Question 986568
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I'm going to assume you mean


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right 0} \frac{e^x\ -\ 1}{x}]


because


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right 0}  \frac{e^{x\,-\,1}}{x}]


doesn't exist (the limit from the left is *[tex \Large -\infty] and the limit from the right is *[tex \Large \infty])


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right 0}  \frac{e^x\ -\ 1}{x}]


Since *[tex \Large e^0\ =\ 1] we have the L'Hôpital indeterminate form *[tex \Large \frac{0}{0}].  Since it is a proper indeterminate form, we can use L'Hôpital's Rule: evaluate the limit of the derivative of the numerator over the derivative of the denominator.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\right 0}  \frac{\frac{d}{dx}(e^x\ -\ 1)}{\frac{dx}{dx}}\ =\ \lim_{x\right 0}  \frac{e^x}{1}\ =\ \frac{e^0}{1}\ =\ 1]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \