How Does the Limit of (x^2-17x+72) / sin(x-9) Evaluate as x Approaches 9?

[Slimsta]

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 I am stuck on where sin(x-9)/x.. does it equal x-9?

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

 

[jbunniii]

Homework Statement

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

I assume you mean

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

Is that correct?

Homework Equations

sinx/x = 1

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

\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1

The Attempt at a Solution

i divided both top and bottom by x.. and then I am stuck on where sin(x-9)/x.. does it equal x-9?

\frac{\sin(x-9)}{x} \neq x - 9

(except when x = 9). I assume you are trying to compute

\lim_{x \rightarrow 9}\frac{\sin(x-9)}{x}

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

\lim_{x \rightarrow 9}\frac{\sin(x-9)}{x} = \frac{\sin(9-9)}{9} = \frac{0}{9} = 0

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

It depends on what x and y are. In general, no.

 

[Bohrok]

\sin(x - y) \ne \sin(x) - \sin(y)
\sin(x \pm y) = \sin x \cos y \pm \sin y \cos x

Try using that identity.

 

[Dick]

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.

 

[Slimsta]

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
$\lim _{x\to 9}\frac{x^2-17 x+72}{\sin (x-9)}=$$\lim _{x\to 9}\frac{(x-9)(x-8)}{\sin (x-9)}=$$\lim _{x\to 9}\frac{u(x-8)}{\sin (u)}=$
$\lim _{x\to 9}\frac{u(x-8) / u}{\sin (u) / u}=$$\lim _{x\to 9}\frac{x-8}$

=9-8=1

right?

 

[Dick]

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.

 

[Slimsta]

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!

 

[jbunniii]

you meant u>9 right?

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

=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

\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)