A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

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

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    satellite u lost 2

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    must be my bed time.

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes you have lost the derivative of constant 1 first time

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    WHERE IS THE DERIVATIVE OF CONSTANT?

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    WHY WE TAKE DERIVATIVE ANY REASON????

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.