1Hey no one interested 2 solve dis one?....come on guys help me out
1) Solve Limx→0[(1-ex)sinx|x|] ,where[.]represents greatest int func
(A)-1 (B)1 (C)0 (D)None of these
2) Solve limx→π2 sinx -(sinx)sinx1-sinx+logesinx ,(here x→ pi/2)
(A)2 (B)0 (C)-1 (D)1
Pls solve!!!
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7 Answers
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1hey ok da ans is (A)-1....But i'm kinda interested in da solution.
106(1-ex)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.
231st sum to fiitjee me aya tha ,
for LHL , 1 > ex , , and sinx / |x| = -sinx/x
so - sinx/x → -1 as x →0-
and 1-ex → 0+ as x →0-
so (1- ex) sinx / |x| tends to 0- as x →0-
hence LHL = -1
for RHL , 1 < ex , and sinx / |x| = sinx/x
so sinx/x → 1 as x →0+
and 1-ex → 0- as x →0+
hence
(1- ex) sinx / |x| tends to 0- as x →0+
so RHL also -1
hence limit = -1