Can somebody explain to me the steps to solve this. Thanks.
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lim x -------> 0+: (1/x) - (1 / (e^x - 1)
lim x -------> 0+: (e^x - 1 - x) / (xe^x - x)
Plugging in 0 for all x values gives us an indeterminate answer of 0/0
Applying L'Hopital's Rule:
lim x -------> 0+: (e^x - 1) / (xe^x + e^x - 1)
Plugging in 0 again for all x values, we still get 0/0, thus allowing us to use L'Hopital's Rule once more:
lim x -------> 0+: (e^x) / (xe^x + 2e^x)
Plugging in 0 for all x values this time around gives us an answer of 1/2.
lim x -------> 0+: (e^x - 1 - x) / (xe^x - x)
Plugging in 0 for all x values gives us an indeterminate answer of 0/0
Applying L'Hopital's Rule:
lim x -------> 0+: (e^x - 1) / (xe^x + e^x - 1)
Plugging in 0 again for all x values, we still get 0/0, thus allowing us to use L'Hopital's Rule once more:
lim x -------> 0+: (e^x) / (xe^x + 2e^x)
Plugging in 0 for all x values this time around gives us an answer of 1/2.
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lim(x->0+) [(1/x) - (1/(e^x - 1))] = lim(x->0+)[(e^x - 1 - x)/(x*(e^x - 1))]. Now use
L'Hopital's Rule twice to end up with lim(x->0+)[e^x/(2*e^x + x*e^x)] = 1/2.
Giving more detail, note that I put the limit in the form 0/0 so that we could then
apply L'Hopital's, (both numerator and denominator are > 0 for x > 0).
The first application yields lim(x->0+)[(e^x - 1)/(e^x - 1 + x*e^x) and the result
of the second application is shown above.
L'Hopital's Rule twice to end up with lim(x->0+)[e^x/(2*e^x + x*e^x)] = 1/2.
Giving more detail, note that I put the limit in the form 0/0 so that we could then
apply L'Hopital's, (both numerator and denominator are > 0 for x > 0).
The first application yields lim(x->0+)[(e^x - 1)/(e^x - 1 + x*e^x) and the result
of the second application is shown above.
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Remember the movie MEAN GIRLS the limit does not exist
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The limit does not exist.