Question

lim x->1 (x^25-1)/(x-1), determine if the limit exists and if it does find its value.

Answer 1

have you learned lhopita rule yet?

Answer 2

l'hopital*

Answer 3

i have no idea what that is

Answer 4

its x^(25)?

Answer 5

yeah in the problem its x^25

Answer 6

\[\frac{x^{25}-1}{x-1}=1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}+x^{13}+ \]\[x^{14}+x^{15}+x^{16}+x^{17}+x^{18}+x^{19}+x^{20}+x^{21}+x^{22}+x^{23}+x^{24} \]There are 25 terms. 25*1 = 25

Answer 7

wooo sweet, can you explain that a little more?

Answer 8

Im thinking if he has not learned l'hopital rule yet, then he shuold be able to do it another way.

Answer 9

The RHS came from Mathematica's "Apart" function.
1^n where n is any positive integer is equal to 1

Answer 10

very cool. what are you doing in class @kjblasion ? are you finding limits by approximation methods, graphically, or are you finding them with algebra only?

Answer 11

oh that makes sense.. but in my class they are making us find the limits through the derivative.

Answer 12

if you take the derivative of the top and bottom as many times as it takes to get it to a form that is determenant, then plug in 1

Answer 13

this is called l'hopital rule.

Answer 14

so you have (25x^24)/1 when x =1 you get 25

Answer 15

L'Hôpital's rule
http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

Divide the derivative of the Numerator by the derivative of the Denominator.\[\frac{\left(25 x^{24}\right)}{1} \] The limit as x->1 with this method is also 25.

Answer 16

I have seen at least one limit problem presented on this site where L'Hôpital's rule was not allowed to be used in the solution.

Answer 17

Oh wow, that makes more sense i was trying to take the quotient rule and I'm pretty sure i screwed up some where. Thanks for helping me solve the problem!!!

Answer 18

You're welcome.

Answer 19

differentiate both the denominator and numerator w.r.t. x until u get the expression free from the form of 0/0 ....
then u can put the limits... which will give the same result.....