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Math Help - Limit and Chain Rule Questions

  1. #1
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    Limit and Chain Rule Questions

    Hi, can someone help me with this?

    lim x->0 \frac{x-tan(4x)}{sin(4x)}

    I tried multiplying by the conjugate then using the trig identity for tan, but I got a bunch of mess I don't know what to do with.

    Also, for

    lim x->0 \frac{x^2-12x+32}{sin(x-4)}

    is the limit -8? just want to make sure that's right.

    and how do I solve this:

    let F(x) = h(g(f(x))), where f(4) = 7, g(7) = 8, f '(4) = 3, g '(7) = 8, and h '(8) = 2

    then find F'(4)

    confused as how to do this

    thanks!
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  2. #2
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    \frac{x - \tan(4x)}{\sin(4x)} = \frac{x}{\sin(4x)} - \frac{\tan(4x)}{\sin(4x)} = \frac{4x}{4\sin(4x)} - \frac{\frac{\sin(4x)}{\cos(4x)}}{\sin(4x)} = \frac{1}{4} \cdot \frac{4x}{\sin(4x)} - \frac{1}{\cos(4x)}
    now take the limit as x approaches 0.


    for the 2nd limit ... if x is approaching 0, then limit is \frac{32}{\sin(-4)}


    last one, use the chain rule ...

    F'(x) = h'[g(f(x))] \cdot g'(f(x)) \cdot f'(x)

    now sub in x = 4 and evaluate.
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  3. #3
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    thanks, still confused about the last one though

    can anyone help me find F'(4) if
    F(x) = h(g(f(x))), where f(4) = 7, g(7) = 8, f '(4) = 3, g '(7) = 8, and h '(8) = 2

    not sure where to sub in 4 or whatever.
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  4. #4
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    F'(4) = h'[g(f(4))] \cdot g'(f(4)) \cdot f'(4)

    F'(4) = h'[g(7)] \cdot g'(7) \cdot 3

    F'(4) = h'[8] \cdot 8 \cdot 3

    F'(4) = 2 \cdot 8 \cdot 3 = 48
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  5. #5
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    ahh of course, how did I not see that??

    thanks!
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