Question

does anyone knows to solve limits

Answer 1

yup.

Answer 2

r u there on fb

Answer 3

lim x tending to 0..............2sinx-sin2x/x^3

Answer 4

can u use L'Hopital's ??
yes, i am on facebook..

Answer 5

plz can u send me a req.

Answer 6

no i dont know hopitals

Answer 7

here, 1st use, sin2x = ... ?

Answer 8

sorry i dont know that

Answer 9

u explain in a simple way

Answer 10

is anyone on skype

Answer 11

sin 2x = 2 sin x cos x

Answer 12

it wd b tooo easy to contact n i have many doubts

Answer 13

hmmmmmm

Answer 14

is amyone on skype

Answer 15

lim (x->0) (2sinx-sin2x)/(x^3)
= lim (x->0) (2sinx-2sinxcosx)/(x^3)
= lim (x->0) 2sinx(1-cosx)/(x^3)
= lim (x->0) 2sinx(2sin^2 (x/2))/(x^3)
= 4 * 1/2 * 1/2
= 1