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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 2 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 2
If
x
is very large, then
2
x
1
+
x
is
Report Question
0%
close to
0
0%
arbitrarily large
0%
lie between
2
and
3
0%
close to
2
Explanation
2
x
1
+
x
=
2
1
+
1
x
=
2
1
+
0
=
2
If
x
→
α
then
1
x
→
0
.
What is
lim
x
→
0
cos
x
π
−
x
equal to?
Report Question
0%
0
0%
π
0%
1
π
0%
1
Explanation
Here, putting
x
=
0
directly we have,
lim
x
→
0
cos
x
π
−
x
=
cos
0
π
−
0
=
1
π
Hence, C is correct.
l
i
m
x
→
0
(
1
+
a
x
)
b
x
is equal to
Report Question
0%
e
a
b
0%
e
a
+
b
0%
non-existent
0%
none of these
Explanation
Now,
l
i
m
x
→
0
(
1
+
a
x
)
b
x
=
l
i
m
x
→
0
{
(
1
+
a
x
)
1
a
x
}
a
b
=
{
l
i
m
x
→
0
(
1
+
a
x
)
1
a
x
}
a
b
=
e
a
b
. [ Since
lim
x
→
0
(
1
+
x
)
1
x
=
e
].
lim
x
→
0
(
(
1
+
x
)
2
e
x
)
4
sin
x
is:
Report Question
0%
e
2
0%
e
4
0%
e
8
0%
e
−
8
Explanation
Now
lim
x
→
0
(
(
1
+
x
)
2
e
x
)
4
sin
x
=
lim
x
→
0
(
{
(
1
+
x
)
1
x
}
x
sin
x
)
8
e
4
x
sin
x
We have
lim
x
→
0
sin
x
x
=
1
and
lim
x
→
0
(
1
+
x
)
1
x
=
e
.
As both the limits of the numerator and denominator exists,and the limit of the numerator is non-vanishing, so we can write the above limits as
=
lim
x
→
0
(
{
(
1
+
x
)
1
x
}
lim
x
→
0
x
sin
x
)
8
lim
x
→
0
e
4
x
sin
x
[Using division property of limits]
=
(
lim
x
→
0
{
(
1
+
x
)
1
x
}
(
lim
x
→
0
x
sin
x
)
)
8
e
(
lim
x
→
0
4
x
sin
x
)
[Using limit property]
=
(
e
1
)
8
e
4
1
=
e
2
If a sequence
<
a
n
>
is such that
a
1
,
a
n
+
1
=
2
+
3
a
n
1
+
2
a
n
and
lim
n
→
∞
a
n
exists, then
a
n
is equal to
Report Question
0%
cos
36
o
0%
2
cos
36
o
0%
sin
18
o
0%
2
sin
36
o
lim
x
→
0
sin
7
x
sin
3
x
equals
Report Question
0%
7
3
0%
10
3
0%
14
3
0%
1
3
Explanation
We have,
lim
x
→
0
sin
7
x
sin
3
x
lim
x
→
0
sin
7
x
sin
3
x
lim
x
→
0
sin
7
x
7
x
×
7
x
sin
3
x
3
x
×
3
x
lim
x
→
0
sin
7
x
7
x
×
7
sin
3
x
3
x
×
3
=
1
×
7
1
×
3
=
7
3
Hence, this is the answer.
lim
x
→
0
(
1
+
sin
x
)
cos
x
is equal to
Report Question
0%
0
0%
e
0%
1
0%
1
e
Explanation
lim
x
→
0
(
1
+
sin
x
)
cos
x
=
(
1
+
sin
0
)
cos
0
=
1
1
=
1
'
Evaluate the following limit :
l
i
m
x
→
0
1
−
cos
2
x
x
2
Report Question
0%
0
0%
1
0%
2
0%
none of these
Explanation
l
i
m
x
→
0
1
−
cos
2
x
x
2
=
l
i
m
x
→
0
2
sin
2
x
x
2
=
2
l
i
m
x
→
0
(
sin
x
x
×
sin
x
x
)
=
2
l
i
m
x
→
0
sin
x
x
×
l
i
m
x
→
0
sin
x
x
=
2
(
1
)
(
1
)
=
2
Evaluate
lim
n
→
∞
1
+
2
+
3
+
.
.
.
+
n
n
2
Report Question
0%
1
2
0%
1
0%
3
2
0%
1
2
2
Explanation
lim
n
→
∞
1
+
2
+
3
+
.
.
.
+
n
n
2
=
lim
n
→
∞
1
n
2
×
n
(
n
+
1
)
2
=
lim
n
→
∞
1
2
(
1
+
1
n
)
=
1
2
Which of the following statement is not correct
Report Question
0%
lim
x
→
c
[
f
(
x
)
+
g
(
x
)
]
=
lim
x
→
c
f
(
x
)
+
lim
x
→
c
g
(
x
)
0%
lim
x
→
c
[
f
(
x
)
−
g
(
x
)
]
=
lim
x
→
c
f
(
x
)
−
lim
x
→
c
g
(
x
)
0%
lim
x
→
c
[
f
(
x
)
.
g
(
x
)
]
=
lim
x
→
c
f
(
x
)
.
lim
x
→
c
g
(
x
)
0%
lim
x
→
c
f
(
x
)
g
(
x
)
=
lim
x
→
c
f
(
x
)
lim
x
→
c
g
(
x
)
Explanation
All the result other than D are correct.
The last result will hold only when
g
(
x
)
≠
0
and
lim
x
→
c
g
(
x
)
≠
0
for
x
belonging to a neighbourhood of
c
.
A, B, C are general results which are always true.
What is the value of
l
i
m
x
→
0
sin
x
tan
3
x
Report Question
0%
1
4
0%
1
3
0%
1
2
0%
1
Explanation
Given,
lim
x
→
0
sin
(
x
)
tan
(
3
x
)
applying L'Hospital rule, we get,
lim
x
→
0
cos
(
x
)
sec
2
(
3
x
)
⋅
3
=
cos
(
0
)
sec
2
(
3
⋅
0
)
⋅
3
=
1
3
If
f
(
x
)
=
{
x
−
5
;
x
≤
1
4
x
2
−
9
;
1
<
x
≤
2
3
x
2
+
4
;
x
>
2
Then
f
(
2
+
)
−
f
(
2
−
)
=
Report Question
0%
0
0%
2
0%
9
0%
4
Explanation
Given:
f
(
x
)
=
{
x
−
5
;
x
≤
1
4
x
2
−
9
;
1
<
x
≤
2
3
x
2
+
4
;
x
>
2
f
(
2
+
)
=
lim
x
→
2
+
f
(
x
)
=
lim
h
→
0
3
(
2
+
h
)
2
+
4
f
(
2
+
)
=
16
f
(
2
−
)
=
lim
x
→
2
−
f
(
x
)
=
lim
h
→
0
4
(
2
−
h
)
2
−
9
f
(
2
−
)
=
7
So,
f
(
2
+
)
−
f
(
2
−
)
=
16
−
7
=
9
If
f
(
x
)
=
{
2
x
+
b
(
x
<
α
)
x
+
d
(
x
≥
α
)
is such that
lim
x
→
α
f
(
x
)
=
L
, then
L
=
Report Question
0%
2
d
−
b
0%
b
−
d
0%
d
+
b
0%
b
−
2
d
Explanation
l
i
m
h
→
0
−
f
(
x
)
=
2
(
α
−
h
)
+
b
=
2
α
+
b
=
L
.
.
.
.
.
.
.
.
.
.
.
.
.
(
1
)
l
i
m
h
→
0
+
f
(
x
)
=
(
α
+
h
)
+
d
=
L
α
=
L
−
d
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(
2
)
Substituting value of euation (2)in (1), we get
2
(
L
−
d
)
+
b
=
L
L
=
2
d
−
b
Hence, option 'A' is correct.
lim
x
→
1
(
1
−
x
)
tan
(
π
x
2
)
=
Report Question
0%
π
0%
2
π
0%
π
2
0%
2
π
Explanation
lim
x
→
1
(
1
−
x
)
tan
(
π
x
2
)
=
lim
x
→
1
1
−
x
cot
(
π
x
2
)
It is of the form
0
0
, so applying L-Hospital's rule
=
lim
x
→
1
−
1
−
π
2
csc
2
(
π
x
2
)
=
2
π
lf
f
(
x
)
=
x
,
x
<
0
;
f
(
x
)
=
0
,
x
=
0
;
f
(
x
)
=
x
2
;
x
>
0
, then
lim
x
→
0
f
(
x
)
is equal to
Report Question
0%
Does not exist
0%
0
0%
−
1
0%
1
Explanation
Given:
f
(
x
)
=
{
x
0
x
2
x
<
0
x
=
0
x
>
0
}
To Find :
lim
x
→
0
f
(
x
)
= ?
Sol: left hand limit
→
lim
x
→
0
−
f
(
x
)
=
lim
x
→
0
x
=
0
right hand limit
→
lim
x
→
0
+
f(x)=
lim
x
→
0
x
2
=
0
,
LHL=RHL
f
(
0
)
=
0
L
H
L
=
R
H
L
=
f
(
0
)
=
0
f
(
x
)
=
2
x
+
1
,
a
=
1
,
l
=
3
and
ϵ
=
0.001
, then
δ
>
0
satisfying
0
<
|
x
−
a
|
<
δ
such that
|
f
(
x
)
−
l
|
<
ϵ
, is
Report Question
0%
0.0005
0%
0.005
0%
0.001
0%
0.0001
Explanation
We have to find
δ
>
0
, which satisfies
0
<
|
x
−
a
|
<
δ
such that
|
f
(
x
)
−
l
|
<
ϵ
∵
|
f
(
x
)
−
l
|
<
ϵ
⇒
−
ϵ
<
f
(
x
)
−
l
<
ϵ
⇒
l
−
ϵ
<
f
(
x
)
<
l
+
ϵ
⇒
3
−
ϵ
<
2
x
+
1
<
3
+
ϵ
⇒
−
ϵ
<
2
x
−
2
<
ϵ
⇒
−
ϵ
2
<
x
−
1
<
ϵ
2
⇒
|
x
−
1
|
<
ϵ
2
⇒
|
x
−
a
|
<
ϵ
2
δ
=
ϵ
2
=
0.001
2
=
0.0005
Hence, option A.
Evaluate:
lim
h
→
0
(
1
h
3
√
8
+
h
−
1
2
h
)
Report Question
0%
1
12
0%
−
4
3
0%
−
16
3
0%
−
1
48
Explanation
Let
L
=
lim
h
→
0
(
1
h
3
√
8
+
h
−
1
2
h
)
=
lim
h
→
0
1
h
(
1
3
√
8
+
h
−
1
2
)
Let
a
=
3
√
8
+
h
⇒
a
3
=
8
+
h
and
b
=
2
⇒
b
3
=
8
Also
a
3
−
b
3
=
(
a
−
b
)
(
a
2
+
b
2
+
a
b
)
=
8
+
h
−
8
=
h
⇒
(1)
∴
L
=
lim
h
→
0
1
h
(
1
a
−
1
b
)
=
−
lim
h
→
0
a
−
b
h
⋅
1
a
b
=
−
lim
h
→
0
a
3
−
b
3
h
⋅
1
a
b
(
a
2
+
b
2
+
a
b
)
=
−
lim
h
→
0
h
h
⋅
1
a
b
(
a
2
+
b
2
+
a
b
)
.....using
(
1
)
As
h
→
0
⇒
a
3
=
8
⇒
a
=
2
=
−
1
2
×
2
(
2
2
+
2
2
+
2
×
2
)
=
=
−
1
48
lf
lim
x
→
a
+
f
(
x
)
=
L
, then for each
ϵ
>
0
, there exists
δ
>
0
so that
Report Question
0%
0
<
|
x
−
a
|
<
δ
⇒
|
f
(
x
)
−
L
|
≥
ϵ
0%
0
<
|
x
−
a
|
<
δ
⇒
|
f
(
x
)
−
L
|
<
ϵ
0%
a
<
x
<
a
+
δ
⇒
f
(
x
)
−
L
<
ϵ
0%
a
−
δ
<
x
<
a
⇒
|
f
(
x
)
−
L
|
<
ϵ
Explanation
It is fundamental concept that, for limit of a function
f
(
x
)
to exist at any point
a
there exists a real number
δ
>
0
, such that
0
<
|
x
−
a
|
<
δ
, for which
|
f
(
x
)
−
L
|
<
ϵ
, where
ϵ
>
0
lim
x
→
0
1
−
cos
x
x
log
(
1
+
x
)
=
Report Question
0%
1
0%
0
0%
−
1
0%
1
2
Explanation
lim
x
→
0
1
−
cos
x
x
log
(
1
+
x
)
lim
x
→
0
2
sin
2
x
2
x
log
(
1
+
x
)
=
lim
x
→
0
2
sin
2
x
2
(
x
2
)
2
×
1
4
1
x
log
(
1
+
x
)
=
lim
x
→
0
1
2
log
(
1
+
x
)
x
=
1
2
lim
x
→
0
(
1
−
e
x
)
sin
x
x
2
+
x
3
=
Report Question
0%
−
1
0%
0
0%
1
0%
2
Explanation
The expression can be simplified as,
(lim
x
→
0
(
1
−
e
x
)
x
)
×
(
s
i
n
x
x
)
(
l
i
m
x
→
0
1
1
+
x
)
Now using,
lim
x
→
0
(
e
x
−
1
)
x
=
1
lim
x
→
0
s
i
n
x
x
=
1
We get,
(
−
1
)
×
1
×
l
i
m
x
→
0
1
1
+
x
=
−
1
lf
f
(
x
)
=
2
x
−
3
,
a
=
2
,
l
=
1
and
ϵ
=
0.001
then
δ
>
0
satisfying
0
<
|
x
−
a
|
<
δ
,
|
f
(
x
)
−
l
|
<
ϵ
, is:
Report Question
0%
0.005
0%
0.0005
0%
0.001
0%
0.0001
Explanation
|
f
(
x
)
−
l
|
<
0.001
=
ϵ
⇒
|
2
x
−
3
−
1
|
<
0.001
⇒
−
0.001
<
2
x
−
4
<
0.001
⇒
−
0.0005
<
x
−
2
<
0.0005
⇒
|
x
−
2
|
<
0.0005
⇒
|
x
−
a
|
<
0.0005
=
δ
Hence,
δ
=
0.0005
>
0
lim
x
→
0
e
x
−
e
sin
x
2
(
x
−
sin
x
)
=
Report Question
0%
−
1
2
0%
1
2
0%
1
0%
3
2
Explanation
We simplify the given expression as,
lim
x
→
0
e
s
i
n
x
(
e
x
−
s
i
n
x
−
1
)
2
(
x
−
s
i
n
x
)
Let
x
−
s
i
n
x
=
y
As
x
→
0
so does y
→
0
Hence, the question transforms into
(
lim
x
→
0
e
s
i
n
x
2
)
×
(
lim
y
→
0
e
y
−
1
y
)
=
1
2
×
1
=
1
2
lim
x
→
0
x
tan
2
x
−
2
x
tan
x
(
1
−
cos
2
x
)
2
=
Report Question
0%
2
0%
−
2
0%
1
2
0%
−
1
2
Explanation
lim
x
→
0
x
tan
2
x
−
2
x
tan
x
(
1
−
cos
2
x
)
2
=
lim
x
→
0
x
2
tan
x
1
−
tan
2
x
−
2
x
tan
x
(
1
−
1
−
tan
2
x
1
+
tan
2
x
)
2
=
lim
x
→
0
2
x
tan
3
x
4
tan
4
x
=
lim
x
→
0
2
x
4
tan
x
=
1
2
lf
lim
x
→
0
(
cos
4
x
+
a
cos
2
x
+
b
x
4
)
is finite then the value of
a
,
b
respectively are
Report Question
0%
5
−
4
0%
−
5
,
−
4
0%
−
4
,
3
0%
4
,
5
Explanation
lim
x
→
0
(
cos
4
x
+
a
cos
2
x
+
b
x
4
)
As
x
→
0
, denominator tends to 0, so the numerator also tends to 0.
⇒
lim
x
→
0
cos
4
x
+
a
cos
2
x
+
b
=
0
at
x
=
0
⇒
cos
4
x
=
1
,
cos
2
x
=
1
⇒
1
+
a
+
b
=
0
⇒
a
+
b
=
−
1
Now, put
a
=
−
4
⇒
b
=
3
Option C satisfies above equation.
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
Explanation
lim
x
→
0
1
−
cos
3
x
x
sin
2
x
=
lim
x
→
0
(
1
−
cos
x
)
(
1
+
cos
x
+
cos
2
x
)
x
sin
2
x
=
lim
x
→
0
2
sin
2
x
2
(
1
+
cos
x
+
cos
2
x
)
4
x
cos
x
sin
x
2
cos
x
2
=
lim
x
→
0
sin
x
2
(
1
+
cos
x
+
cos
2
x
)
4
×
x
2
cos
x
2
×
cos
x
=
3
4
...... As
{
lim
x
2
→
0
sin
x
2
x
2
=
0
}
lim
x
→
π
2
(
π
2
−
x
)
sec
x
c
o
s
e
c
x
=
Report Question
0%
1
0%
0
0%
-1
0%
1
π
Explanation
lim
x
→
π
2
(
π
2
−
x
)
sec
x
csc
x
=
lim
h
→
0
(
π
2
−
π
2
+
h
)
sec
(
π
2
−
h
)
csc
(
π
2
−
h
)
=
lim
h
→
0
h
csc
h
sec
h
=
lim
h
→
0
h
cos
h
sin
h
=
1
Solve :
lim
x
→
0
sin
x
sin
(
π
3
+
x
)
sin
(
π
3
−
x
)
x
Report Question
0%
3
4
0%
1
4
0%
4
3
0%
0
Explanation
Solve :
lim
x
→
0
sin
x
sin
(
π
3
+
x
)
sin
(
π
3
−
x
)
x
=
lim
x
→
0
sin
x
x
×
lim
x
→
0
sin
(
π
3
+
x
)
×
lim
x
→
0
sin
(
π
3
−
x
)
=
1
×
sin
π
3
×
sin
π
3
=
1
×
√
3
2
×
√
3
2
=
3
4
lim
x
→
0
(
1
−
cos
2
x
)
sin
5
x
x
2
sin
3
x
=
Report Question
0%
10
3
0%
3
10
0%
6
5
0%
5
6
Explanation
Using,
1
−
c
o
s
2
x
=
2
s
i
n
2
x
,
The expression transforms to,
l
i
m
x
→
0
2
s
i
n
2
x
s
i
n
5
x
x
2
s
i
n
3
x
Rewriting the expression in a different form,
l
i
m
x
→
0
2
s
i
n
2
x
x
2
×
s
i
n
5
x
5
x
×
3
x
s
i
n
3
x
×
5
3
Therefore, the limit to the expression is
10
3
Hence, option 'A' is correct.
The value of
lim
α
→
β
[
sin
2
α
−
sin
2
β
α
2
−
β
2
]
is:
Report Question
0%
0
0%
1
0%
sin
β
β
0%
sin
2
β
2
β
Explanation
lim
α
→
β
[
sin
2
α
−
sin
2
β
α
2
−
β
2
]
=
lim
α
→
β
[
sin
(
α
−
β
)
sin
(
α
+
β
)
(
α
−
β
)
(
α
+
β
)
]
=
lim
α
→
β
[
sin
(
α
−
β
)
(
α
−
β
)
]
×
lim
α
→
β
[
sin
(
α
+
β
)
(
α
+
β
)
]
=
lim
α
−
β
→
0
[
sin
(
α
−
β
)
(
α
−
β
)
]
.
(
sin
2
β
2
β
)
.......
[
∵
α
→
β
⟹
α
−
β
→
0
]
=
1.
sin
2
β
2
β
.....
[
∵
lim
x
→
0
sin
x
x
=
1
]
=
sin
2
β
2
β
Evaluate:
lim
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
Explanation
Given,
lim
x
→
0
sec
4
x
−
sec
2
x
sec
3
x
−
sec
x
sec
x
=
1
cos
x
=
lim
x
→
0
1
cos
4
x
−
1
cos
2
x
1
cos
3
x
−
1
cos
x
=
lim
x
→
0
cos
2
x
−
cos
4
x
cos
x
−
cos
3
x
cos
3
x
cos
x
cos
4
x
cos
2
x
=
lim
x
→
0
−
2
sin
3
x
sin
(
−
x
)
−
2
sin
(
2
x
)
sin
(
−
x
)
cos
3
x
cos
x
cos
4
x
cos
2
x
=
lim
x
→
0
sin
3
x
3
x
×
3
x
sin
2
x
2
x
×
2
x
cos
3
x
cos
x
cos
4
x
cos
2
x
as
cos
0
=
1
and
lim
x
→
0
sin
x
x
=
1
=
3
x
2
x
=
3
2
lim
x
→
π
2
c
o
s
e
c
x
−
cot
x
x
=
Report Question
0%
1
0%
2
π
0%
1
2
0%
1
5
Explanation
lim
x
→
π
2
csc
x
−
cot
x
x
=
lim
x
→
π
2
1
−
cos
x
x
sin
x
=
2
π
lim
x
→
π
4
sec
x
.
tan
(
4
x
−
π
)
sin
(
4
x
−
π
)
=
Report Question
0%
√
2
0%
1
√
2
0%
−
√
2
0%
−
1
√
2
Explanation
lim
x
→
π
4
sec
x
.
tan
(
4
x
−
π
)
sin
(
4
x
−
π
)
=
lim
x
→
π
4
sec
x
.
tan
(
4
x
−
π
)
(
4
x
−
π
)
sin
(
4
x
−
π
)
(
4
x
−
π
)
=
√
2
lim
x
→
π
6
3
sin
x
−
√
3
cos
x
6
x
−
π
=
Report Question
0%
√
3
0%
1
√
3
0%
−
√
3
0%
−
1
√
3
Explanation
lim
x
→
π
6
3
sin
x
−
√
3
cos
x
6
x
−
π
=
2
√
3
lim
x
→
π
6
√
3
2
sin
x
−
1
2
cos
x
6
x
−
π
=
2
√
3
6
lim
x
→
π
6
cos
π
6
sin
x
−
sin
π
6
cos
x
x
−
π
6
=
1
√
3
lim
x
→
π
6
sin
(
x
−
π
6
)
(
x
−
π
6
)
=
1
√
3
lim
x
→
0
t
a
n
x
0
x
=
Report Question
0%
0
0%
1
0%
π
180
0%
π
Explanation
lim
x
→
0
t
a
n
x
0
x
=
π
180
lim
x
→
0
tan
(
π
180
x
)
(
π
180
)
x
=
π
180
lim
x
→
0
3
sin
x
−
sin
3
x
x
3
=
Report Question
0%
0
0%
1
0%
(
π
180
)
3
0%
4.
(
π
180
)
3
Explanation
lim
x
→
0
3
sin
x
−
sin
3
x
x
3
=
lim
x
→
0
4
sin
3
x
x
3
=
lim
x
→
0
4
sin
3
(
π
180
x
)
x
3
=
(
π
180
)
3
lim
x
→
0
4
sin
3
(
π
180
x
)
x
3
(
π
180
)
3
=
4
(
π
180
)
3
Solve:
lim
x
→
0
3
tan
x
−
tan
3
x
2
x
3
Report Question
0%
1
4
0%
3
4
0%
4
0%
−
4
Explanation
lim
x
→
0
3
tan
x
−
tan
3
x
2
x
3
lim
x
→
0
3
(
x
+
x
3
3
+
2
x
5
15
+
.
.
.
.
.
.
)
−
(
3
x
+
(
3
x
)
3
3
+
2
(
3
x
)
5
15
+
.
.
.
.
)
2
x
3
=
1
−
9
2
=
−
4
lim
x
→
0
a
x
−
b
x
x
=
Report Question
0%
log
(
a
b
)
0%
log
(
b
a
)
0%
log
(
a
b
)
0%
log
b
a
Explanation
lim
x
→
0
a
x
−
b
x
x
=
lim
x
→
0
a
x
−
1
−
b
x
+
1
x
=
lim
x
→
0
(
a
x
−
1
)
−
(
b
x
−
1
)
x
=
log
a
−
log
b
......... Using formula
=
log
(
a
b
)
Hence,
lim
x
→
0
a
x
−
b
x
x
=
log
(
a
b
)
lim
x
→
0
(
sin
x
−
x
x
)
(
sin
1
x
)
is:
Report Question
0%
does not exist
0%
is equal to 0
0%
is equal to 1
0%
exists and different from 0 and 1
Explanation
s
i
n
x
=
x
−
x
3
3
!
+
x
5
5
!
.
.
.
.
(
1
)
s
i
n
x
−
x
=
−
x
3
3
!
+
x
5
5
!
−
x
7
7
!
s
i
n
x
−
x
x
=
−
x
2
3
!
+
x
4
5
!
−
x
6
7
!
.
.
.
.
=
0
(
a
s
x
f
o
u
n
d
t
o
0
)
sin
(
1
x
)
is an oscillatory function means same finite no.
0
×
f
i
n
i
t
e
n
o
.
=
0
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
Explanation
lim
x
→
π
4
cos
x
−
sin
x
(
π
4
−
x
)
(
cos
x
+
sin
x
)
=
lim
x
→
π
4
1
−
tan
x
(
π
4
−
x
)
(
1
+
tan
x
)
.................(dividing the num and deno by cosx)
=
lim
x
→
π
4
tan
(
π
4
−
x
)
(
π
4
−
x
)
=
lim
h
→
0
tan
h
h
=
1
lim
x
→
π
2
1
−
sin
θ
cos
θ
(
π
2
−
θ
)
=
Report Question
0%
1
0%
−
1
0%
−
1
2
0%
1
2
Explanation
P
u
t
θ
=
π
2
−
ϕ
L
i
m
ϕ
→
0
1
−
c
o
s
ϕ
s
i
n
ϕ
ϕ
ϕ
2
=
1
2
lim
x
→
π
2
1
−
sin
x
(
π
−
2
x
)
2
=
Report Question
0%
1
3
0%
1
4
0%
1
6
0%
1
8
Explanation
lim
x
→
π
2
1
−
sin
x
(
π
−
2
x
)
2
=
lim
h
→
0
1
−
sin
(
π
2
−
h
)
(
π
−
2
(
π
2
−
h
)
)
2
......(x-->h-pi/2)
=
lim
h
→
0
1
−
cos
h
4
h
2
=
1
8
lim
x
→
∞
x
+
sin
x
x
+
cos
x
=
Report Question
0%
1
0%
−
1
0%
1
3
0%
0
Explanation
lim
x
→
∞
x
+
sin
x
x
+
cos
x
=
lim
x
→
∞
1
+
sin
x
x
1
+
cos
x
x
=
1
+
Any finite number between -1 and 1
∞
1
+
Any finite number between -1 and 1
∞
=
1
+
0
1
+
0
=
1
f
(
x
)
=
{
3
x
2
sin
2
x
2
,
x
<
0
x
2
+
2
x
+
C
1
−
3
x
2
,
x
≥
0
,
x
≠
1
√
3
0
,
x
=
1
√
3
lim
x
→
0
+
f
(
x
)
=
Report Question
0%
C
4
0%
2C
0%
C
3
0%
C
Let
f
(
x
)
=
{
x
2
−
1
,
0
<
x
<
2
2
x
+
3
,
2
≤
x
<
3
,
The quadratic equation whose roots are
lim
x
→
2
−
f
(
x
)
and
lim
x
→
2
+
f
(
x
)
is
Report Question
0%
x
2
−
10
x
+
21
=
0
0%
x
2
−
6
x
+
9
=
0
0%
x
2
−
14
x
+
49
=
0
0%
x
2
+
6
x
+
9
=
0
Explanation
For
0
<
x
<
2
,
f
(
x
)
=
x
2
−
1
lim
x
→
2
−
f
(
x
)
=
3
For,
2
<
x
<
3
f
(
x
)
=
2
x
+
3
lim
x
→
2
+
f
(
x
)
=
7
Thus the equation with roots, 3 and 7 is
x
2
−
10
x
+
21
=
0
The value of
√
e
e
upto four decimal places is?
Report Question
0%
1.6484
0%
1.5237
0%
1.2589
0%
1.9190
Explanation
(
e
)
1
2
=
1
+
1
2
+
(
1
2
)
2
2
!
+
(
1
2
)
3
3
!
+
(
1
2
)
4
4
!
=
1
+
0
⋅
5
+
0
⋅
125
+
0
⋅
02083
+
0
⋅
00260
=
1
⋅
6484
If
f
(
x
)
=
x
2
+
6
x
sin
x
, then
lim
x
→
0
−
f
(
x
)
=
Report Question
0%
2
0%
4
0%
6
0%
8
Explanation
−
lim
x
→
0
f
(
x
)
=
x
2
+
6
x
sin
x
=
lim
h
→
0
f
(
h
)
=
h
2
+
6
(
0
−
h
)
sin
(
0
−
h
)
=
lim
h
→
0
f
(
h
)
=
h
−
6
−
sin
(
h
)
/
h
=
−
6
−
1
=
6
Hence, the answer is "C".
lim
x
→
0
log
(
1
+
a
x
)
−
log
(
1
+
b
x
)
x
=
Report Question
0%
a
+
b
0%
a
−
b
0%
−
(
a
+
b
)
0%
a
b
Explanation
lim
x
→
0
log
(
1
+
a
x
)
−
log
(
1
+
b
x
)
x
=
a
.
lim
x
→
0
log
(
1
+
a
x
)
a
x
−
b
.
lim
x
→
0
log
(
1
+
b
x
)
b
x
=
a
−
b
Let
f
be a continuous function on [1,3]. lf
f
takes only rational values for all
x
and
f
(
2
)
=
10
then
f
(
1.5
)
is equal to
Report Question
0%
8
0%
f
(
1
)
+
f
(
3
)
3
0%
20
0%
10
Explanation
Since
f
is continuous, so it must take all real values between
f
(
1
)
and
f
(
3
)
.
But since
f
takes only rational values so
f
must be a constant function.
Hence
f
(
1.5
)
=
f
(
2
)
=
10
Evaluate:
lim
x
→
0
sin
3
x
2
cos
(
2
x
2
−
x
)
Report Question
0%
0
0%
−
1
0%
4
0%
−
6
Explanation
Given,
lim
x
→
0
(
sin
(
3
)
x
2
cos
(
2
x
2
−
x
)
)
=
sin
(
3
)
⋅
0
2
cos
(
2
⋅
0
2
−
0
)
=
0
lim
x
→
0
−
3
sin
(
2
x
2
)
x
2
=
A
then the value of
A
is
Report Question
0%
2
0%
4
0%
6
0%
8
Explanation
lim
x
→
0
−
3
sin
(
2
x
2
)
x
2
=
lim
x
→
0
−
2
∗
3
sin
(
2
x
2
)
2
x
2
=
6
Hence, answer is option
C
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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