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Math Help - find the limit

  1. #1
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    find the limit

    \lim_{x\to0} \frac{sin(cos x)}{sec x}

    I can just plug in zero but my teacher says that we can't use the direct substitution property with trig functions even if you can get the correct answer, which is lame. So how do I solve this? And yes its sin(cos x) not sin x(cos x)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by yoman360 View Post
    \lim_{x\to0} \frac{sin(cos x)}{sec x}

    I can just plug in zero but my teacher says that we can't use the direct substitution property with trig functions
    What's wrong with using direct substitution with trig functions?!?!

    \lim_{x\to0}\frac{\sin\left(\cos x\right)}{\sec x}=\frac{\sin\left(\cos 0\right)}{\sec 0}=\sin\left(1\right)
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    What's wrong with using direct substitution with trig functions?!?!

    \lim_{x\to0}\frac{\sin\left(\cos x\right)}{\sec x}=\frac{\sin\left(\cos 0\right)}{\sec 0}=\sin\left(1\right)
    Because the teacher says the definition for direct substitutution property can only be used for polynomial and rational functions.

    Theorem: (Direct Substitution Property) Suppose f(x) is a polyno-
    mial or a rational function and a is in the domain of f.

    http://web.viu.ca/wattsv/math121/Ove...bstitution.pdf

    He said that "using the direct substitutution property for trig functions won't be theromatically (if thats a word) correct."

    Seriously someone needs to revise the direct substitutution property so that it will be "theromatically" correct for trig functions.
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  4. #4
    ynj
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    Quote Originally Posted by yoman360 View Post
    \lim_{x\to0} \frac{sin(cos x)}{sec x}

    I can just plug in zero but my teacher says that we can't use the direct substitution property with trig functions even if you can get the correct answer, which is lame. So how do I solve this? And yes its sin(cos x) not sin x(cos x)
    \lim_{x\rightarrow 0}\frac{\sin (\cos x)}{\sec x}=\lim_{x\rightarrow 0}\sin (\cos x)\cos x=\lim_{x\rightarrow 0}\sin (\cos x)\lim_{x\rightarrow 0}\cos x
    =\sin (\cos \lim_{x\rightarrow 0}x)\cos 0=\sin (\cos 0)\cos 0=\sin 1
    Note that there is also another theorem:
    if f(x)is continuous at x_0, then
    \lim_{x\rightarrow x_0} f(x)=f(x_0),not just your teacher have said!
    Last edited by ynj; August 31st 2009 at 08:44 PM.
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  5. #5
    ynj
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    Also note that \cos ,\sin\circ\cosare both continuous at 0
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  6. #6
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    Quote Originally Posted by ynj View Post
    \lim_{x\rightarrow 0}\frac{\sin (\cos x)}{\sec x}=\lim_{x\rightarrow 0}\sin (\cos x)\cos x=\lim_{x\rightarrow 0}\sin (\cos x)\lim_{x\rightarrow 0}\cos x
    =\sin (\cos \lim_{x\rightarrow 0}x)\cos 0=\sin (\cos 0)\cos 0=\sin 1
    Note that there is also another theorem:
    if f(x)is continuous at x_0, then
    \lim_{x\rightarrow x_0} f(x)=f(x_0),not just your teacher have said!
    Thank you so much
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  7. #7
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    Your teacher's point was that you had not yet learned that theorem and so shouldn't use it until you had. That is what was "theoretically" bad. (No, "theromatically", as good as it sounds, is NOT a word!)
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