For f(x) to be continuous at x=0:
x→0limf(x)=f(0)
From the given function:
f(0)=23k
When x=0:
f(x)=xsin3x
Finding the limit:
x→0limxsin3x
=x→0lim3x3sin3x
=3⋅x→0lim3xsin3x
Let u=3x. When x→0, then u→0.
=3⋅u→0limusinu
=3⋅1
=3
Applying the continuity condition:
x→0limf(x)=f(0)
3=23k
6=3k
k=2
Therefore, the value of k is 2.