I'm a complete beginner so please be gentle. So we have:
Which derivative is described by the following expression? $$\lim_{x\to2} \frac{x^2-4}{x-2}$$
What I do know is that since the limit is $\lim_{x\to2}$, it means that the derivative is at $x = 2$ (as $x$ gets closer and closer to $2$, which lets us estimate the average rate of change in a more precise way.
So I have this big-time beginner confusion here.
It says that we can conclude that the quotient expression is $x^2-4$, which is the equivalent of $(x)^2 - (2)^2$ and so, therefore, we conclude that the function is $g(x)=x^2$.
Now I don't even know where to start, I don't get this at all so if someone could just, if it isn't too much trouble direct me somewhere that explains that, that would be lovely.
The correct answer is: $g'(2)$ where $g(x)=x^2$
I know this is a little bit far streched, I don't expect to get a complete tutorial here but a little bit of help and maybe some guidance to the right direction would be really appreciated.
Thanks guys!
The definition of the derivative you've presented is this in its general form:
$$f'(c)=\lim_{x\to c} {f(x)-f(c)\over x-c}$$
This is just a reframed definition of slope. Recall the definition of slope is $\Delta y \over \Delta x$, or in words, 'rise over run'. In this case, $f(x)-f(c)$ is the rise, because it's the change in $y$-values. $x-c$ is the run because it's the change in the $x$-values.
A derivative, then, is the rise over run as an $x$-value $c$ approaches $x$ infinitely closely, thus giving the rate of change in that instant.
Now applying this to this situation, $c$, in this case, is $2$ and $f(x)=x^2$. Thus:
$$\lim_{x\to c} {f(x)-f(c)\over x-c} \rightarrow \lim_{x\to 2} {f(x)-f(2)\over x-2} = \lim_{x\to 2} {\left(x\right)^2-\left(2\right)^2\over x-2} = \lim_{x\to 2} {x^2-4\over x-2}=f'(2)$$