Question

lim(xto0)2[(e^x-(1+x))/x^2]=

Answer 1

\[\lim_{x\to 0} e^x-\frac{(1+x)}{x^2}\]
?

Answer 2

satellite u lost 2

Answer 3

ooh
i know
\[lim_{x\to 0} \frac{2e^x-(1+x)}{x^2}\]

Answer 4

i bet that is it. l'hopital's rule problem. of the form \[\frac{0}{0}\]

Answer 5

take derivative top and bottom separate. you will need to do it twice because after the first time you still get \[\frac{0}{0}\]

Answer 6

the top is 2e^0-1-0=2(1)-1-0=2-1=1

Answer 7

first time
\[\frac{2e^x-1}{2x}\]

Answer 8

still \[\frac{0}{0}\]

Answer 9

second time
\[\frac{2e^x}{2}\] let x ->0 get 1

Answer 10

oops i am totally wrong. sorry. first time it is \[\frac{1}{0}\] no limit. so sorry.

Answer 11

must be my bed time.

Answer 12

yes you have lost the derivative of constant 1 first time

Answer 13

by first time i mean before you start. it is not \[\frac{0}{0}\] to begin with so no l'hopital. sorry

Answer 14

WHERE IS THE DERIVATIVE OF CONSTANT?

Answer 15

Mmmm....looking carefully at the question as posted, we are looking for the limit of 2(e^x - 1 - x)/x^2. As x goes to zero we get 2(e^0 - 1 - 0)/0^2 = 2(1 - 1)/0 = 0/0, so use l'Hopital twice to get Lim as x goes to 0 of e^x, which is 1.

Answer 16

did you try graphing it?
both sides of 0 f(x) goes to positive infinity

Answer 17

WHY WE TAKE DERIVATIVE ANY REASON????