infty)$.
We must find $f'$ such that for all $x_{0}$, $f'(x_{0})$ defined as: $\lim_{x\rightarrow x_{0}}\dfrac{f(x)-f(x_{0})}{x-x_{0}}$ exists. Note that, $$x-x_{0}=(\sqrt{x}-\sqrt{x_{0}})(\sqrt{x}+\sqrt{x_{0}}).$$ Then, \begin{align*}f'(x_{0})=\lim_{x\rightarrow x_{0}}\dfrac{f(x)-f(x_{0})}{x-x_{0}} &=\lim_{x\rightarrow x_{0}}\dfrac{\sqrt{x}-\sqrt{x_{0}}}{(\sqrt{x}-\sqrt{x_{0}})(\sqrt{x}+\sqrt{x_{0}})}\\ &=\lim_{x\rightarrow x_{0}}\dfrac{1}{\sqrt{x}+\sqrt{x_{0}}}\\ &=\dfrac{1}{2 \sqrt{x_{0}}}. \end{align*}
When we try to find the derivative of a function $x_o$, what we are basically trying to find is the slope at the certain point of the $f(x)$.So when you try to find the derivative of a function at a certain point, this $x_o$ must be in the domain of the function. Else, you cannot find the derivative on a point outside the domain cause you cannot find the slope at this point.
The fact that $x$ approaches $x_o$ happens just for the fact I explained before: You want to find the slope. As you already know, you can find the slope of a straight line by knowing two points of this line. But what it means to find the slope at a certain point? It basically means to find the slope from $x$ to $x_o$, where $x$ is very close to $x_o$, but not equal to it. It is like thinking that the function at the interval $(x,x_o)$ is a straight line, cause the interval is very small. If $x=x_o$, then you cannot find any slope.
The substitution of $x_o$ depends on what kind of limit you have. If the limit ends up to be of the form $\frac{0}{0}$ then you must try other tricks.
I hope I helped you!