Posted by Grace on Wednesday, January 18, 2012 at 6:51pm.
I do not think you have given us the whole problem.
In the problem before it says
r(x)= (x^5 - c^5)/(x-c)
Maybe it's referring to that?
Does that help?
now we are cooking
I wish I could say that's why I don't understand this problem. But sadly, I still don't get it. Haha.
Limit:
Hide steps
lim_(x->c) (x^5-c^5)/(x-c)
Factor the numerator and denominator:
= lim_(x->c) ((c-x) (-c^4-c^3 x-c^2 x^2-c x^3-x^4))
/((c-x) (-1))
Cancel terms, assuming c-x!=0
(c^4+c^3 x+c^2 x^2+c x^3+x^4)
The limit of c^4+c^3 x+c^2 x^2+c x^3+x^4 as x approaches c is 5 c^4
Limit:
Hide steps
lim_(x->c) (x^5-c^5)/(x-c)
Factor the numerator and denominator:
= lim_(x->c) ((c-x) (-c^4-c^3 x-c^2 x^2-c x^3-x^4))
/((c-x) (-1))
Cancel terms, assuming c-x!=0
(c^4+c^3 x+c^2 x^2+c x^3+x^4)
The limit of c^4+c^3 x+c^2 x^2+c x^3+x^4 as x approaches c is 5 c^4
I got that from
http://www.wolframalpha.com/input/?i=limit+%28x^5+-+c^5%29%2F%28x-c%29++as+x-%3Ec
click on "show steps" at upper right of box
Thanks so much!
I just divided the original numerator by (x-c) and got
x^4 + cx^3 +c^2x^2 +c^3 x + c^4
which is 5x^4 or 5c^4 as x-->c
Ah. Alright, I think I understand.
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