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Math Help - trig limit(2)

  1. #1
    Member akhayoon's Avatar
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    trig limit(2)

    \lim_{x\to1}\frac{sin(x^2-1)}{x-1}

    I tried making t=x-1

    but I was left with sin(t(x+1)) which I didn't know what to do with
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  2. #2
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    Quote Originally Posted by akhayoon View Post
    \lim_{x\to1}\frac{sin(x^2-1)}{x-1}

    I tried making t=x-1

    but I was left with sin(t(x+1)) which I didn't know what to do with
    t = x - 1 we get \lim_{t\to 0} \frac{\sin t(t+2)}{t}.

    We get, \lim_{t\to 0}\frac{\sin t^2 \cos 2t + \cos t^2\sin 2t}{t}

    Replace \cos 2t and \cos t^2 by 1 because that is their limit and we have:

    \lim_{t\to 0}\frac{\sin t^2}{t} + \frac{\sin 2t}{t} = \lim_{t\to 0}\frac{\sin t^2}{t^2}\cdot t + \frac{\sin 2t}{2t} \cdot 2 = 2
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  3. #3
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    There are many ways to approach these limits.

    You could let t=x^{2}-1

    x=\sqrt{t+1}

    \lim_{t\rightarrow{0}}\frac{sin(t)}{\sqrt{t+1}-1}

    Now, multiply by the conjugate:

    \lim_{t\rightarrow{0}}\frac{\sqrt{t+1}sin(t)}{t}+\  lim_{t\rightarrow{0}}\frac{sin(t)}{t}

    We can see the familiar sin(t)/t limit. Which equals 1.

    So you have the limit 1+1=2
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  4. #4
    Member akhayoon's Avatar
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    sorry,I don't understand how you got \sqrt{t+1}-1

    nor do I understand how you manipulated the rest of the equation to make those to limits at the end
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  5. #5
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    Here's another attempt.

    Quote Originally Posted by akhayoon View Post
    \lim_{x\to1}\frac{sin(x^2-1)}{x-1}
    \lim_{x \to 1} \frac{{\sin \left( {x^2 - 1} \right)}}<br />
{{x - 1}} = \lim_{x \to 1} \frac{{\sin \left( {x^2 - 1} \right)}}<br />
{{x^2 - 1}} \cdot (x + 1).

    You can see the limit of the first factor exists and it's equal to 1, just make u=x^2-1 to check it.

    From here you can easily see that the requested value of the limit is 2.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akhayoon View Post
    \lim_{x\to1}\frac{sin(x^2-1)}{x-1}

    I tried making t=x-1

    but I was left with sin(t(x+1)) which I didn't know what to do with
    \lim_{x\to{1}}\frac{\sin(x^2-1)}{x-1}=\lim_{x\to{1}}\frac{\sin(x^2-1)-\sin(1^2-1)}{x-1}=\bigg(\sin(x^2-1)\bigg)'\bigg|_{x=1}=2

    That was a tad simpler I think
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