- #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)?

- Thread starter Slimsta
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- #1

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

x>9

sinx/x = 1

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

- 3,394

- 180

I assume you mean## Homework Statement

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

x>9

[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## Homework Equations

sinx/x = 1

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

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

(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.and does sin(x-y) = sin(x)-sin(y)?

- #3

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[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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- #5

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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?

- #6

Dick

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- #7

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oh i get it..

thanks for help!!

- #8

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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. Thusyou 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?

[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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