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Limits of sinx/x

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data
    lim x^2-17x+72 / sin (x-9)
    x>9

    2. Relevant equations
    sinx/x = 1


    3. The attempt at a solution
    i divided both top and bottom by x.. and then im stuck on where sin(x-9)/x.. does it equal x-9?

    and does sin(x-y) = sin(x)-sin(y)?
     
  2. jcsd
  3. Oct 8, 2009 #2

    jbunniii

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    I assume you mean

    [tex]\lim_{x \rightarrow 9} \frac{x^2 - 17x + 72}{\sin(x - 9)}[/tex]

    Is that correct?

    Actually, this equation doesn't have a solution. Do you mean

    [tex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/tex]

    [tex]\frac{\sin(x-9)}{x} \neq x - 9[/tex]

    (except when [itex]x = 9[/itex]). I assume you are trying to compute

    [tex]\lim_{x \rightarrow 9}\frac{\sin(x-9)}{x}[/tex]

    Is that correct? If so, you can just plug in 9 for x, because the denominator isn't zero in that case. Therefore,

    [tex]\lim_{x \rightarrow 9}\frac{\sin(x-9)}{x} = \frac{\sin(9-9)}{9} = \frac{0}{9} = 0[/tex]

    It depends on what x and y are. In general, no.
     
  4. Oct 8, 2009 #3
    [tex]\sin(x - y) \ne \sin(x) - \sin(y)[/tex]
    [tex]\sin(x \pm y) = \sin x \cos y \pm \sin y \cos x[/tex]

    Try using that identity.
     
  5. Oct 8, 2009 #4

    Dick

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    Uh, x^2-17x+72=(x-9)*(x-8). So you've got lim x->9 (x-9)(x-8)/sin(x-9). I suggest you substitute u=x-9 and take the limit as u->0.
     
  6. Oct 8, 2009 #5
    you meant u>9 right?

    if so, i get
    [tex]$\lim _{x\to 9}\frac{x^2-17 x+72}{\sin (x-9)}=$[/tex][tex]$\lim _{x\to 9}\frac{(x-9)(x-8)}{\sin (x-9)}=$[/tex][tex]$\lim _{x\to 9}\frac{u(x-8)}{\sin (u)}=$[/tex]
    [tex]$\lim _{x\to 9}\frac{u(x-8) / u}{\sin (u) / u}=$[/tex][tex]$\lim _{x\to 9}\frac{x-8}$[/tex]

    =9-8=1

    right?
     
  7. Oct 8, 2009 #6

    Dick

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    Right, but I meant what I said. If x->9 then u=(x-9)->0. sin(u)/u->1 as u->0. If u doesn't approach 0 then sin(u)/u probably doesn't approach 1. I think that's what jbunniii was saying.
     
  8. Oct 8, 2009 #7
    oh i get it..
    thanks for help!!
     
  9. Oct 8, 2009 #8

    jbunniii

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    This is right, but if you're going to make a variable substitution, it's much better form to make the substitution everywhere, not just in some of the terms. Thus

    [tex]\lim_{x \rightarrow 9}\frac{(x-9)(x-8)}{\sin(x-9)} = \lim_{u \rightarrow 0}\frac{(u)(u+1)}{\sin(u)} = \lim_{u \rightarrow 0}\left(\frac{u}{\sin(u)}\right) \cdot \lim_{u \rightarrow 0} (u+1)[/tex]
     
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