Question

In calculus,

I need to use L'Hopital's rule to
figure out the derivative of
lim (x^2)(e^-x)
x->0
can someone explain how to use the rule on a non fractional function?

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 :)

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`