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Math Help - Determine whether or not f'(0) exists.

  1. #1
    Member mybrohshi5's Avatar
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    Determine whether or not f'(0) exists.

    Let:

    Determine whether or not exists.

    Yes or No?

    I think it is yes because the derivative of 0 is just 0 so f'(0) would still exist at 0?

    Am i correct or no?

    Thank you

    Chad
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  2. #2
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    Quote Originally Posted by mybrohshi5 View Post
    Let:

    Determine whether or not exists.

    Yes or No?

    I think it is yes because the derivative of 0 is just 0 so f'(0) would still exist at 0?

    Am i correct or no?

    Thank you

    Chad
    Remember that the derivative is defined in terms of a limit, so you have:

    f'(0)= \lim_{t \rightarrow 0} \frac{f(0+t) - f(0)}{t} = \lim_{t \rightarrow 0} \frac{f(t)}{t}= \lim_{t \rightarrow 0} [\sin (\frac{5}{t} )] which doesn't exist, therefore f'(0) doesn't exist.
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  3. #3
    Member mybrohshi5's Avatar
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    thank you.

    so then for



    will it also be no because as the lim as x-->0 3x^3 sin 1/x will not exists so, f'(0) would not exist correct?
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  4. #4
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    Quote Originally Posted by mybrohshi5 View Post
    thank you.

    so then for



    will it also be no because as the lim as x-->0 3x^3 sin 1/x will not exists so, f'(0) would not exist correct?
    Be careful:

    f'(0)= \lim_{t \rightarrow 0} \frac{3t^3 \sin( \frac{1}{t} ) }{t} =\lim_{t \rightarrow 0} 3t^2 \sin (\frac{1}{t} ) and since \sin (\frac{1}{t} ) is bounded, the limit is 0 so f'(0)=0
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  5. #5
    Member mybrohshi5's Avatar
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    can i ask a quick question about these two problems?

    you said sin(1/x) is bounded so therefore it does exist. why is sin(1/x) bounded and sin(5/x) not bounded which therefore does not make sin(5/x) exist?

    From looking at the graphs of both they both seem to be bounded.

    I guess im just kind of confused on why one exists and the other does not and i have a test coming up this week so im just trying to make sure i understand everything clearly.

    Any more help or explanations would be much appreciated.

    Thank you

    Chad
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  6. #6
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    Krizalid's Avatar
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    those are bounded.

    read again post #2, it's different from post #4.
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  7. #7
    Member mybrohshi5's Avatar
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    ok i just looked over both of those posts.

    so for xsinx^(5/x) why was the lim x-->0 f(0 + t) - f(0) / t used

    and then for 3x^3sin(1/x) why was the lim x-->0 3t^3sin(1/t)/t used?

    I mean like why are two different things used for similar functions?
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  8. #8
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    Quote Originally Posted by mybrohshi5 View Post
    ok i just looked over both of those posts.

    so for xsinx^(5/x) why was the lim x-->0 f(0 + t) - f(0) / t used

    and then for 3x^3sin(1/x) why was the lim x-->0 3t^3sin(1/t)/t used?

    I mean like why are two different things used for similar functions?
    \lim_{t \rightarrow 0} \frac{f(0+t)-f(0)}{t}= \lim_{t \rightarrow 0} \frac{3t^3 \sin ( \frac{1}{t} )}{t}. It's the same. The first limit doesn't exist because although  \sin( \frac{1}{t} ) is bounded the limit doesn't exist, but for the second limit, the term 3t^2 dominates and it goes to 0
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