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Math Help - limits of a vector function

  1. #1
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    limits of a vector function

    Hello, the question is:

    limit t-->0+ (cos(t), sin(t), tln(t))

    (the limit as t approaches 0 from the right)

    Thanks,
    -cb
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  2. #2
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    Quote Originally Posted by sprinks13 View Post
    Hello, the question is:

    limit t-->0+ (cos(t), sin(t), tln(t))

    (the limit as t approaches 0 from the right)

    Thanks,
    -cb
    Just pass the limit on each term.
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  3. #3
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    yup, I got (1,0,?) i can't figure out the tln(t) part.

    Doesn't ln(t) approach infinity when t approaches 0? And then it is 0+ multiplied by infinity?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sprinks13 View Post
    yup, I got (1,0,?) i can't figure out the tln(t) part.

    Doesn't ln(t) approach infinity when t approaches 0? And then it is 0+ multiplied by infinity?
    note that t \ln t = \frac {\ln t}{\frac 1t}. use L'Hopital's rule (i know everyone hates me for saying that, but i don't care) (and lnt goes to -infinity as t goes to 0)
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    (i know everyone hates me for saying that, but i don't care)
    I am working on one approach without using L'Hopital.
    (This is not for the original poster. Do not confuse yourself. This is just for fun for other posters).
    ---
    Directly by definition.
    ---
    We want to show,
    \lim_{t\to 0^+} t\ln t = \lim_{t\to 0^+} t\int_1^t \frac{1}{x} dx = \lim_{t\to 0^+} - t\int_t^1 \frac{1}{x} dx

    Ignore the minus sign for now. Now try to show that \lim_{t\to 0^+} t\int_t^1 \frac{1}{x} \ dx = 0.

    Let t be small enough so that we are working on the interval [t,1].

    Now, \frac{1}{x} \leq \frac{1}{e^x} on [t,1] because e^x \geq x on (0,1]. Thus, \int_t^1 \frac{1}{x} dx \leq \int_1^t \frac{1}{e^x} dx = \frac{1}{e^t}-1.

    This means,
    \left| t\int_t^1 \frac{1}{x} dx \right| \leq \frac{t}{e^t} - t \leq \frac{t}{e^t}.
    But,
    \lim_{t\to 0^+} \frac{t}{e^t} = 0.

    No L'Hopital!
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