# Thread: Help With Trig Limit

1. ## Help With Trig Limit

How do you solve the limit: lim x --> 0 sin (9x)/ sin(5x)

I really don't have much of a clue where to start, so if someone could explain every step of how to solve this, I'd be real grateful.

How do you solve the limit: sin (9x)/ sin(5x)

I really don't have much of a clue where to start, so if someone could explain every step of how to solve this, I'd be real grateful.
Where is x going towards in this problem?

3. as $\displaystyle x\to0,$ otherwise what'd be the sense on taking other value.

How do you solve the limit: lim x --> 0 sin (9x)/ sin(5x)

I really don't have much of a clue where to start, so if someone could explain every step of how to solve this, I'd be real grateful.
Hint

$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{\sin 9x}} {{\sin 5x}} = \frac{9} {5} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{\sin 9x}} {{9x}} \cdot {\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin 5x}} {{5x}}} \right)^{ - 1}} = \ldots$

5. What I'm kind of confused on is how u factored out the 5/9 form the sin (x) equations. Could you please explain that?

Oh wait! Is the sin 9x/ 9x thing taken from the sin (x)/ x is equal to 1? But then don't you have to multiply the top and bottom both by 9x in order for it to remain equivalent? Same with the 5x?

What I'm kind of confused on is how u factored out the 5/9 form the sin (x) equations. Could you please explain that?

Oh wait! Is the sin 9x/ 9x thing taken from the sin (x)/ x is equal to 1? But then don't you have to multiply the top and bottom both by 9x in order for it to remain equivalent? Same with the 5x?
$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{\sin 9x}} {{\sin 5x}} = \frac{9} {5} \cdot \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin 9x}} {{9x}} \cdot \frac{{5x}} {{\sin 5x}}} \right] =$

$\displaystyle = \frac{9} {5} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{\sin 9x}} {{9x}} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{5x}} {{\sin 5x}} =$

$\displaystyle = \frac{9} {5} \cdot \mathop {\lim }\limits_{x \to 0} \frac{{\sin 9x}} {{9x}} \cdot {\left( {\mathop {\lim }\limits_{x \to 0} \frac{{\sin 5x}} {{5x}}} \right)^{ - 1}} = \ldots$

Now use this

$\displaystyle {\color{red}\boxed{{\color{black}\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{ax}} = 1}}}$