LIMITS PART 2(DNT WRY DIS IS DA LAST PART)

1st one is -1

ans is -1.............

hey ok da ans is (A)-1....But i'm kinda interested in da solution.

(1-e^x)sinx/|x| = (1-(1+x+x²/2))*(x -x³/6)|x|
= (-x-x²/2 - c')(x-x³/6 + c)|x|
= x|x|(1-x²/6 + c)(-x-x²/2-c')
= (-x+x³/6 -cx -x-x²/2-c'+x³/6 + x⁴/12 + c'x²/2 -cx +cx²/2 - cc')x|x|
= (-2x -x²/2)x|x|
HERE c and c' are higher powers of x.
you could have neglected powers such as x²/2 and x³/6

if we take x-->0_- then we get lim x-->0_- (2x+x²/2) < 0 hence [.] = -1

if we take x--> 0₊ then we get lim x--> 0₊ (-2x-x²/2) < 0 hence [.] = -1

so limit = -1.

1st sum to fiitjee me aya tha ,

for LHL , 1 > e^x , , and sinx / |x| = -sinx/x

so - sinx/x → -1 as x →0^-

and 1-e^x → 0⁺ as x →0^-
so (1- e^x) sinx / |x| tends to 0^- as x →0^-

hence LHL = -1

for RHL , 1 < e^x , and sinx / |x| = sinx/x

so sinx/x → 1 as x →0⁺
and 1-e^x → 0^- as x →0⁺
hence
(1- e^x) sinx / |x| tends to 0^- as x →0⁺

so RHL also -1

hence limit = -1