Question
left hand derivative of f(x) = [x]sin( pie x) at x=k where k is an integer,
Left hand derivative of given function at x = k (k is an integer) is
limh -> 0 {f(k-h) - f(k)}/ {k-h-k}
= limh -> 0 {[k-h] sin π(k-h) - [k] sin πk}/{-h} now 2 cases arise for even and odd k
first case
= limh ->0 (k-1) sin (-πh) - 0 / (-h) if k is even , here we used sin πk = 0 and sin (2π-x) = sin (-x)
= π(k-1) here we used sin (-πx)/(-πx) = 1
second case
= limh->0 (k-1) sin (πh) - 0 / (-h) if k is odd, here we used sin πk = 0 and sin (3π-x) = sin x
= -π(k-1)
Left hand derivative of given function at x = k (k is an integer) is
limh -> 0 {f(k-h) - f(k)}/ {k-h-k}
= limh -> 0 {[k-h] sin π(k-h) - [k] sin πk}/{-h} now 2 cases arise for even and odd k
first case
= limh ->0 (k-1) sin (-πh) - 0 / (-h) if k is even , here we used sin πk = 0 and sin (2π-x) = sin (-x)
= π(k-1) here we used sin (-πx)/(-πx) = 1
second case
= limh->0 (k-1) sin (πh) - 0 / (-h) if k is odd, here we used sin πk = 0 and sin (3π-x) = sin x
= -π(k-1)
so for even and odd k answer will be different.
Read 1 Solution.
limh--0 f (k-h) - f(k)/-k use the follwing data sin πk =0 as k are integers
[k-h]=k-1
after that u will get 0/0 form again use L.Hospital rule .then put the limit and get the answers .
answer will have 2 values ax cosπ=-1 wheras cos2π =1