The points where `f(x) = lim_(x rarr oo) ( sin (pi x/2))^2n` is discontinuous are:

The points where f(x)=lim_(x rarr oo)(sin(pi(x)/(2)))^(2)n is discontinuous are:

Let f(x)=lim_(n rarr oo)(sin x)^(2n)

lim_(x rarr oo)(2+sin x)/(x^(2)+3)

Find the number of integers lying in the interval (0,4) where the function f(x)=lim_(n rarr oo)(cos(pi x)/(2))^(2n) is discontinuous

lim_(x rarr0)x^(2)sin(pi/x)=

lim_(x rarr 0) x^(2). sin (pi/x) is

lim_ (x rarr oo) x ^ (2) sin ((pi) / (x)) =

" f) "lim_(x rarr pi)(sin x)/(pi-x)=?

Let f(x)=lim_(n to oo) sinx/(1+(2 sin x)^(2n)) then f is discontinuous at