# Limits of sinx/x

## Homework Statement

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

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

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jbunniii
Homework Helper
Gold Member

## 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 im 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.

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

Try using that identity.

Dick
Homework Helper
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.

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
Homework Helper
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.

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
Homework Helper
Gold Member
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

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