Limits and L' hospital rule... I keep getting the wrong answer please help!?

Question

limit as X approaches 0:

(5/x)- (5/( e^x-1))

i keep getting -5 and this is not a possible answer an someone help!

Answer 1

(5/x)- (5/( e^x-1))
= (5/x) (e^x-1-x)/(e^x-1)
= (5/x) (x^2/2! + O(x^3) ) / ( x/1! + O(x^2) )
--> 5/2

or using l'hopital
lim (5/x)- (5/( e^x-1))
= 5 lim (e^x-1-x ) / [x(e^x-1)]
= 5 lim [ (e^x-1) / [ e^x-1 + xe^x ]
= 5 lim e^x / [e^x + e^x + xe^x ]
= 5/2

Answer 2

infinity-infinity is indefinite

equate the denominators
(5e^x-5-5x)/(x(e^x-1))

it will be 0/0 indefinite.
L' hospital rule
(5e^^x-5)/(e^x-1+xe^x)
0/0 indefinite

L' hospital rule again
5e^x/(e^x+e^x+xe^x)

or

5/(2x)

The limit gives
5/(2+0)

=5/2

Answer 3

5/x-5/(e^x-1)=5(e^x-x-1)/(x(e^x-1))
f(x)=5(e^x-x-1)
g(x)=x(e^x-1)
Differentiating:
f'(x)=5(e^x-1)
g'(x)=e^x-1+x*e^x
So we still have 0 on top and bottom, differentiate again:
f''(x)=5e^x
g''(x)=e^x+e^x+x*e^x
limit = 5/(1+1+0)=5/2
Hope that helps!