Limits and continuity?

Question

Hello guys

err i know i have to do my homework on my own but really believe me i have done trying this thing heap of times but i cant find do with it
pls help me ,,,,, i have find answer of

lim x-> 1

[(x) + (x)^2 + (x)^3 + ......... + (x)^n - n ] / (x - 1 )


i know the answer is n*(n+1) / 2 but how do you get it ???
can some one help me with evrything step by step ????


thanx a lottt in advance

Answer 1

L'Hopital's rule

Plug 1 into numerator and notice it is 1+1+1+ ... +1-n = 0 and into the denominator 1-1 =0 so the limit is of the form 0/0 so you differentiate top and bottom to get 1+2x+3x^2+ ... nx^(n-1) - 0 over 1 then limit as x ->1 is (1+2+3..+n)/1 = famous traingular number formula n(n+1)/2

There is also probably some factoring method...... like expression = [(x^(n+1)-x)/(x-1)-n] /(x-1) = [x^(n+1) - nx -x -1]/ (x-1)^2 but then ... Hmmm

Answer 2

[(x) + (x)^2 + (x)^3 + ......... + (x)^n - n ] / (x - 1 )
I do not understand the -n that you have at the enn of your power series. It should look like this:
[x+x^2 +x^3+ ..... x^n + x^(n+1) +........]/(x-1)
This then is a power series which diverges to infinty for x=>1.
Thus as x--> 1 we get infinity/0 which is indeterminate.

Use L'Hospitals rule and you get infinity/1 = infinity for the limit

Now if x<1, then the series will converge to a value of 1/(1-r) where r is the common multiplier.For example if we let x=1/2 then the series is 1/2+1/4 +1/8 +1/16 +... 1/2n +..., so r = 1/2 and this series would converge to 1/(1-1/2)= 2.

So, our expression, if |x|<1, becomes [1/(1-x)]/(x-1)]=
1/[(1-x)(x-1)] = 1/(-x^2 + 2x - 1) so lim x --> 0 is -1

But maybe I.m wrong because as I said at the start, I do not understand the -n at the end of your power series.

Answer 3

/1st/ really s = 1 +2 +3 +4 + + +n = n(n+1)/2, because 1st+last =n+1, also 2nd+last_but_1 = 2+(n-1) =n+1, and so on to the middle, number of pairs being =n/2; thus s=n(n+1)/2;
\2nd\ now l’hopital: (1 +2x +3x^2 +4x^3 + + +nx^(n-1))/1 = {at x=1}= n(n+1)/2;

Answer 4

First of all, you have to know that (to be shown later)
S(n)=1+(x) + (x)^2 + (x)^3 + ......... + (x)^n = (1-x^(n+1))/(1-x) (*)
Then
(x) + (x)^2 + (x)^3 + ......... + (x)^n = (1-x^(n+1))/(1-x) - 1

Then applying the L'hopital's rule and taking derivatives of the numerator and the denominator 2 times with respect to x, we have

lim x-> 1 [(1-x^(n+1))/(1-x) - 1 -n ] /(x-1) =
lim x-> 1 (-1)*[(1-x^(n+1)) - (1 -n)*(1-x) ] /((x-1)^2) =
lim x-> 1 [(n+1)*x^(n+1)- (1 -n) ] /(2*(x-1)) =
lim x-> 1 (n+1)*n*x^(n) /2 =(n+1)*n /2, QED.

In order to see that,
S(n)=1+(x) + (x)^2 + (x)^3 + ......... + (x)^n (*)
multiply it by x so
x*S(n) = (x) + (x)^2 + (x)^3 + ......... + (x)^n+(x)^(n+1)
and subtract it from (*) to get
S(n) - x*S(n) = 1- (x)^(n+1) and, simplifying,

S(n) = (1-x^(n+1))/(1-x).

Hope this helps.

Answer 5

Try this: for x<1 you have a geometric series whose sum is
x^0 + x^1 +...+ x^2 = [1-x^(n+1)]/(1-x)