`lim_(x->1) (1+sin((3pix)/(1+x^2)))/(1+cospix)` is (a) 0 (b) 1 (c) 2 (d) 3

lim_(xto1) (1+sinpi((3x)/(1+x^(2))))/(1+cospix) is equal to

Let L = lim_(x->0) (sin^(-1)x-tan^(-1)x)/(x^2), M = lim_(x->1) (1+sin((3pix))/(1+x^2))/(1+cos pix) nd N is the number of integral points where f(x) = [x^2]-[x]^2, AA x in [-7/2,7/2] s discontinuous (where [.] denotes greatest integer function). Then which of the following hold(s) good? A) 6L-M=3 B 6L+N gt 9 C) L lt M D) L ge M

lim_(xto1) (1-x^(2))/(sin2pix) is equal to

lim_(x rarr0)(sin x)/(x(1+cos x)) : (a)0 (b)1/2 (c)1 (d)-1

Find the value of lim_(x to 1) (1+sin pix)^(cos pix)

lim_(x rarr0)((1)/(x sin^(-1)x)-(1-x^(2))/(x^(2)))

(lim)_(x rarr)(sin^(4)x-sin^(2)x+1)/(cos^(4)x-cos^(2)x+1) is equal to (a) 0( b) 1 (c) (1)/(3)(d)(1)/(2)

(lim)_(x->1)(1-x)tan((pix)/2)\ a. pi/2 b. pi c. 2/pi d. 0

lim_(x rarr0)(cos^(-1)((1-x^(2))/(1+x^(2))))/(sin^(-1)x)

The value of lim_(x->1)(2-x)^(tan((pix)/2)) is e^(-2/pi) (b) e^(1/pi) (c) e^(2/pi) (d) e^(-1/pi)