Interrelation between continuity and differentiability? I need a "big picture" of how those things relate to each other and probably a list of fundamental theorems that glue them up together. 
My current (quite limited) understanding:
1) The fact that function $f(x)$ has a derivative at point $P$ does mean that it behaves nearly like a straight line infinitely close to the $P$ and thus might be approximated with such a straight line with infinitely small error (which becomes exactly 0 on at least two points $P, P_{neighbor}$ since both are shared by given line and original $f(x)$). 
2) As such, $f(x)$ has to be continuous (not necessarily uniformly continuous) at the $P$, otherwise as long as $P_{neighbor} \rightarrow P$, $f(x)$ kind of "jumps" rather then being smooth and thus can not be estimated with a straight line to an infinitely small tolerance (but still could be approximated with a line, which obviously won't be precise and practically valuable, but still theoretically possible). 
3) As such the fact that $f(x)$ is indeed continuous at $P$ guarantees that it could be "goodly" approximated with a straight line? 
Am I right or wrong? 

UPDATE:
I think the given answer beneath helped me to discover one mistake I wasn't able to see before. Let me try to formulate a new statement and you, guys, please correct me in case I am still mistaking things. 
There are 2 different ways to approximate any (or only continuous?) $f(x)$: 
1) when you pick up any 2 points belong to $f(x)$ and graph what is called a "secant line"; thus $[P_1, P_2]$ interval emerges where your linear estimation could be as precise as one wish (as long as $P_2$ approaches $P_1$).
2) But when things come down to the limit of such an constantly decreasing distance between both points, $P_2$ vanish and merges with $P_1$ such that only single point exits. Since this very moment, given line ceases to be a secant and becomes an "tangent line", which guaranteed to share at least single $P_1$ point with original $f(x)$ (and thus the error there must be precisely zero). Such an approach approximates $f(x)$ around $P_1$: to be more concrete, on some $[P_1 - \delta, P_1 + \delta]$. So whenever $$x \in [P_1 - \delta, P_1 + \delta]$$
there exists $\epsilon \gt 0$ such that: $$|L(x) - f(x)| \lt \epsilon$$
4) Now my question: $[P_1 - \delta, P_1 + \delta]$ actually ships two points, which makes me come back to the initial point: there is a secant line, crossing two points belonging for original $f(x)$. 
 A: Let me give you a List, with countereaxmples and theorems glueing the three concepts together:
First of all differentiability always implies continuity.
The converse is not true, take for example $f(x)=|x|$. Here you cannot find a unique tangent of the graph at zero.
Uniform continuity is a bit more tricky. It always implies continuity, but the converse may only be true in certain situations. One theorem is as follows:
If $f:K\rightarrow \mathbb{R}$ is continuous and $K\subset\mathbb{R}$ is compact, then $f$ is uniformely continuous. 
You cannot neglect the compactness assumption in this theorem. Take for example $f(x)=e^x$ on the whole real numbers. Assume it is uniformely continuous. The for every $\varepsilon>0$, you find a $\delta>0$ such that 
$$|x-y|<\delta\Rightarrow |e^x-e^y|<\varepsilon.$$
Now take $x\in\mathbb{R}$ to be arbitrary and $y=x+\frac{\delta}{2}$. Then 
$$\varepsilon>|e^x-e^y|=|e^x-e^{x+\frac{\delta}{2}}|=|e^x||1-e^{\frac{\delta}{2}}|\rightarrow \infty$$
for $x\rightarrow \infty$.
Now to your understanding:
1) Your understanding is not quite correct, since there is no concept of neigbour, such that the error of the linear approximation would be zero. The definition for differentiability is as follows: $f$ is differentiable in $x$ if the limit
$$\lim_{y\rightarrow x}\frac{f(y)-f(x)}{y-x}$$
does exist.
Now take for example $f(x)=x^2$. The derivative at zero is $f'(0)=0$. So your linear function approximating $f$ at $x=0$ is the nullfunction. If you now pick any $x\neq 0$, then $f(x)\neq 0$. Therefore $\frac{f(x)-0}{x-0}\neq 0$. Hence you cannot find a neighbouring point such that the error is zero.
To 2): Your right, differentiability implies continuity, but this kind of $f$ jumping is not really good. There can be very strange examples of functions, which are not continuous, for example
$$f(x)=\left\{\begin{array}{cc}\sin\frac{1}{x},&x\neq 0\\0,&x=0\end{array}\right.$$
Here you also see, that you cannot find a line approximating $f$ at zero in any good sense, such that the error would be controllable.
To 3). $f(x)=|x|$ is continuous but not differentiable as mentioned above. You cannot find any good linear approximations, since you cannot find a unique tangent at zero.
Let me give you one last advice. Try to stop relying on your graphical intuition and start working with the abstract definitions. Then your understanding will improve.
