What derivative is described by the expression $\lim_{x\to2} \frac{x^2-4}{x-2}$?
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$$g'(2)=\lim_{x\to2}\frac{g(x)-g(2)}{x-2}=\lim_{x\to2}\frac{x^2-2^2}{x-2}=\lim_{x\to2}\frac{x^2-4}{x-2}.$$
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user472288
Updated on March 06, 2020Comments
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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!
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Masacroso over 5 yearsit is not clear what thing you dont understand. Do you knows the definition of derivative at a point?
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Nap D. Lover over 5 yearsStare 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?
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user472288 over 5 yearsThanks 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.
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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 $x^2 - 4$.
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user472288 over 5 yearsThanks Andrew! I missed the part of what indicated the function to be $x^2$ /awkward.