If `f'(2^+)=0` and `f'(2^-)=0` then is `f(x)` continuous at `x=2?`

A function f(x) is defined as follows f(x)={(sin x)/(x),x!=0 and 2,x=0 is,f(x) continuous at x=0? If not,redefine it so that it become continuous at x=0 .

If f(x)=(1-cos2x)/(x^(2))quad , x!=0 and f(x) is continuous at x=0 ,then f(0)=?

If f(x)=(1-cos2x)/(x^(2)) , x!=0 and f(x) is continuous at x=0 ,then f(0)= (a) 1 (b) 2 (c) 3 (d) None of these

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

Value of a so that f(x) = (sin^(2)ax)/(x^(2)), x != 0 and f(0) = 1 is continuous at x = 0 is

[" 14.If "f(x)=(1*cos2x)/(x^(2))quad :x!=0" and "f(x)" is continuous at "x=0" ,then "f(0)=?],[[" (a) "1," (b) "2," (c) "3," (d) None of these "]]

If f(x) ={ (x.ln(cosx))/ln (1+x^2) x!=0 , 0, x=0 then a) f is continuous at x =0 b) f is continuous at x =0 but not differentiable at x=0 c) f is differemtiable at x =0 d) f is not continuous at x =0

If f(x) = (2x+ tanx)/(x) , x!=0 , is continuous at x = 0, then f(0) equals