I do not understand the alternative definition of derivative and how to use it to find tangent slope. Find the slope of the tangent line at $x=3$ for $|2x-6|$
The alternative definition of derivative is $$f(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$
If I want to find the derivative at $3$. I would plug $3$ into $c$.
This results in: $$\lim_{x\to3}\frac{|2x-6|-0}{x-3}$$ 
This answer results in $0/0$ because it is a sharp turn.
I want to find the tangent slope of points around it to prove it is a sharp turn. I am going to find the tangent slope at $2.9$ and $3.1$. How come it works if I plug in $2.9$ and $3.1$ into $x$?
$$\frac{|2(2.9)-6|-0}{2.9-3}=-2$$
$$\frac{|2(3.1)-6|-0}{3.1-3}=2$$
I check desmos and both of the slope is correct. My question is why? The formula of the derivative is $$f(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$ To find the tangent slope, I should do $$f(2.9)=\lim_{x\to 2.9}\frac{f(x)-f(2.9)}{x-2.9}$$
$$f(2.9)=-.2$$
$$f(2.9)=\lim_{x\to 2.9}\frac{|2(x)-6|-.2}{x-2.9}$$
$$f(2.9)=\lim_{x\to 2.9}\frac{|2(2.9)-6|-.2}{2.9-2.9}=0$$
Why does the first way work, but not the second? Why does the second method answer indicate a sharp turn.
I was watching the youtube video. https://www.youtube.com/watch?v=rowIURKDnag&index=5&list=PLRWGJgOxXM4T2ZOuFMxtdKqiKra2_BLKY. I do not understand how she did the problem at time stamp 12:08.
 A: 
This results in: $$\lim_{x\to3}\frac{\lvert 2x-6\rvert - 0}{x-3}$$
This answer results in $0/0$ because it is a sharp turn.

No, actually it is not "because it is a sharp turn."
The only way you can get $\frac00$ out of
$\frac{\lvert 2x-6\rvert-0}{x-3}$ is if
you simply substitute $x = 3.$
At that point, the fact that $\lvert 2x-6\rvert$ has a "sharp turn" is completely irrelevant.
If you try to find derivatives this way, by simply setting $x$ to the same value as $c$ in the definition, you will always get $\frac00$,
because $f(c) - f(c) = 0$ and $c - c = 0.$
This will happen with any function $f$, even a nice one such as
$f(x) = x^2$ or even a constant function such as $f(x) = 5.$
Let's try this graphically.
When $x \neq c,$ the expression
$$\frac{f(x)-f(c)}{x-c}$$
gives you the slope of the unique line through the two points
$(x,f(x))$ and $(c,f(c))$ on the graph of $y = f(x).$
This is called a secant line because it goes through two distinct points of the function's graph, like a secant through two points on a circle.
The idea of the "limit" in the definition
$$\lim_{x\to c} \frac{f(x)-f(c)}{x-c}$$
is that we can force the slope of the secant line to be as close to the true "tangent" slope at $(c,f(c))$ as we can possibly want, just by putting a restriction on $x$ that it must be within a certain neighborhood around $c.$
(A "neighborhood" is always the set of all points within some distance of $c,$ not including $c$ itself.)
But for your particular example, where $f(x) = \lvert 2x-6\rvert,$
what you find when you set $x = 2.9$ is the same result you will find when you set $x$ to any value less than $3.$ The slope of the secant line is $-2.$
What you find when you set $x = 3.1$ is the same result you will find when you set $x$ to any value greater than $3.$ The slope is $2.$
So no matter how tightly you restrict the neighborhood around $c=3,$
you'll always have secant lines with slope $-2$ and others with slope $2.$
Hence you never get a single limit value, hence no limit,
hence no derivative.
If you actually evaluate the derivative at $c = 2.9$ or $c = 3.1,$
or any value other than $c= 3,$
what you will find is that the derivative is always $-2$ when $c < 3$
and always $2$ when $c > 3.$
This is your "sharp corner," and it is true that a function with a "sharp corner" will not have a derivative at that point.
But that's not because of some $\frac00$ calculation,
it's because you cannot get the secants for $x < c$ to converge together with the secants for $x > c$ to the same real number.
A: Yes by the definition the derivative at that point is given by the limit
$$\lim_{x\to3}\frac{|2x-6|-0}{x-3}$$
but
$$\lim_{x\to3^+}\frac{|2x-6|-0}{x-3}=\lim_{x\to3^+}\frac{2(x-3)}{x-3}=2\neq \lim_{x\to3^-}\frac{|2x-6|-0}{x-3}=\lim_{x\to3^-}\frac{-2(x-3)}{x-3}=-2$$
and therefore the function is not differentiable at $x=3$ since the given limit doesn't exist at that point.
