A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

lim (1+1/x)^x x-> The answer is e but I do not know how to get there

  • This Question is Closed
  1. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well, that is the definition for 'e' :)

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

    I do not understand what you mean

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

    sorry I did not mean to repeat

  4. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i mean that 'e' is defined as the limit as x -> inf; (1+1/x)^x

  5. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    try using the definition of a derivative..

  6. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f(x+h) - f(x) lim h->inf; ----------- h

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

    ok, I will try it

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

    one definition is it is the solution to \[\int^x_1 \frac{1}{t}\,dt = 1\]

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

    @amistre64: Its not right, the definition is \[h \rightarrow 0\] As to make the secant a tangent!

  10. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol...same diff :)

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

    Ok, thanks amistre64, but I am not grasping the point

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

    What do you mean ami

  13. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i mean.... i might be wrong ;)

  14. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what math is this for hix?

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

    calculus I

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

    another definition of e is \[e= \lim_{x\to\infty} (1+\frac{1}{x})^x\] in which case the answer is immediate.

  17. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well, calc1 at this semester should be talking about limits right?

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

    i have a final tomorrow

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

    or you can take the log, compute the limit, get 1, and then 'exponentiate' to get e

  20. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm.... my finals were 2 weeks ago; just starting the summer semester here

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

    I have to go to class, but please post anything that can help, and I will check it later. Thank you all very much. see ya

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

    \[lim_{x\to\infty} (1+\frac{1}{x})^x\] \[ln(1+\frac{1}{x})^x = x ln(1+\frac{1}{x})\] which is of the form \[\infty \times 0\] rewrite as \[\frac{ln(1+\frac{1}{x})}{\frac{1}{x}}\]

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

    Then use l'hopital. But this is a lot of unnecessary work since you have the definition of e to begin with.

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

    Let \[y = \lim_{x\to\infty}(1+1/x)^x\] we have to put it in indeterminate form in order to employ l'hopital's rule so we take natural log of both sides \[y = \lim_{x\to\infty}(1+1/x)^x \rightarrow \ln{y} = \lim_{x\to\infty}\ln((1+1/x)^x) \] Using log rule of exponents we have \[\lim_{x\to\infty}x*\ln((1+1/x))\rightarrow \infty-0\] Rewriting the whole expression \[\lim_{x\to\infty}\ln((1+1/x))/(1/x)\rightarrow 0/0 \] which is indeterminate, using L'Hopital's rule we have \[\lim_{x\to\infty}(-x/(x^2(x+1))/(-1/x^2)\rightarrow \lim_{x\to\infty} x/(x+1) \rightarrow \infty/\infty \] which permits us to use L'Hopital's rule again so we have \[\ln{y} = \lim_{x\to\infty}(1/1) = 1\] Raising both sides to e yields \[y = e^1=e\] Done.

  25. 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.