Question

In calculus,

Answer 1

Okay from what I know that equation would basically be over one, on L'hopital's rule you take the derivative of the numerator and denominator to find the limit, if you take the derivative of 1 then you get 0 and the limit couldnt exist, so in this problem the limit does not exist :) <span />

Answer 2

Well, the expression above IS fractional to begin with, hmmm but if you have a non-fractional expression, you make it a fraction, by multiplying with something or using a negative exponent, or some reciprocal.

$\bf \lim\limits_{x\to 0}~x^2 e^{-x}\implies \lim\limits_{x\to 0}~x^2\cdot \cfrac{1}{e^x}\implies \lim\limits_{x\to 0}~ \cfrac{x^2}{e^x}\stackrel{LH}{\implies }\lim\limits_{x\to 0}~\cfrac{2x}{e^x} \\\\\\ \cfrac{2(0)}{e^0}\implies \cfrac{0}{1}\implies 0$