Question #f9d14

Question

1679 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-06T14:10:39

#sin(6x) = sin(4x+2x) = sin(4x)cos(2x)+cos(4x)sin(2x)#

So,

#(sin(6x)-sin(4x))/x^2 = (sin(4x)cos(2x)+cos(4x)sin(2x)-sin(4x))/x^2#

# = (sin(4x)/x (cos(2x)-1)/x) + (sin(2x)/x cos(4x)/x)#

As #xrarr0^+#, we get #((4)(0))+((2)(oo)) = oo#

As #xrarr0^-#, we get #((4)(0))+((2)(-oo)) = -oo#

Therefore

#lim_(xrarr0)(sin(6x)-sin(4x))/x^2# does not exist.