## rainbow22 3 years ago lim x->0 [(3sin4x)/(sin 3x)] = 4 lim x->0 [(sin 4x/4x)x(3x/sin3x)] How is this true?

1. hartnn

to get 4x in the denominator, 4x is multiplied in numerator and denominator. got some idea?

2. Algebraic!

Multiply the whole thing by (12x/12x). rearrange.

3. dumbcow

you can also use L'Hopitals Rule -Take derivative of both top and bottom, then reevaluate the limit

4. rainbow22

But it would be undefined if I took the limit of the bottom

5. Algebraic!

Always with l'Hopital's. You'd think it was the only thing anyone ever retained from calc. class.

6. dumbcow

haha..yeah pretty much

7. rainbow22

I haven't learned it yet :(

8. Algebraic!

They're learning about sinc(x) here anyway ... that's a few lectures before the class on l'Hopital's.

9. hartnn

|dw:1347162711646:dw| got it now?

10. hartnn

just use sin t / t when limit t->0 = 1 formula.

11. Algebraic!

@hartnn what did you multiply by? It's cut off in the sketch.

12. rainbow22

Why is it that .... the 4 is moved outside the limit?

13. hartnn

multiplied by 4x in num and denom, and it can be seen,iguess

14. hartnn

the 4 is constant and constants can be moved outside the limit

15. rainbow22

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16. rainbow22

Sorry, I meant 4x

17. rainbow22

so 12x/sin3x

18. hartnn

now 12x / sin 3x can again be written as 4 * (3x/sin 3x) and using same limit again its 4*1= 4 ok?

19. Algebraic!

|dw:1347163221313:dw|

20. rainbow22

Oh, I get it now! Thank you guys SO MUCH!