A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

TrojanPoem

  • one year ago

Limits question :

  • This Question is Closed
  1. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @phi

  2. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    limit sin x/ x x-> 0 this one is 0/0 but assume x = 1 /t limit tsin1/t 1/t -> 0 so t->0

  3. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Now i got the second in my exam limit tsin1/t which is obviously not 0/0

  4. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    My teacher told me you can assume t= 1/x and it will be sinx/x which makes me insane.

  5. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    limit as x->0 of (sin x) / x is a famous limit The derivation of this limit that I have seen rely on geometric arguments https://www.khanacademy.org/math/differential-calculus/limits_topic/squeeze_theorem/v/proof-lim-sin-x-x

  6. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    but mustn't it be 0/0 in any case to apply this

  7. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I assume you mean the problem is limit t sin1/t as t->infinity rewrite as x=1/t , x-> 0 and the problem is limit sin(x)/x as x->0 which = 1

  8. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Lol , I never thought sinx/ x was proved with a triangle tanx = sinx/x LAWL !

  9. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    my mistake t -> infinite

  10. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The other way to find the limit of sin x / x is to use the series definition of sin x https://en.wikipedia.org/wiki/Trigonometric_functions#Series_definitions sin x = x - x^3/3! + x^5/5! - ... now divide by x sin x / x = 1 - x^2/3! + x^4/5! - .... now let x->0 to get limit sin x / x = 1

  11. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The series definition is out of my curriculum :o

  12. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You know what was wrong ? it was written as 1/t -> 0 and I dumbly turned it into t-> 0 so it was meaningless .

  13. TrojanPoem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks phi.

  14. 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.