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Math Help - Limits Problem, can someone help with this?

  1. #1
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    Limits Problem, can someone help with this?

    Lim
    x-> Pi/4

    1+tan(x)
    ---------
    csc(x)+2

    I'm guessing you can take the lim of 1, lim of tan(x) and divide it by the lim of csc(x) + 2?
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  2. #2
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    Why can't you just substitute x = \frac{\pi}{4}?

    You should know that \tan{\frac{\pi}{4}} = 1 and \csc{\frac{\pi}{4}} = \frac{1}{\sin{\frac{\pi}{4}}} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Why can't you just substitute x = \frac{\pi}{4}?

    You should know that \tan{\frac{\pi}{4}} = 1 and \csc{\frac{\pi}{4}} = \frac{1}{\sin{\frac{\pi}{4}}} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}.
    Thank you, I don't know why I didn't think of that. I'm taking Calc for the first time as a summer class at my university and the professor is just flying by the course so I'm trying my best to keep up out of class
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  4. #4
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    As long as the denominator is not 0, taking the limit of each part separately is just using the various "laws of limits" and is valid. If each part is a continuous function (and the denominator is not 0 so the whole combination is a continous function) then that is exactly the same as "substitute x= \frac{\pi}{4}".
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  5. #5
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    Here's another one I'm having trouble with


    lim (s^3-y^3)/(s^4-y^4)
    s->y


    My professor didn't teach us the L'Hopital's rule, so I have absolutely no idea how to go about this one.
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Why can't you just substitute s=y?
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Why can't you just substitute s=y?
    This is actually a web assignment due and if I just substitue in y I get 0/0 which can't be done.
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  8. #8
    Senior Member eumyang's Avatar
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    Try factoring?

    s^3 - y^3 = (s - y)(s^2 + sy + y^2)
    and
    \begin{aligned}<br />
s^4 - y^4 &= (s^2 - y^2)(s^2 + y^2) \\<br />
&= (s - y)(s + y)(s^2 + y^2)<br />
\end{aligned}

    Then the (s - y) factors cancel out.
    \lim_{s \to y}\frac{s^3 - y^3}{s^4 - y^4}
    = \lim_{s \to y}\frac{s^2 + sy + y^2}{(s + y)(s^2 + y^2)}
    = \frac{y^2 + y^2 + y^2}{(2y)(y^2 + y^2)}
    = \frac{3y^2}{4y^3}
    = \frac{3}{4y}

    EDIT: Typing in LaTeX is #($*&#) difficult, grr...
    Last edited by eumyang; July 6th 2010 at 12:35 PM.
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  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Nikhiln25 View Post
    This is actually a web assignment due and if I just substitue in y I get 0/0 which can't be done.

    Ohhh really?!

    What you write first time was:

    lim_{s\to y}(s^3-y^3)/(s^4/y^4)


    Bravo!
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  10. #10
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Ohhh really?!

    What you write first time was:

    lim_{s\to y}(s^3-y^3)/(s^4/y^4)


    Bravo!
    lol I realized and I edited sorry about that, I thought I was quick enough to update
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  11. #11
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    Quote Originally Posted by eumyang View Post
    Try factoring?

    s^3 - y^3 = (s - y)(s^2 + sy + y^2)
    and
    \begin{aligned}<br />
s^4 - y^4 &= (s^2 - y^2)(s^2 + y^2) \\<br />
&= (s - y)(s + y)(s^2 + y^2)<br />
\end{aligned}

    Then the (s - y) factors cancel out.
    \frac{s^2 + sy + y^2}{(s + y)(s^2 + y^2)}
    My answer has to be in terms of y according to my assignment and if I substitute y in from here I end up getting y/2 as the limit which isn't showing up as right
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  12. #12
    Senior Member eumyang's Avatar
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    Quote Originally Posted by Nikhiln25 View Post
    lol I realized and I edited sorry about that, I thought I was quick enough to update
    ^^ I was doing the same thing, editing while you were replying. Look at my post again.
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  13. #13
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    Quote Originally Posted by eumyang View Post
    ^^ I was doing the same thing, editing while you were replying. Look at my post again.
    Yeah that answer worked, and it makes a lot more sense now. I emailed my professor for help earlier and he told me to use synthetic division which got me lost even more. Thank you.
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  14. #14
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    Can someone tell me how to solve these kind of problems? I have a couple of these in my homework, and had one on a quiz today but couldn't really figure out how to solve them

    1)
    lim sin4x/9x
    x->0

    2)
    lim
    x->0 tan4x/2x

    I tried to put it into wolfram alpha but that used the L'Hopital's rule which made no sense to me since the professor didn't teach L'Hopital's rule to my class
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  15. #15
    MHF Contributor Also sprach Zarathustra's Avatar
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    ...

    Here something useful...

    lim_{x\to 0} \frac{sin x}{x}=1

    in your case (q. a):


    lim_{x\to 0} \frac{sin 4x}{9x}=\frac{4}{9}lim_{x\to 0} \frac{sin 4x}{4x}

    Try apply this also for b.
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