If the function `f: Rsetminus` `{0}vec` given by `f(x)=1/x-2/(e^(2x)-1)` is continuous at `x=0,` then find the value of `f(0)`

The function f:R-{0}rarr R given by f(x)=(1)/(x)-(2)/(e^(2)x-1) can be made continuous at x=0 by defining f(0) as

if f(x)=(e^(x^(2))-cosx)/(x^(2)) , for xne0 is continuous at x=0 , then value of f(0) is

If the function f defined as f(x)=(1)/(x)-(k-1)/(e^(2x)-1),x!=0, is continuous at x=0, then the ordered pair (k,f(0)) is equal to:

The function f(x)=((3^(x)-1)^(2))/(sin x*ln(1+x)),x!=0, is continuous at x=0, Then the value of f(0) is

If f(x) = (e^(x^2)-cos x)/(x^(2)), "for" x != 0 is continuous at x = 0, then value of f(0) is

If the function f(x) = (x(e^(sinx) -1))/( 1 - cos x ) is continuous at x =0 then f(0)=

If the function, f(x) = {((1-cos(ax))/x^2, ; "when" x != 0), (1, ; "when" x = 0):} be continuous at x=0, then find the values of a.

If f(x) (2^(x)-1)/(1-3^(x)) , x != 0 is continuous at x = 0 then : f(0) =