# Limits of sinx/x

1. Oct 8, 2009

### Slimsta

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. Oct 8, 2009

### jbunniii

I assume you mean

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

Is that correct?

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

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

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

(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,

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

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

3. Oct 8, 2009

### 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.

4. Oct 8, 2009

### 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.

5. Oct 8, 2009

### Slimsta

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?

6. Oct 8, 2009

### 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.

7. Oct 8, 2009

### Slimsta

oh i get it..
thanks for help!!

8. Oct 8, 2009

### jbunniii

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

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