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Thread: trig limit(2)

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

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

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

    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 $\displaystyle \lim_{t\to 0} \frac{\sin t(t+2)}{t}$.

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

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

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

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

    $\displaystyle x=\sqrt{t+1}$

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

    Now, multiply by the conjugate:

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

    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 $\displaystyle \sqrt{t+1}-1$

    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
    $\displaystyle \lim_{x\to1}\frac{sin(x^2-1)}{x-1}$
    $\displaystyle \lim_{x \to 1} \frac{{\sin \left( {x^2 - 1} \right)}}
    {{x - 1}} = \lim_{x \to 1} \frac{{\sin \left( {x^2 - 1} \right)}}
    {{x^2 - 1}} \cdot (x + 1).$

    You can see the limit of the first factor exists and it's equal to 1, just make $\displaystyle u=x^2-1$ 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
    $\displaystyle \lim_{x\to1}\frac{sin(x^2-1)}{x-1}$

    I tried making t=x-1

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

    That was a tad simpler I think
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