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Math Help - Trig limits

  1. #1
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    Trig limits

    Find \lim_{x\rightarrow1} \frac{sin(x-1)}{x^2+x-2}

    = \lim_{x\rightarrow1}\frac{sin(x-1)}{(x+2)(x-1)}

    do the (x-1) terms cancel out? If so, doesn't that leave me with \lim_{x\rightarrow1}\frac{sinx}{x+2} ?

    then by direct sub \lim_{x\rightarrow1}\frac{sinx}{x+2} = \frac{sin1}{3} ?
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  2. #2
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    Re: Trig limits

    Quote Originally Posted by Foxlion View Post
    Find \lim_{x\rightarrow1} \frac{sin(x-1)}{x^2+x-2}

    = \lim_{x\rightarrow1}\frac{sin(x-1)}{(x+2)(x-1)}

    do the (x-1) terms cancel out? If so, doesn't that leave me with \lim_{x\rightarrow1}\frac{sinx}{x+2} ?

    then by direct sub \lim_{x\rightarrow1}\frac{sinx}{x+2} = \frac{sin1}{3} ?
    you should already be familiar with this well known trig limit ...

    \lim_{\theta \to 0} \frac{\sin{\theta}}{\theta} = 1


    \lim_{x \to 1} \frac{\sin(x-1)}{(x-1)(x+2)} =

    \lim_{x \to 1} \frac{\sin(x-1)}{x-1} \cdot \frac{1}{x+2} =

    \lim_{x \to 1} \frac{\sin(x-1)}{x-1} \cdot \lim_{x \to 1} \frac{1}{x+2} =

    1 \cdot \frac{1}{3} = \frac{1}{3}
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  3. #3
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    Re: Trig limits

    Hello, skeeter!

    You should already be familiar with this well known trig limit:

    . . \lim_{\theta \to 0} \frac{\sin{\theta}}{\theta} = 1

    I'm afraid that he will ask:

    . . Do the \theta cancel out?
    . . If so, doesn't that leave me with: . \lim_{\theta\to0}\,\sin \:=\:1\,?

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  4. #4
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    Re: Trig limits

    Quote Originally Posted by Soroban View Post
    Hello, skeeter!


    I'm afraid that he will ask:

    . . Do the \theta cancel out?
    . . If so, doesn't that leave me with: . \lim_{\theta\to0}\,\sin \:=\:1\,?

    That's funny...
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  5. #5
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    Re: Trig limits

    Quote Originally Posted by Soroban View Post
    Hello, skeeter!


    I'm afraid that he will ask:

    . . Do the \theta cancel out?
    . . If so, doesn't that leave me with: . \lim_{\theta\to0}\,\sin \:=\:1\,?

    Isn't this the same reason why \displaystyle \begin{align*} \frac{\sin{x}}{n} = \textrm{six} = 6 \end{align*}?
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  6. #6
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    Re: Trig limits

    Quote Originally Posted by Prove It View Post
    Isn't this the same reason why \displaystyle \begin{align*} \frac{\sin{x}}{n} = \textrm{six} = 6 \end{align*}?
    Exactly!, we'll call it a corollary prop of the alphabetical cancellation rule...
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