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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


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    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


    simplify and what is ...

  5. Sep 18, 2007 #4


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    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


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    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]


    [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


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    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


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    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.)

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