If `g(x)=x f(x), where f(x)={x sin(1/x); x != 0 0; x=0},` then at `x = 0`

g(x)=xf(x) , where f(x)=x"sin"(1)/(x),x!=0=0,x=0 . At x=0 :

Write the value of lim_(x rarr0)f(x), where f(x)={x(sin((1)/(x))+log(x^(2)));x!=0;0,x=0

Let g(x)=xf(x) , where f(x)={{:(x^(2)sin.(1)/(x),":",x ne0),(0,":",x=0):} . At x=0 ,

Let g(x)=xf(x) , where f(x)={{:(x^(2)sin.(1)/(x),":",x ne0),(0,":",x=0):} . At x=0 ,

If f(x)=x sin ""1/x"for "x ne 0, f(0)=0" then at x"=0, f(x) is

Let g(x)=f(f(x)) where f(x)={1+x;0<=x<=2} and f(x)={3-x;2

Let g(x) = f(f(x)) where f(x) = { 1 + x ; 0 <=x<=2} and f(x) = {3 - x; 2 < x <= 3} then the number of points of discontinuity of g(x) in [0,3] is :

If f(x)=(sin(x^(2)))/(x),x!=0,f(x)=0,x=0 then at x=0,f(x) is