What is the limit of ((1/x)-((1)/(e^(x)-1)) as x approaches 0^+?

1 Answer
Steve M
Aug 6, 2017

# lim_(x rarr 0^+) 1/x-(1)/(e^x-1) = 1/2#

Explanation:

Let:

# f(x) = 1/x-(1)/(e^x-1) #
# " " = ((e^x-1) - (x)) / (x(e^x-1))#
# " " = (e^x-1 - x) / (xe^x-x)#

Then we seek:

# L = lim_(x rarr 0^+) f(x) #
# \ \ = lim_(x rarr 0^+) (e^x-1 - x) / (xe^x-x) #

As this is of an indeterminate form #0/0# we can apply L'Hôpital's rule.

# L = lim_(x rarr 0^+) (d/dx(e^x-1 - x)) / (d/dx(xe^x-x)) #
# \ \ = lim_(x rarr 0^+) (e^x-1) / (xe^x+e^x - 1) #

Again, this is of an indeterminate form #0/0# we can apply apply L'Hôpital's rule again:

# L = lim_(x rarr 0^+) (d/dx(e^x-1)) / (d/dx(xe^x+e^x - 1)) #
# \ \ = lim_(x rarr 0^+) (e^x) / (xe^x+e^x+e^x) #
# \ \ = (e^0) / (0+e^0+e^0) #
# \ \ = 1/2 #

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