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August 2, 2014

August 2, 2014

Posted by **Grace** on Wednesday, January 18, 2012 at 6:51pm.

lim r(x)=5c^4

(x->c)

- ? -
**Damon**, Wednesday, January 18, 2012 at 6:58pmI do not think you have given us the whole problem.

- Precalc -
**Grace**, Wednesday, January 18, 2012 at 7:12pmIn the problem before it says

r(x)= (x^5 - c^5)/(x-c)

Maybe it's referring to that?

Does that help?

- Precalc -
**Damon**, Wednesday, January 18, 2012 at 7:17pmnow we are cooking

- Precalc -
**Grace**, Wednesday, January 18, 2012 at 7:22pmI wish I could say that's why I don't understand this problem. But sadly, I still don't get it. Haha.

- Precalc -
**Damon**, Wednesday, January 18, 2012 at 7:27pmLimit:

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

- Precalc -
**Damon**, Wednesday, January 18, 2012 at 7:29pmLimit:

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

- Precalc -
**Damon**, Wednesday, January 18, 2012 at 7:29pmI got that from

http://www.wolframalpha.com/input/?i=limit+%28x^5+-+c^5%29%2F%28x-c%29++as+x-%3Ec

- Precalc -
**Damon**, Wednesday, January 18, 2012 at 7:30pmclick on "show steps" at upper right of box

- Precalc -
**Grace**, Wednesday, January 18, 2012 at 7:30pmThanks so much!

- Precalc -
**Damon**, Wednesday, January 18, 2012 at 7:46pmI 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

- Precalc -
**Grace**, Wednesday, January 18, 2012 at 8:04pmAh. Alright, I think I understand.

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