Question

let f(x) = {[1-cos (cx)]/[x sin x] , x not equals 0 and
2, x = 0}
find c so that f is continuous at x=0 (no L'hospital allowed)

Answer 1

whats your attempt ?

Answer 2

is lim x->0 (1-cox cx)/cx = 1 ?

Answer 3

no, solve that by multiplying numerator and denominator by
(1+cos cx)

Answer 4

so on the numerator it will be 1-cos^cx = sin ^ cx ?

Answer 5

yes

Answer 6

lim x->0 sin^2cx/[x sin x (1+cos cx)] then what ?

Answer 7

but you will need this limit
lim x->0 (1-cox cx)/cx^2

right ??

Answer 8

btw: lim x->0 (1-cox cx)/cx = 0 ??

Answer 9

yes

Answer 10

c lim x->0 [(1-cos cx)/cx ]x lim x->0 sin x

Answer 11

so it will evaluate to zero ... but the answer given is c = +- 2

Answer 12

c lim x->0 [(1-cos cx)/cx ] . lim x->0 sin x

Answer 13

so, 0/0 form,

thats why i said, that you will need
lim x->0 (1-cox cx)/cx^2

Answer 14

could you pls show 1-2 steps ? i will understand then

Answer 15

yes,



the denominator limit = 1
you just need numerator limit