What derivative is described by the expression $\lim_{x\to2} \frac{x^2-4}{x-2}$?

2,350

Solution 1

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:

$$\lim_{x\to c} {f(x)-f(c)\over x-c} \rightarrow \lim_{x\to 2} {f(x)-f(2)\over x-2} = \lim_{x\to 2} {\left(x\right)^2-\left(2\right)^2\over x-2} = \lim_{x\to 2} {x^2-4\over x-2}=f'(2)$$

Solution 2

I don't really see where you are confused, as you said all the right things. You have, if $g(x)=x^2$, $$ \lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}\frac{g(x)-g(2)}{x-2}=g'(2). $$

Solution 3

Let $g$ be your function (that is, $g(x)=x^2$). Then by definition you have$$g'(2)=\lim_{x\to2}\frac{g(x)-g(2)}{x-2}=\lim_{x\to2}\frac{x^2-2^2}{x-2}=\lim_{x\to2}\frac{x^2-4}{x-2}.$$

Share:
2,350

Related videos on Youtube

user472288
Author by

user472288

Updated on March 06, 2020

Comments

  • user472288
    user472288 over 3 years

    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!

    • Masacroso
      Masacroso over 5 years
      it is not clear what thing you dont understand. Do you knows the definition of derivative at a point?
    • Nap D. Lover
      Nap D. Lover over 5 years
      Stare at the definition of derivative $f'(c)=\lim_{x\to c} \frac{f(x)-f(c)}{x-c}$ and then stare at the same thing but with $f(x)=x^2$ and $c=2$. Do you see?
  • user472288
    user472288 over 5 years
    Thanks Andrew, ofcourse I know what a slope is and what a derivative is as I've explained in the OP but I didn't quite understand how $g(x)-g(2)$ is the same as of $(x^2)-4$ for some reason I can't see it.
  • Andrew Li
    Andrew Li over 5 years
    @user472288 Okay, so remember that the function is $x^2$. In the definition of a derivate, $c$ is a constant, in this case 2. So basically it's $f(x)-f(2)$ which is $x^2 - 4$.
  • user472288
    user472288 over 5 years
    Thanks Andrew! I missed the part of what indicated the function to be $x^2$ /awkward.