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!

Finding the limit of trig functions

  1. Sep 18, 2007 #1
    I'm having a hard time find the limits of these trig functions. Please help me with it. Thanks in advance.

    1. [tex]\lim_{x \rightarrow \pi}(\frac{\sqrt{x}}{csc x})[/tex]

    From this function I know that [tex]csc x= \frac{1}{sin x}[/tex] which cannot equal 0.

    X, therefore, cannot equal [tex]\pi n[/tex] where n is any integer.

    Am I doing something wrong here? When I graph the function, there don't seem to be any asymptotes at all. I don't get this. I thought that the limit at pi was suppose to be equal to negative infinity or positive infinity but it isn't...
     
    Last edited: Sep 18, 2007
  2. jcsd
  3. Sep 18, 2007 #2

    dynamicsolo

    User Avatar
    Homework Helper

    Given that [tex]csc x= \frac{1}{sin x}[/tex],

    how could you re-write this expression: [tex](\frac{\sqrt{x}}{csc x})[/tex] ?

    Is there any problem in applying the Limit Laws to it?
     
  4. Sep 18, 2007 #3
    [tex]\lim_{x->\pi}\frac{\sqrt{x}}{\csc{x}}[/tex]

    [tex]\lim_{x->\pi}\frac{\frac{\sqrt{x}}{1}}{\frac{1}{\sin{x}}}[/tex]

    simplify and what is ...

    [tex]\frac{\sin{x}}{x}[/tex]
     
  5. Sep 18, 2007 #4

    dynamicsolo

    User Avatar
    Homework Helper

    and obtain [tex]\lim_{x->\pi}\sqrt{x}}\sin{x}[/tex],

    which should be decently-behaved...



    Is this a separate question?
     
  6. Sep 18, 2007 #5
    if you divide the limit question by x, you obtain sinx\x which allows you to have an answer other than 0
     
  7. Sep 18, 2007 #6

    dynamicsolo

    User Avatar
    Homework Helper

    What is this in reference to? I'm not following you.

    The compound fraction you wrote:

    [tex]\frac{\frac{\sqrt{x}}{1}}{\frac{1}{\sin{x}} }[/tex]

    becomes

    [tex]\sqrt{x}}\sin{x}[/tex] .

    Where did (sin x)/x come from??
     
  8. Sep 18, 2007 #7
    Okay, I got [tex]\lim_{x \rightarrow \pi}x\sqrt{x}[/tex] but then if I use direct substitution I get [tex] \pi\sqrt{\pi}[/tex] which is not the correct answer...
     
  9. Sep 18, 2007 #8
    that's exactly what i get eventually, idk what to tell you :(
     
  10. Sep 19, 2007 #9
    Well, my book says the correct answer is 0 but I don't know how they arrived at that.
     
  11. Sep 19, 2007 #10

    dynamicsolo

    User Avatar
    Homework Helper

    Ah, here's the problem.

    Forget what I had here before! That's what I get for trying to analyze something like this in the wee hours...

    We're getting distracted by that (x/sin x). You're evaluating the limit at *x = pi*, not x = 0! Your expression becomes [sqrt(pi)]/(pi/0) = [sqrt(pi)]/inf. = 0 . So there *is* no contradiction! It just makes the algebra easier if you multiply the original function through by sin x , rather than playing with the compound fraction. But the results are equivalent.

    I couldn't understand how the original function was getting turned into

    x·sqrt(x) , which behaves nothing like sqrt(x)/(csc x) ...
     
    Last edited: Sep 19, 2007
  12. Sep 19, 2007 #11
    if it's zero, rather than putting it under x to get rid of sinx, just plug it in when you have sinx times sqrtX
     
  13. Sep 19, 2007 #12

    dynamicsolo

    User Avatar
    Homework Helper

    I think this illustrates the perils of too-late-night calculus. I've completely revised my previous post. There is no difficulty and the answer *is* zero.
     
    Last edited: Sep 19, 2007
  14. Oct 24, 2007 #13
    Maybe you could set it up as lim (h-->0) of [f(x+h)-f(x)]/h
    the first principles

    or you can just know that its -cot(x)sec(x) or something like that, i dont have my limit laws sheet infront of me right now, but hopefully that can help
     
  15. Oct 24, 2007 #14
    Why? That would give you a derivative, but this question does not ask for a derivative...

    No it is not. The problem is asking for an evaluation of the limit, the asnwer should be a number, not a function.
     
  16. Sep 6, 2009 #15
    wait why i dont get why the answer couldnt be that x^(1/2)*sinx as x approaches pi so that sinx goes to zero by direct substitution and 0*x^(1/2)=0.
     
  17. Sep 6, 2009 #16
    Assuming we are all clear that the original limit's argument simplifies to [tex]\sqrt{x} \sin x[/tex], the substitution of pi into the argument only works if the argument is continuous at pi. Just seeing that sin(pi) = 0 is not enough. One must also check to see what the behavior of the radical factor is. Fortunately the limit is:

    [tex]\lim_{x \rightarrow \pi} \sqrt{x} \sin x = \sqrt{\pi} \sin \pi = \sqrt{\pi} \cdot 0 = 0[/tex].

    (Aplogies in advance for potentially violating the "don't give out solutions" forum policy, but I am assuming the solution has already been found and I mean only to comment on being careful with the logic at arriving at that solutin.)

    --Elucidus
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?