# Limit (sin(4x)/sin(6x)) as x->0

1. Aug 29, 2014

Hi all,

I'm just beginning calculus and I'm having trouble figuring this one out.

I need to solve this one without using l'hospital's rule.

1. The problem statement, all variables and given/known data

Find

Limit (sin(4x)/sin(6x)) as x->0

2. Relevant equations

We know that limit (sin(x)/x) as x -> 0 equals 1

Thanks!!! Much appreciated!

2. Aug 29, 2014

### nrqed

In order to use the limit you now, you must rewrite the expression so that you have a factor
$\frac{\sin(4x)}{4x}$ right? And you need to rewrite the $\frac{1}{\sin(6x)}$ in what form?

3. Aug 29, 2014

OH. so i'd go

= limit (sin(4x)/1) * Limit (1/sin(6x))

= Limit (4sin(4x)/4) * Limit (6/6sin(6x))

= 4/6

=2/3

Yeah?

4. Aug 29, 2014

### nrqed

Almost! You need to get $\frac{\sin(4x)}{4x}$ in order to take the limit, not
$\frac{\sin(4x)}{4}$. And same thing for the sin(6x).

So you need to correct one step. But your final answer is correct. Good job!

5. Aug 29, 2014

### Midhun

This can be done by
L HOSPITAL's rule

6. Aug 29, 2014

### Midhun

Differentiate with respect to x on denomenator and numerator and you get the answer

7. Aug 29, 2014

### Midhun

Ans is 2/3

8. Aug 29, 2014

### nrqed

The question asks to NOT use L'Hospital's rule. Which is why I suggested this approach.