# Solving trigonometric limits

1. Feb 21, 2012

### mtayab1994

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

solve the following limit: $$\lim_{x\rightarrow0}\frac{x+sinx}{2x-sin3x}$$

3. The attempt at a solution

I know the principle of solving all sorts of limits but not trig limits, and since we just started trig limits it's still not clear to me, but i think to solve it we have to separate it then solve but i don't know. Any help on how to start please?

2. Feb 21, 2012

### Dick

Do you know l'Hopital's rule?

3. Feb 21, 2012

### mtayab1994

No haven't learned it can you fill me on it please?

4. Feb 21, 2012

### SammyS

Staff Emeritus
Are you familiar with Taylor series ?

Added in Edit:

Better yet:
multiply by (1/x) over (1/x) .​

5. Feb 21, 2012

### Dick

You may not have covered it yet so you may be expected to do it a different way. Divide the numerator and denominator by x. Now you probably do know the limit sin(x)/x. Try to use that to find the limits of the terms.

6. Feb 21, 2012

### mtayab1994

No sorry i'm not familiar with taylor series, but i know that for l'hospital's rule we have to take the derivative of the numerator and the derivative of the denominator; but i have never learned derivatives so i can't do that. Is it possible to solve it some other way?

7. Feb 21, 2012

### mtayab1994

Ok the derivative of x+sinx is cosx+1 and how about the bottom?

8. Feb 21, 2012

### mtayab1994

ok i counted and found the limit to be -2 is that correct?

9. Feb 21, 2012

### Dick

Yes, if you used l'Hopital then you might check whether that is allowed. Otherwise you should try doing it using that the limit of sin(x)/x is 1.

10. Feb 21, 2012

### mtayab1994

Yea thank you very much I get the trick. It's the same for almost all of them.

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