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Math Help - Limits and Continuity

  1. #1
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    Limits and Continuity

    Please help me solve lim (sin 2x)/x

    x approaching zero.

    I'm supposed to determine the limit graphically and then confirm algebraically.

    Thank you.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by macgyver View Post
    Please help me solve lim (sin 2x)/x

    x approaching zero.

    I'm supposed to determine the limit graphically and then confirm algebraically.

    Thank you.

    Here's the graph:



    Can you determine \lim_{x\to 0}\frac{\sin(2x)}{x} now?

    Here's a hint on how to verify it algebraically.

    \lim_{x\to 0}\frac{\sin(2x)}{x}=\lim_{x\to 0}\frac{2\sin(2x)}{2x}=2\lim_{x\to 0}\frac{\sin(2x)}{2x}

    This limit should look pretty familiar...

    Does this help?

    --Chris
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  3. #3
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    Hello,

    I don't know what you mean by "graphically"(draw a graph? computing areas?).
    For the "algebraically" one, do you know lim((sin x)/x)=1?
    If so, transform as (sin 2x)/x=2(sin 2x)/(2x) and let 2x go to 0.

    Bye.

    P.S. Seems I was slower than Chris.
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    Here's the graph:



    Can you determine \lim_{x\to 0}\frac{\sin(2x)}{x} now?

    Here's a hint on how to verify it algebraically.

    \lim_{x\to 0}\frac{\sin(2x)}{x}=\lim_{x\to 0}\frac{2\sin(2x)}{2x}=2\lim_{x\to 0}\frac{\sin(2x)}{2x}

    This limit should look pretty familiar...

    Does this help?

    --Chris
    Thank you very much for responding Chris. I only get part of it though. could you please tell me how to determine the limit just by looking at the graph? and also how do I solve the rest of the equation. I am very new to calculus so I would really appreciate your help.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by macgyver View Post
    Thank you very much for responding Chris. I only get part of it though. could you please tell me how to determine the limit just by looking at the graph? and also how do I solve the rest of the equation. I am very new to calculus so I would really appreciate your help.
    To find the limit graphically, look at the value of the graph at x=0. In this case it appears to be 2, although it doesn't exist at 0 [otherwise we would be dividing by zero, which is illegal].

    To finish it off, this is what we would do algebraically:

    \lim_{x\to 0}\frac{\sin(2x)}{x}=\lim_{x\to 0}\frac{2\sin(2x)}{2x}=2\lim_{x\to 0}\frac{\sin(2x)}{2x}

    This limit has the form of \lim_{u\to 0}\frac{\sin(u)}{u}=1

    So let u=2x. Thus, as x\to 0, u\to 0 as well.

    So our limit transforms from 2\lim_{x\to 0}\frac{\sin(2x)}{2x} to 2\lim_{u\to 0}\frac{\sin u}{u}=2(1)=\color{red}\boxed{2}

    Does this make sense?

    --Chris
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  6. #6
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    yes thank you that helped a lot.

    could u help me with this other one please?

    f(x)=(e^-x)/(x)

    I'm supposed to
    a: use graphs and tables to find lim x approaching infinity f(x)
    b: lim x approaching negative infinity f(x)
    c: Identify all horizontal asymptotes
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