If `f(x)=xcos(1/x), x!=0, f(x)=k, x=0`, then find the value of `k` if `f(x)` is continous at `x=0`

If f(x)=x cos((1)/(x)),x!=0,f(x)=k,x=0 then find the value of k if f(x) is continous at x=0

If f(x)=x cos((1)/(x)),x!=0,f(x)=k,x=0 then find the value of k if f(x) is continous at x=0

let f(x)=[tan((pi)/(4)+x)]^((1)/(x)),x!=0 and f(x)=k,x=0 then the value of k such that f(x) hold continuity at x=0

If f(x)={{:(,x([(1)/(x)]+[(2)/(x)]+.....+[(n)/(x)]),x ne 0),(,k,x=0):} and n in N . Then the value of k for which f(x) is continuous at x=0 is

If f(x)={{:(,x([(1)/(x)]+[(2)/(x)]+.....+[(n)/(x)]),x ne 0),(,k,x=0):} and n in N . Then the value of k for which f(x) is continuous at x=0 is

Let f(x)={(xsin(1/x), x ne 0),(k, x=0):} find k if f is continuous at x=0,

The function f(x)= xcos(1/x^2) for x != 0, f(0) =1 then at x = 0, f(x) is

Let f(x) = {:{((1-cos2x)/(2x^2), x 0):} . Find k if f is continous at x =0