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Math Help - L'hospitals rule mistake?

  1. #1
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    L'hospitals rule mistake?

    lim x-> 0 of sin5x/sinx
    Is the answer 5?

    Basically since its 0/0 i used L'hospitals rule and took the derivative of top and bottom to get:
    5cos5x/cosx
    which it then 5(1)/1... so =5?

    But then i put it into this website that calculates it for you and it said the answer is 1 soo im a tad confused.
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by kensington View Post
    lim x-> 0 of sin5x/sinx
    Is the answer 5?

    Basically since its 0/0 i used L'hospitals rule and took the derivative of top and bottom to get:
    5cos5x/cosx
    which it then 5(1)/1... so =5?

    But then i put it into this website that calculates it for you and it said the answer is 1 soo im a tad confused.
    it's 5 where did you calculate ?

    try here Wolfram|Alpha—Computational Knowledge Engine
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  3. #3
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    \displaystyle \lim_{x \to 0} \frac{\sin(5x)}{\sin{x}} = 5
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    Member mfetch22's Avatar
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    Quote Originally Posted by kensington View Post
    lim x-> 0 of sin5x/sinx
    Is the answer 5?

    Basically since its 0/0 i used L'hospitals rule and took the derivative of top and bottom to get:
    5cos5x/cosx
    which it then 5(1)/1... so =5?

    But then i put it into this website that calculates it for you and it said the answer is 1 soo im a tad confused.
    I beleive that your answer is correct, and that either the website made a mistake or you might've typed it in wrong (since using calculator syntax can be a cause for error in a problem like this one). I support your answer, not only in agreeing with your methodology, but I ploted y =\frac{sin(5x)}{sin(x)} and it most certainly appears to approach 5 as x \to 0. I could be wrong though. Lets see what other people say.
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  5. #5
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    Quote Originally Posted by yeKciM View Post
    it's 5 where did you calculate ?

    try here Wolfram|Alpha—Computational Knowledge Engine
    I used Mathway: Math Problem Solver and i know i typed it in right because it showed me the question and then the answer.. wierd.
    But i was ready to give up because i didnt see how they could have gotten 1
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  6. #6
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by kensington View Post
    I used Mathway: Math Problem Solver and i know i typed it in right because it showed me the question and then the answer.. wierd.
    But i was ready to give up because i didnt see how they could have gotten 1
    to try it in wolfram type

    lim_(x->0) sin(5x)/sin(x)

    many times (for "easy" calculating) you have steps if you need to check your work...
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  7. #7
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    Quote Originally Posted by kensington View Post
    lim x-> 0 of sin5x/sinx
    Is the answer 5?

    Basically since its 0/0 i used L'hospitals rule and took the derivative of top and bottom to get:
    5cos5x/cosx
    which it then 5(1)/1... so =5?

    But then i put it into this website that calculates it for you and it said the answer is 1 soo im a tad confused.
    Alternatively...

    \displaystyle\huge\frac{ Sin(5x)}{Sin(x)}=\frac{5x}{5x}\frac{Sin(5x)}{Sin(x  )}=5\frac{x}{Sin(x)}\frac{Sin(5x)}{5x}

    As "u" approaches zero, Sinu\rightarrow\ u

    Now use \displaystyle\huge\lim_{u\rightarrow0}\frac{Sinu}{  u}=1
    Last edited by Archie Meade; August 15th 2010 at 03:33 AM.
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  8. #8
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    Quote Originally Posted by Archie Meade View Post
    Alternatively...

    \displaystyle\huge\frac{ Sin(5x)}{Sin(x)}=\frac{5x}{5x}\frac{Sin(5x)}{Sin(x  )}=5\frac{x}{Sin(x)}\frac{Sin(5x)}{5x}

    Now use \displaystyle\huge\lim_{u\rightarrow0}Sinu=u
    Use instead \lim_{u\to 0} \frac{sin u}{u}= 1.

    \lim_{u\to 0} Sin u= 0, not u.
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