4663 1
- Himanshu Giria 3->A(1)
Anindita Roy 3) does not exist as lhl is not equal to rhl 
Raunaq Saha Ya I got the answer of 3 . It will be (d)
In the shastras answer of 1 is given as (b)log1
Q1: 
a)-1
b)loge1
c)1
d)none
q2:
a)12sin3a
b)12cosec2a
c)sin3a
d)cosec3a
q3:
a)1
b)-1
c)0
d)none
q4:if 0<x<y then 
a)e
b)x
c)y
d)none
q5:
a)1
b)sinxx
c)xsinx
d)none
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8 Answers
1357first one
log(sinx)/1/x
Applying L'Hospital
cotx/(-1/x2)=-x2/tanx=-(x)(x/tanx)=0 as x->0
1357For second one i m getting the (1-cosa)/(1-cota)
I think there's a mistake in the question..
it should be
lim (x->a)(cosx-cosa)/(cotx-cota)
- Raunaq Saha ya sorry the limit is x->a but the rest of the expression is as given in the shastras question bankUpvote·0· Reply ·2014-06-06 08:47:09
85Sir how does the first one exist ?
M not understanding how lhl exists .. log of -ve is not defined . So sir where m i wrong ?
- Manish Shankar thats why we are trying to find out 0+
- Anindita Roy But sir if lhl does not exist at all that means we cant approach frm -ve side to 0 at all then how can lim x ->0 exist... evn though rhl exists .. lhl can never be equal to rhl
Sahil Jain It's a one sided limits. Lhl has no significance as log x is not defined for x <0
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