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Math Help - [SOLVED] Limits with trig functions

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Question [SOLVED] Limits with trig functions

    i really hate finding the limits of equations with [for ex.] sinx or xsin x...it throws me off and i dont know what to do...im going ovver some problems over some practice problems because my test is either tomorrow or thursday.

    lim x->0 x cscx + 1 / x cscx

    the only thing i would have to guess is thati might have to do something with 1/sinx. i may be wrong.

    Thank you in advance
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  2. #2
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    An important trig limit that you need to use in this problem is \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 See if you can apply that to your problem.
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    wait wait wait, i think i got it...since cscx = 1/sin x, you can [since it is multiplied by x again...] you have wht you have provided...and then that equals one...meaning 1+1/1 = 2, the limit!


    Quote Originally Posted by Jameson View Post
    An important trig limit that you need to use in this problem is \lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1 See if you can apply that to your problem.
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  4. #4
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    i really hate finding the limits of equations with [for ex.] sinx or xsin x...it throws me off and i dont know what to do...im going ovver some problems over some practice problems because my test is either tomorrow or thursday.

    lim x->0 x cscx + 1 / x cscx

    the only thing i would have to guess is thati might have to do something with 1/sinx. i may be wrong.

    Thank you in advance
    Is the expression \frac{\text{cosec} (x) + 1}{x \, \text{cosec} x} ? If so, then it can be re-written as \frac{1 + \sin x}{x} and the limit clearly does not exist.
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