## xilinix 3 years ago lim -> 0 sin(2x) /(x*cos(x) ) help

1. myko

sin (2x) = 2sin(x)cos(x) so sin(2x) /(x*cos(x) =2sin(x)cos(x)/x*cos(x)=2sinx/x which as x->0 goes to 1

2. xilinix

3. myko

ups sry, forgot about 2 multuplying: lim 2sinx/x=2*1=2 x->0

4. xilinix

don't undestand you

5. myko

sin (2x) can be written like: sin (2x) = 2sin(x)cos(x) substituting this expretion in the original one: sin(2x) /(x*cos(x) =2sin(x)cos(x)/x*cos(x)=2sinx/x taking limits, and noticing that lim as x->0 of sinx/x=1: lim 2sinx/x=2*1=2 x->0

6. xilinix

|dw:1352667129469:dw|

7. myko

no

8. xilinix

why not? l'Hôpital rule

9. ccappiello

yes, using l'hospital's rule works

10. Varuns

Don't break your head ! You know that sin(x)=x as x tends to zero. So sin(2x) = 2x. In the denominator, keep x as it is and cos(x)=1. Hence the answer

11. xilinix

thanks, but the image is ok?

12. Varuns

|dw:1352669528245:dw|

13. lopus

claro que es posible es solamente derivar arriba y abajo como lo muestras en la imagen y ya