# Limit of a trigonometric function

1. Jul 14, 2009

### Jimmy84

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

Find the limit as x approaches 0 of

f(x) = 3x / sin 5x

2. Relevant equations

I might have to use the fact that "the lim of x approaches 0 of sin x/x = 1"

3. The attempt at a solution

I dont know how to start to solve the expresion, I would appreciate any help.
thanks.

2. Jul 14, 2009

### Dick

Substitute u=5x. So u also approaches 0. Then use lim u/sin(u)=1 as u->0.

3. Jul 14, 2009

### rl.bhat

You can rewrite f(x) as
f(x) = 3/5*5x/sin5x and the find the limit.

4. Jul 15, 2009

### n!kofeyn

A good explanation of this problem lies in this https://www.physicsforums.com/showthread.php?t=305154".

I think the best advice is to combine the above two suggestions. Rewrite the limit as
$$\lim_{x\to0} \frac{3x}{\sin 5x} = \lim_{x\to0} \frac{3}{5} \frac{5x}{\sin 5x} = \frac{3}{5}\lim_{x\to0} \frac{5x}{\sin 5x}$$

Now let u=5x, and follow the example shown in the posted link.

Last edited by a moderator: Apr 24, 2017
5. Jul 16, 2009

### Jimmy84

thanks a lot guys, the explanation you gave in the other post were useful for me, my book didnt explain that way of substitution a lot.

right now im having a difficult time subtituting this:

1.) f(x) 3x^2 / 1 - cos 1/2 x^2

I think that for this problem I gotta use the fact that x - cos x / x = 0

and
2.) g(x) sin x /x^2

at first sight, these problems arent substutited in the same was as you described in the previous post though. Sometimes when I try to substitute terms I get all confused and end up mading up answers for the trigonometric functions, I was wondering if there are certain substitution steps that I could use for most of the trigonometric functions? like the one you described in the previous posts.

take care and thanks a lot.

6. Jul 16, 2009

### rock.freak667

is this supposed to be

$$\lim_{x \rightarrow 0} \frac{3x^2}{1-cos\frac{x^2}{2}}$$

$$\lim_{x \rightarrow 0} \frac{sinx}{x^2}$$

if that is the question then try using the fact that 1/x2 = (1/x)*(1/x)
and

$$\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a}f(x) * \lim_{x \rightarrow a} g(x)$$

7. Jul 17, 2009

### Jimmy84

the second problem is right, the first one is 3x^2 / (1-cos 1/2 x)^2
sorry im having problems with the notation.

8. Jul 17, 2009

### HallsofIvy

Staff Emeritus
Then both multiply and divide the numerator by 1/4 to get
[tex]\frac{12\frac{x^2}{4}}{\left(1- cos \frac{x}{2}\right)^2}= 12\left(\frac{\frac{x}{2}}{1- cos(\frac{x}{2})}\right)^2[/itex]
and use the fact that
[tex]\lim_{a\rightarrow 0}\frac{1- cos(u)}{u}= 0[/itex]
But note that your fraction is the reciprocal of that- that's very important here!