## anonymous 5 years ago lim(xto0)2[(e^x-(1+x))/x^2]=

1. anonymous

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

2. anonymous

satellite u lost 2

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. myininaya

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

7. anonymous

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

8. anonymous

still $\frac{0}{0}$

9. anonymous

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

10. anonymous

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

11. anonymous

must be my bed time.

12. anonymous

yes you have lost the derivative of constant 1 first time

13. anonymous

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

14. anonymous

WHERE IS THE DERIVATIVE OF CONSTANT?

15. anonymous

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.

16. myininaya

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

17. anonymous

WHY WE TAKE DERIVATIVE ANY REASON????