# Limit of trigonometry

1. Apr 29, 2013

### songoku

1. The problem statement, all variables and given/known data
$$\lim_{x \to 0} \frac{tan (cos 4x - 1)}{3x ~ sin (\frac{4}{3} x)}$$

2. Relevant equations
limit for trigonometry

3. The attempt at a solution
can I do it like this:

$$\frac{tan (cos 4x - 1)}{3x ~ sin (\frac{4}{3} x)}$$

$$= \frac{- tan (2 sin^{2} 2x)}{3x ~ sin (\frac{4}{3} x)}$$

and then using the property of trigonometry limit, it becomes:

$$= \frac{-2 . 4}{3 . \frac{4}{3}}$$

$$=-2$$

2. Apr 29, 2013

### jing2178

What do you know about sin(x), cos(x) and tan(x) when x is very small?

3. Apr 29, 2013

### songoku

I am not sure what you mean, maybe like this:

a. when x is very small (close to zero):
the value of sin x is close to 0
the value of cos x is close to 1
the value of tan x is close to 0

or

b. when x is very small (close to zero):
sin x ≈ x
cos x ≈ 1 - 1/2 x2 ≈ 1
tan x ≈ x

but I still don't know what the properties related to the question

4. Apr 29, 2013

### Mentallic

You're looking to use the properties of b.

If $\cos(x)\approx 1-x^2/2$ then what is $\cos(4x)$ approximately equal to?

What's $\sin(4x/3)$ approximately equal to?

Finally, you'll need to also convert the tan function as well in the same fashion.

5. May 2, 2013

### songoku

Oh I see. I don't know before that the properties can be used in limit as well.

Thanks a lot for all the help