`lim_(x->0^+) (1/x - 1/(e^x - 1))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t...

`lim_(x->0^+) (1/x - 1/(e^x - 1))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Chapter 4, 4.4 - Problem 51 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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lemjay | High School Teacher | (Level 3) Senior Educator

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`lim_(x->0^+) (1/x-1/(e^x-1))`

Plug-in x=0 to the function.

`lim_(x->0^+) (1/x-1/(e^x-1)) = 1/0-1/(e^0-1)=oo - oo`

Since the result is indeterminate, to take its limit apply L'Hospital's Rule. To do so, expression the function as one fraction.

`lim_(x->0^+) (1/x-1/(e^x-1)) = lim_(x->0^+) ((e^x-1)/(x(e^x-1)) - x/(x(e^x-1)))=lim_(x->0^+) (e^x-x-1)/(x(e^x-1))`

Then, take the derivative of the numerator and denominator.

`lim_(x->0^+) ((e^x-x-1)')/((x(e^x-1))')=lim_(x->0^+) (e^x-1-0)/(x*e^x + 1*(e^x-1)) = lim_(x->0^+) (e^x-1)/(xe^x + e^x-1)`

And, plug-in x=0.

`lim_(x->0^+)(e^x-1)/(xe^x+e^x-1)= (e^0-1)/(0*e^0+e^0-1) = 0/0`

Since the result it still indeterminate, apply L'Hospitals Rule again. Take the derivative of the numerator and denominator.

`lim_(x->0^+) ((e^x-1)')/((xe^x+e^x-1)') = lim_(x->0^+) (e^x-0)/(x*e^x+1*e^x+e^x-0) = lim_(x->0^+) e^x/(xe^x + 2e^x) = lim_(x->0^+)1/(x+2)`

And, plug-in x=0.

`lim_(x->0^+)(1/x+2)=1/(0+2)=1/2`

Therefore, `lim_(x->0^+) (1/x -1/(e^x-1)) = 1/2` .

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