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Calculus1
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Find the following limit
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\[\lim_{x \rightarrow 0+}\frac{ x^{2} - 4x}{ 1-e^{2x} }\]
We have to use l'hopital rule
so first get the derivative of the numerator and denominator. \[=\frac{ 2x-4 }{ -2e ^{2x} }\] and then plug in zero into x
Btw we have to use l'hospital rule because limit as x->0 for the numerator is 0 and same with the denominator \[\lim_{x \rightarrow 0} = \frac{ 0 }{ 0 }\]
thank you I understand it better now!
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