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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 11 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 11
The value of
l
i
m
x
→
0
27
x
−
9
x
−
3
x
+
1
√
5
−
√
4
+
c
o
s
x
Report Question
0%
√
5
(
l
o
g
3
)
2
0%
8
√
5
l
o
g
3
0%
16
√
5
l
o
g
3
0%
8
√
5
(
l
o
g
3
)
2
lim
x
→
0
sin
4
x
tan
7
x
=
Report Question
0%
1
0%
4
7
0%
7
4
0%
0
The value of
l
i
m
x
→
0
27
x
−
9
x
−
3
x
+
1
√
5
−
√
4
+
c
o
s
x
is
Report Question
0%
√
5
(
l
o
g
3
)
2
0%
8
√
5
l
o
g
3
0%
16
√
5
l
o
g
3
0%
8
√
5
(
l
o
g
3
)
2
lim
x
→
0
|
cos
(
sin
(
3
x
)
)
|
−
1
x
2
equals
Report Question
0%
−
9
2
0%
−
3
2
0%
3
2
0%
9
2
lim
x
→
π
4
cos
x
−
sin
x
(
π
4
−
x
)
(
cos
x
+
sin
x
)
=
?
Report Question
0%
2
0%
1
0%
0
0%
3
lim
x
→
0
{
l
o
g
e
(
1
+
x
)
x
2
+
x
−
1
x
}
is equal to?
Report Question
0%
1
2
0%
−
1
2
0%
1
0%
None of these
Let [x] denote the greatest integer less than or equal to x. Then :
l
i
m
x
→
0
t
a
n
(
π
s
i
n
2
x
)
+
(
|
x
|
−
s
i
n
)
(
x
[
x
]
)
2
x
2
:
Report Question
0%
does not exist
0%
equal
π
0%
equal 0
0%
equal
π
+ 1
l
i
m
x
→
π
2
c
o
t
x
−
c
o
s
x
(
π
−
2
x
)
3
equals :
Report Question
0%
1
8
0%
1
4
0%
1
24
0%
1
16
lim
n
→
∞
n
2
(
x
1
n
−
x
1
n
+
1
)
,
x
>
0
is equal to
Report Question
0%
0
0%
e
x
0%
l
o
g
e
x
0%
N
o
n
e
o
f
t
h
e
s
e
If
lim
x
→
0
[
1
+
a
x
+
b
x
2
]
(
2
/
x
)
=
e
3
, then
Report Question
0%
a
=
3
,
b
=
0
0%
a
=
3
2
,
b
≠
1
0%
a
=
3
2
,
b
=
R
0%
a
=
2
,
b
=
3
If [.] deotes the greatest integer function then
l
i
m
x
→
π
/
2
[
x
−
π
2
c
o
s
x
]
is equal to
Report Question
0%
1
0%
−
1
0%
2
0%
−
2
If
l
i
m
x
→
0
x
(
1
+
a
c
o
s
x
)
−
b
s
i
n
x
x
3
=
1
then value of a + b
Report Question
0%
-4
0%
-6
0%
1
0%
None of these
x
l
i
m
→
a
(
sin
x
−
a
2
tan
π
x
2
a
)
Report Question
0%
a
/
π
0%
−
a
/
π
0%
π
/
a
0%
−
π
/
a
Value of
l
i
m
x
→
0
3
√
1
+
tan
x
−
3
√
1
−
tan
x
x
is
Report Question
0%
1
2
0%
−
2
3
0%
1
3
0%
0
Value of
l
i
m
x
→
π
2
tan
x
.
ℓ
n
s
i
n
x
is
Report Question
0%
0
0%
1
2
0%
3
4
0%
None of these
The value of
lim
x
→
0
(
sin
x
)
1
x
+
(
1
+
x
)
(
sin
x
)
)
=
0
, where
x
>
0
, is :
Report Question
0%
0
0%
−
1
0%
1
0%
2
The value of
l
i
m
x
→
0
[
x
s
i
n
x
]
,
where [.] represents the greatest inter function , is
Report Question
0%
1
0%
0
0%
−
1
`
0%
none of these
l
i
m
x
→
0
(
1
+
t
a
n
x
1
+
s
i
n
x
)
c
o
s
e
c
x
is equal to
Report Question
0%
e
0%
1
e
0%
1
0%
None of these
The value of
lim
x
→
0
(
1
x
2
−
cot
x
)
equals
Report Question
0%
1
0%
0
0%
∞
0%
D
o
e
s
n
o
t
e
x
i
s
t
L
t
θ
⟶
0
3
t
a
n
θ
−
t
a
n
3
θ
2
θ
3
=
Report Question
0%
-4
0%
1/4`
0%
3/4
0%
4
L
t
x
→
0
s
e
c
x
−
1
(
s
e
c
x
+
1
)
2
=
Report Question
0%
1/8
0%
11/4
0%
3.2
0%
2
L
t
x
→
0
(
1
+
s
i
n
x
)
c
o
t
x
=
Report Question
0%
e
0%
e
2
0%
e
3
0%
e
4
For
a
>
1
then
lim
x
→
∞
a
x
−
a
−
x
a
x
+
a
−
x
=
?
Report Question
0%
1
0%
1
/
2
0%
1
/
3
0%
1
/
15
The value of
lim
h
→
0
{
1
h
.
(
8
+
h
)
1
/
3
−
1
2
h
}
equals
Report Question
0%
0
0%
−
4
3
0%
−
16
3
0%
−
1
48
l
i
m
x
→
0
(
c
o
s
e
c
x
)
1
l
o
g
x
is equal to
Report Question
0%
0
0%
1
0%
1
e
0%
None of these
l
i
m
h
→
0
(
2
+
h
)
c
o
s
(
2
+
h
)
−
2
c
o
s
2
h
=
Report Question
0%
cos2-2sin2
0%
cos2+2sin2
0%
sin2-2cos2
0%
sin2+2cos2
l
i
m
x
→
0
1
e
2
t
a
n
(
π
4
+
x
)
1
/
x
Report Question
0%
0
0%
1
0%
-1
0%
e
lim
x
→
1
1
+
sin
π
(
3
x
1
+
x
2
)
1
+
cos
π
x
is equal to
Report Question
0%
0
0%
1
0%
2
0%
4
l
i
m
x
→
0
−
x
(
[
x
]
+
∣
x
∣
)
s
i
n
[
x
]
∣
x
∣
is equal to
Report Question
0%
−
s
i
n
1
0%
0
0%
1
0%
s
i
n
1
The value of
lim
x
→
∞
(
|
x
2
|
+
x
)
log
(
x
cot
−
1
x
)
is :
Report Question
0%
1
3
0%
−
1
3
0%
2
3
0%
−
2
3
lim
x
→
π
2
sin
x
cos
−
1
[
1
4
(
3
sin
x
−
sin
3
x
)
]
, where [.] denotes greatest integer function is :
Report Question
0%
2
π
0%
1
0%
4
π
0%
d
o
e
s
n
o
t
e
x
i
s
t
lim
x
→
π
2
(
1
+
cos
x
1
−
cos
x
)
sec
x
=
Report Question
0%
e
0%
e
2
0%
e
3
0%
e
/
4
I
m
(
1
1
−
cos
θ
+
i
sin
θ
)
is equal to
Report Question
0%
1
2
tan
θ
2
0%
1
2
cot
θ
2
0%
−
1
2
tan
θ
2
`
0%
−
1
2
cot
θ
2
The value of
lim
x
→
0
(
e
x
+
e
−
x
−
2
x
2
)
1
/
x
2
equals
Report Question
0%
e
1
/
2
0%
e
1
/
4
0%
e
1
/
3
0%
e
1
/
12
lim
x
→
0
3
√
1
+
sin
x
−
3
√
1
−
sin
x
x
=
Report Question
0%
0
0%
1
0%
3
/
2
0%
2
/
3
L
t
x
→
π
4
√
2
−
cos
x
−
sin
x
(
4
x
−
π
)
2
=
?
Report Question
0%
1
16
√
2
0%
1
32
√
2
0%
1
16
0%
1
8
lim
x
→
0
ln
(
sin
3
x
)
ln
(
sin
x
)
is equal to
Report Question
0%
0
0%
1
0%
2
0%
none of these
L
t
x
→
∞
5
x
s
i
n
(
a
5
x
)
=
Report Question
0%
0
0%
5
0%
log5
0%
None of these.
L
i
m
x
→
0
sec
4
x
−
sec
2
x
sec
3
x
−
sec
x
=
Report Question
0%
3/2
0%
2/3
0%
1/3
0%
3/4
the value of
l
i
m
x
→
0
s
i
n
α
X
−
s
i
n
β
X
e
α
X
−
e
β
X
equals
Report Question
0%
0
0%
1
0%
-1
0%
α
−
β
lim
h
→
0
sin
(
a
+
3
h
)
−
3
sin
(
a
+
2
h
)
+
3
sin
(
a
+
h
)
−
sin
a
h
3
is equal to
Report Question
0%
cos
a
0%
−
cos
a
0%
sin
a
0%
sin
a
cos
a
lim
x
→
−
∞
x
4
sin
1
x
+
x
2
1
+
|
x
|
3
=
Report Question
0%
1
0%
-1
0%
2
0%
-3
L
i
m
x
→
∞
(
sin
√
x
+
1
−
sin
√
x
)
=
Report Question
0%
2
0%
-2
0%
0
0%
1
lim
x
→
0
1
−
cos
3
x
x
sin
2
x
=
Report Question
0%
1
/
2
0%
3
/
2
0%
3
/
4
0%
1
/
4
The value of
l
i
m
x
→
0
(
(
s
i
n
x
)
1
/
x
+
(
1
x
)
s
i
n
x
)
equals
Report Question
0%
0
0%
1
0%
∞
0%
-1
L
t
x
→
0
c
o
s
5
x
c
o
s
3
x
x
(
s
i
n
5
x
s
i
n
3
x
)
=
Report Question
0%
−
4
0%
4
0%
−
1
4
0%
1
4
lim
x
→
π
/
2
[
x
tan
x
−
(
π
2
)
sec
x
]
is equal to
Report Question
0%
1
0%
-1
0%
0
0%
None of these
The value of
lim
x
→
0
(
(
sin
x
)
1
/
x
+
(
1
x
)
sin
x
)
Report Question
0%
0
0%
1
0%
∞
0%
−
1
The value of
θ
,
i
s
l
i
m
o
→
o
c
o
s
2
{
1
−
c
o
s
2
(
1
−
c
o
s
2
.
.
.
.
.
(
c
o
s
2
{
1
−
c
o
s
2
θ
}
)
)
}
s
i
n
(
π
(
√
θ
+
4
−
2
θ
)
Report Question
0%
√
2
4
0%
√
2
0%
1
0%
2
L
i
m
x
→
∞
(
s
i
n
√
x
+
1
−
s
i
n
√
x
)
=
Report Question
0%
2
0%
-2
0%
0
0%
1
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Answered
1
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Incorrect : 0
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