Limits of sinx/x

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  • #1
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Homework Statement


lim x^2-17x+72 / sin (x-9)
x>9

Homework Equations


sinx/x = 1


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)?
 

Answers and Replies

  • #2
jbunniii
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Homework Statement


lim x^2-17x+72 / sin (x-9)
x>9
I assume you mean

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

Is that correct?

Homework Equations


sinx/x = 1
Actually, this equation doesn't have a solution. Do you mean

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

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?
[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]

and does sin(x-y) = sin(x)-sin(y)?
It depends on what x and y are. In general, no.
 
  • #3
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[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.
 
  • #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.
 
  • #5
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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.
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?
 
  • #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.
 
  • #7
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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.
oh i get it..
thanks for help!!
 
  • #8
jbunniii
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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?
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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