1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limits question!

  1. Sep 15, 2008 #1
    1. The problem statement, all variables and given/known data
    limit as x approaches 0 of (sin^2 (x))/x^2


    2. Relevant equations i generally know how to solve the equation, but i'm not sure what to do about the top. is sin^2 (x) the same as sinx^2?



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 15, 2008 #2

    danago

    User Avatar
    Gold Member

    sin^2 (x) = (sin x)^2
     
  4. Sep 15, 2008 #3
    You have probably learned a rule for the situation where both the numerator and denominator go to zero (or both to infinity). That is what I would use.
     
  5. Sep 15, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    sinx^2 can mean two different things. It could mean sin(x^2) or (sin(x))^2. Which do you mean? Try not to write ambiguous forms when posting a question. As danago said, sin^2(x) is the latter. And bdforbes is referring to l'Hopital's rule. If you haven't learned that then you probably proved directly that sin(x)/x approaches 1 as x approaches 0. Use that.
     
  6. Sep 15, 2008 #5
    ok, so it would = (sin x)^2/ x^2, then the limit is 1. because you can cancel the ^2 on top and bottom to = (sin x)/x, which is 1
     
  7. Sep 15, 2008 #6

    danago

    User Avatar
    Gold Member

    Woa hold up. You cant just cancel the indexes from the top and the bottom. Id probably solve this by using the result that Dick mentioned above with one of the general limit rules.
     
  8. Sep 15, 2008 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    2^2/1^2=4. 2/1=2. You CAN'T cancel an exponent. You can write it as (sin(x)/x)^2 and draw conclusions from that.
     
  9. Sep 15, 2008 #8
    yeah, i remember the preofesor mentioning l hopitals rule, but i don't really know it yet, i'll look it up and get back w/u all later
     
  10. Sep 16, 2008 #9
    Use these two pieces of information:

    1) (sinx)^2/x^2 = (sinx/x) * (sinx/x)

    2) Given two functions, f(x) and g(x, and any point p, the lim [f(x) * g(x)] = lim f(x) * lim g(x)
    x-> p x-> p x-> p
     
  11. Sep 16, 2008 #10
    Provided the latter two limits exist
     
  12. Sep 16, 2008 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    which, fortunately, they do in this case.
     
  13. Sep 16, 2008 #12
    Now the OP has only to Taylor expand sin(x) and the problem will be solved.
     
  14. Sep 17, 2008 #13
    I don't see a need for that. It seems to me that it's sufficient to rewrite (sinx)^2/x^2 as (sinx/x) * (sinx/x) and to remember that the limit of a product of functions is the product of each limit.
     
  15. Sep 17, 2008 #14
    Yes but what is the limit of sinx/x? I propose a Taylor expansion or l'Hopitals rule.
     
  16. Sep 17, 2008 #15
    0 < cosx < (sinx)/x < 1/cosx is valid for -pi/2 < x < pi/2. Using the squeeze theorem here is extremely quick to find the limit of sinx/x as x approaches 0.
     
  17. Sep 17, 2008 #16
    lim(sinx)/x=1 should exists before we get (sinx)'=cosx. Thus L'hospital's rule is not a proper way to prove(though it's a way to calculate) that sinx/x->1, as x->0.

    It works.
     
  18. Sep 17, 2008 #17

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Gack! You don't have to know the limit exists to apply l'Hopital. All you have to know is that it has the form 0/0. Why I'm contributing again to this bloated useless thread, I don't know. But that is soooo wrong.
     
  19. Sep 18, 2008 #18
    I disagree. If [tex]\stackrel{lim}{x\rightarrow0}\frac{f'(x)}{g'(x)}[/tex] exists, and [tex]\stackrel{lim}{x\rightarrow0}f(x)=0[/tex] and [tex]\stackrel{lim}{x\rightarrow0}g(x)=0[/tex], then it is true that [tex]\stackrel{lim}{x\rightarrow0}\frac{f(x)}{g(x)} = \stackrel{lim}{x\rightarrow0}\frac{f'(x)}{g'(x)}[/tex]. Do you not agree with this? The problem at hand satisfies these conditions, so I don't see why we couldn't use it.
     
  20. Sep 18, 2008 #19
    Sorry if I misunderstood someone's words (English is not my mother tongue:confused:) . It seems to me that someone is PROVING that lim sinx/x = 1 using L'hospital's rule, which I think is wrong.

    If I do not know lim sinx/x = 1 already, how do I know (sinx)' EQUALS TO cosx ( by what means I can get this? If you have one without using "lim sinx/x = 1", tell me plz, just curious about it )? If I do not know that (sinx)' equals to cosx, how do I apply L'hospital's rule?

    As I said before, L'hospitals is okay to CALCULATE sinx/x but is not a PROPER way to PROVE it. They are absolutely different things.

    Applying the L'hospital's rule to the original question is nothing wrong, since we simply take "lim sinx/x = 1" as true, and moreover, "(sinx)'=cosx" as true.

    Anything wrong? Sorry again if I misunderstood something

    just to make it more clear. Try to think that we are asked to prove that (sinx)'=cosx. We want to know what is the result of lim (sin(x+h)-sinx )/h when h->0, and you are not going to use L'hospital's rule to prove it, right? (If using L'hospital's rule, we of course get the correct answer cosx! but unfortunately, we are using "(sinx)'=cosx" already)
     
    Last edited: Sep 18, 2008
  21. Sep 18, 2008 #20
    Actually what we want to know is lim sinx/x. We are allowed to assume (sinx)'=cosx, this isn't high school maths. Then l'Hopitals rule DOES prove the limit.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Limits question!
  1. Limit question ! (Replies: 2)

  2. Limit question (Replies: 10)

  3. Limits question (Replies: 6)

  4. Limit question (Replies: 3)

  5. Limits question (Replies: 5)

Loading...