# Limit of trig function

1. Aug 28, 2014

### jog511

1. The problem statement, all variables and given/known data
lim x->0 sin4x/2x

2. Relevant equations
lim x->0 sinx/x =1

3. The attempt at a solution
can I write lim x->0 sin4x/2x as sinx/x * 4/2 = 1*2 or am I missing a step ?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 28, 2014

### nrqed

]No, one cannot do that. There is no rule that says one can pull out a factor from a trig function like this. You have to rewrite you expression in the form
$\lim_{y \rightarrow 0} \, C \frac{\sin(y)}{y}$ where C is some constant and y is something that depends on x, y=y(x). Then this limit will simply give C.

So what you have to do is to identify what choice of y(x) will bring your limit in the form I just described.

3. Aug 28, 2014

### jog511

like 4/4

4. Aug 28, 2014

### HallsofIvy

Staff Emeritus
I have no idea what you mean by that. But surely you know that sin(2x) is NOT equal to 2sin(x)?

5. Aug 28, 2014

### nrqed

Set aside the limit for now. Try writing the expression in the form $C \, \sin(y)/y$. What is y(x)? What is C?

6. Aug 28, 2014

### jog511

sin4x/2x * 4/4 = 4*sin4x/4*2x

7. Aug 28, 2014

### LCKurtz

8. Aug 28, 2014

### jog511

I'll have to get back to the forum on this

9. Aug 29, 2014

### jog511

C = 4,
y(x) = 4x

10. Aug 29, 2014

### nrqed

Well, if you write $C \sin(y)/y$ using what you just wrote here, do you get the function that was given in the question?

11. Aug 29, 2014

### jog511

I believe I get sin4x/4x

12. Aug 29, 2014

### jog511

I don't understand this concept. Even Calculus by Larson does not explain it well.

13. Aug 29, 2014

### nrqed

Hold on. If y =4x and C = 4 then
$C \sin(y)/y = 4 \frac{ \sin(4x) }{4x}$
right? This is not the initial expression.

14. Aug 29, 2014

### nrqed

You need the form
$C \frac{\sin(4x)}{4x}$ WHat is the constant C equal to?

15. Aug 29, 2014

### nrqed

The key point is that we do not know the limit of $\frac{\sin(4x)}{2x }$ but we do know how to take the limit of $\frac{\sin(4x)}{4x}$ So you need to rewrite your initial function in the form
$\text{ constant } \times \frac{\sin(4x)}{4x}$ Then you will be able to take the limit.

16. Aug 29, 2014

### jog511

c in the original equation was 1

17. Aug 29, 2014

### nrqed

What I meant is that we must write

$\frac{sin(4x)}{2x} = C \frac{\sin(4x)}{4x}$
What is the value of C?

18. Aug 29, 2014

### jog511

It has to be 2

19. Aug 29, 2014

### nrqed

Correct.

So you have shown that the initial function may be written as
$2 \frac{\sin(4x)}{4x}$
which, defining y = 4x, may be written as
$2 \frac{\sin(y)}{y}$
Now, when x goes to zero, y also goes to zero so you are all set to take the limit.

20. Aug 29, 2014

### jog511

the limit is 2.