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Math Help - Vector Function Help

  1. #1
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    Vector Function Help

    I need some help understanding how:

    Find {\lim_{t \to 0} r(t)}, where r(t) = (1+t^3) i +  (te^{-t}) j + (\frac{sin(t)}{t}) k

    = {\lim_{t \to 0} r(t)} = [{\lim_{t \to 0} (1+t^3)}] i + [{\lim_{t \to 0} (te^{-t})}] j + [{\lim_{t \to 0} (\frac{sin(t)}{t})}] k

    = i + k (form the book)

    I would think that one would have to replace the t by 0, getting:

    = {\lim_{t \to 0} r(t)} = [{\lim_{t \to 0} (1+0^3)}] i + [{\lim_{t \to 0} ((0)e^{0})}] j + [{\lim_{t \to 0} (\frac{sin(0)}{0})}] k

    = {\lim_{t \to 0} r(t)} = [{\lim_{t \to 0} (1)}] i + [{\lim_{t \to 0} (0)}] j + [{\lim_{t \to 0} (D.N.E)}] k

    = i

    So I guess what I'm asking is why is it equal to i + k and not i?
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  2. #2
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    [{\lim_{x \to 0} (\frac{sin(x)}{x})}] = 1.

    It's just one of those things, a "special limit" to basically just memorize. At least that's how I look at it.

    If you want to see a proof of this particular limit, this is a good place to start:

    http://www.csun.edu/~ac53971/courses.../xtra_sine.pdf.

    Also, if a limit does not exist, as you posited for your k vector, that doesn't mean it's zero - zero is a number that exists just like 1, 2, 3, etc. If you find that a limit does not exist but you need to evaluate it, you need to figure out a way TO evaluate it.
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  3. #3
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    Thanks, but what if you have other values like:

    1) {\lim_{t \to 0} \frac{e^{t}-1}{t}}

    2) {\lim_{t \to 0} \frac{t^2}{sin^{2}t}}

    3) {\lim_{t \to 0} \frac{\sqrt{1+t}-1}{t}}

    will these be equal to some specific number, or will they all be undefined? And or those specific properties only applicable on trigonometric functions?
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  4. #4
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    Quote Originally Posted by lllll View Post
    Thanks, but what if you have other values like:

    1) {\lim_{t \to 0} \frac{e^{t}-1}{t}}

    2) {\lim_{t \to 0} \frac{t^2}{sin^{2}t}}

    3) {\lim_{t \to 0} \frac{\sqrt{1+t}-1}{t}}

    will these be equal to some specific number, or will they all be undefined? And or those specific properties only applicable on trigonometric functions?
    There are a number of different approaches that can be taken.

    1) From l'Hopital's rule:

    \lim_{t \rightarrow 0} \frac{e^t - 1}{t} = \lim_{t \rightarrow 0} \frac{e^t}{1} = 1.

    You could also substitute the power series for e^t:

    \lim_{t \rightarrow 0} \frac{e^t - 1}{t} = \lim_{t \rightarrow 0} \frac{(1 + t + \frac{t^2}{2!} + ...) - 1}{t} = \lim_{t \rightarrow 0} (1 + \frac{t}{2!} + ....) = 1.



    2) From l'Hopital's Rule:

    \lim_{t \rightarrow 0} \frac{t^2}{\sin^2 t} = \lim_{t \rightarrow 0} \frac{2t}{2 \sin t \cos t} = \lim_{t \rightarrow 0} \frac{2t}{\sin (2t)} = \lim_{t \rightarrow 0} \frac{2}{2 \cos (2t)} = 1.

    Alternatively, using known limits:

    \lim_{t \rightarrow 0} \frac{t^2}{\sin^2 t} = \lim_{t \rightarrow 0} \left( \frac{t}{\sin t} \right) \lim_{t \rightarrow 0} \left( \frac{t}{\sin t} \right) = 1 \times 1 = 1.



    3) \lim_{t \rightarrow 0} \frac{\sqrt{1+t} - 1}{t} = \lim_{t \rightarrow 0} \frac{(\sqrt{1+t} - 1)}{t} \times \frac{(\sqrt{1+t} + 1)}{(\sqrt{1+t} + 1)}


    = \lim_{t \rightarrow 0} \frac{(1 + t) - 1}{t (\sqrt{1+t} + 1)} = \lim_{t \rightarrow 0} \frac{1}{\sqrt{1+t} + 1} = \frac{1}{1 + 1} = \frac{1}{2}.
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