`lim_(x->pi/2)(cosx)/(x-pi/2)`

If l_1=lim_(xrarr-2)(x+|x|),l_2=lim_(xrarr-2)(2x+|x|) and l_(3)=lim_(xrarr pi//2)(cosx)/(x-pi//2) , then

If l_1=lim_(xrarr-2)(x+|x|),l_2=lim_(xrarr-2)(2x+|x|) and l_(3)=lim_(xrarr pi//2)(cosx)/(x-pi//2) , then

Evaluate: lim_(x->pi/2)(cosx)^(cosx)

Evaluate: ("lim")_(x->pi/2)(cosx)^(cosx)

(lim)_(x->0)(cosx)/(pi-x)

lim_(x-pi/2) (cot x - cosx)/(pi-2x)^3 equals: (1) 1/8 (2) 1/4 (3) 1/24 (4) 1/16

Let f(x) =(kcosx)/(pi-2x) if x!=pi/2 and f(x)=3 if x=pi/2 then find the value of k if lim_(x->pi/2) f(x)=f(pi/2)

The value of lim_(xto pi//2)(cot x-cosx)/(pi-2x)^3 is

Consider a function, f(x)=2/3ln^(3/2) (sinx+cosx) for x in (0,pi/2) The value of Lim_(x->pi/2) f(x)/(pi/2-x)^(3/2)