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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 10 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 10
l
i
m
n
→
∞
Σ
n
r
=
1
π
n
s
i
n
(
π
r
n
)
is equal to
Report Question
0%
1
0%
2
0%
3
0%
4
Evaluate :
lim
x
→
0
(
e
x
ℓ
n
(
3
x
−
1
)
−
(
3
x
−
1
)
x
sin
x
e
x
ℓ
n
x
)
is equal to
Report Question
0%
1
e
ℓ
n
3
0%
e
ℓ
n
3
0%
3
0%
1
3
L
t
x
→
0
s
i
n
x
−
x
+
x
3
/
6
x
5
=
Report Question
0%
1
120
0%
1
110
0%
1
100
0%
none of these
lim
x
→
0
(
cos
α
)
x
−
(
sin
α
)
x
−
cos
2
α
(
x
−
4
)
,
α
∈
(
0
,
π
2
)
is equal to
Report Question
0%
cos
4
α
.
log
(
cos
α
)
−
sin
4
α
.
log
(
sin
α
)
0%
sin
4
α
.
log
(
cos
α
)
−
cos
4
α
.
log
(
sin
α
)
0%
sin
4
α
.
log
(
cos
α
)
+
cos
4
α
.
log
(
sin
α
)
0%
N
o
n
e
o
f
t
h
e
a
b
o
v
e
lim
x
→
0
3
sin
(
x
9
)
−
sin
(
x
9
)
x
3
=
q
Report Question
0%
0
0%
4
(
π
200
)
3
0%
π
200
0%
π
100
L
t
x
→
0
t
a
n
x
−
x
x
2
t
a
n
x
equals:
Report Question
0%
1
0%
1/2
0%
1/3
0%
None of these
evaluate
l
i
m
x
→
0
x
−
∫
x
0
c
o
s
t
2
d
t
x
3
−
6
x
Report Question
0%
3
0%
−
1
0%
0
0%
1
lim
x
→
0
{
tan
(
π
4
+
x
)
}
1
x
=
Report Question
0%
e
0%
e
2
0%
e
−
1
0%
e
−
2
lim
x
→
∞
(
x
+
1
2
x
+
1
)
x
2
equals?
Report Question
0%
0
0%
e
0%
1
0%
∞
If
g
(
x
)
=
x
[
x
]
f
o
r
x
>
2
t
h
e
n
L
i
m
x
→
2
+
g
(
x
)
−
g
(
2
)
x
−
2
Report Question
0%
-1
0%
0
0%
1/2
0%
1
l
i
m
x
→
∞
2
tan
−
1
x
π
equals
e
L
then
L
is equal to
Report Question
0%
2
π
0%
−
2
π
0%
−
π
2
0%
1
lim
x
→
x
/
2
[
1
−
tan
(
x
2
)
]
[
1
−
sin
x
]
[
1
+
tan
(
x
2
)
]
[
π
−
2
x
]
3
is
Report Question
0%
1
8
0%
0
0%
1
32
0%
∞
lim
θ
→
0
4
θ
(
tan
θ
−
2
θ
tan
θ
)
(
1
−
cos
2
θ
)
2
is
Report Question
0%
1
/
√
2
0%
1
/
2
0%
1
0%
2
the value of
l
i
m
x
→
∞
x
5
5
x
is-
Report Question
0%
0
0%
1
0%
e
5
0%
e
−
5
If
k
is an integer such that
lim
n
→
∞
[
(
cos
k
π
4
)
2
−
(
cos
k
π
6
)
2
]
=
0
then :
Report Question
0%
k
is divisible neither by
4
nor by
8
.
0%
k
must be divisible by
12
but not necessarily by
24
.
0%
k
must be divisible by
24
.
0%
either
k
is divisible by
24
or
k
is neither divisible by
4
nor by
6
.
lim
x
→
5
sin
2
(
x
−
5
)
tan
(
x
−
5
)
(
x
2
−
25
)
(
x
−
5
)
=
Report Question
0%
1
/
10
0%
1
0%
0
0%
−
6
The value of
l
i
m
x
→
0
cos
e
c
4
x
x
2
∫
0
I
n
(
1
+
4
t
)
1
+
t
2
d
t
is
Report Question
0%
1
0%
2
0%
3
0%
4
lim
n
→
∞
n
2
(
x
1
/
n
−
x
1
/
(
n
+
n
)
)
,
x
>
0
, is equal to
Report Question
0%
0
0%
e
x
0%
log
e
x
0%
n
o
n
e
o
f
t
h
e
s
e
lim
x
→
0
l
n
(
s
i
n
3
x
)
l
n
(
s
i
n
x
)
is equal to
Report Question
0%
0
0%
1
0%
2
0%
Non existent
l
i
m
x
→
n
/
2
cot
x
−
cos
x
(
π
−
2
x
)
3
equals
Report Question
0%
1
24
0%
1
16
0%
1
8
0%
1
4
l
i
m
x
→
2
3
√
60
+
x
2
−
4
sin
(
x
−
2
)
equals
Report Question
0%
1
4
0%
0
0%
1
12
0%
Does not exist
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
lim
x
→
0
2
(
√
3
sin
(
π
6
+
x
)
−
cos
(
π
6
+
x
)
)
x
√
3
(
√
3
cos
x
−
sin
x
)
Report Question
0%
−
1
/
3
0%
2
/
3
0%
4
/
3
0%
−
4
/
3
lim
x
→
π
/
4
2
√
2
−
(
cos
x
+
sin
x
)
2
1
−
sin
2
x
is equal to
Report Question
0%
1
0%
2
√
2
0%
4
√
2
3
0%
d
o
e
s
n
o
t
e
x
i
s
t
Let
f
be a differentiable function such that
f
′
(
x
)
=
7
−
3
4
f
(
x
)
x
,
(
x
>
0
)
and
f
(
1
)
≠
4
.
Then
lim
x
→
0
+
x
f
(
1
x
)
:
Report Question
0%
Exists and equals 4
0%
Does not exist
0%
Exist and equals
0%
Exists and equals
4
7
Explanation
f
′
(
x
)
=
7
−
3
4
f
(
x
)
x
(
x
>
0
)
Given
f
(
1
)
≠
4
lim
x
→
0
+
x
f
(
1
x
)
=
?
d
y
d
x
+
3
4
y
x
=
7
(This is LDE)
IF
=
e
∫
3
4
x
d
x
=
e
3
4
l
n
|
x
|
=
x
3
4
y
.
x
3
4
=
∫
7.
x
3
4
d
x
y
.
x
3
4
=
7.
x
7
4
7
4
+
C
f
(
x
)
=
4
x
+
C
.
x
−
3
4
f
(
1
x
)
=
4
x
+
C
.
x
3
4
lim
x
→
0
+
x
f
(
1
x
)
=
lim
x
→
0
+
(
4
+
C
.
x
7
4
)
=
4
If
L
i
m
x
→
0
(
1
+
a
3
)
+
8
e
1
/
x
1
+
(
1
−
b
3
)
+
e
1
/
x
=
2
then
Report Question
0%
a
=
1
,
b
=
2
0%
a
=
1
,
b
=
−
3
1
/
3
0%
a
=
2
,
b
=
3
1
/
3
0%
none of these
l
i
m
x
→
π
2
t
a
n
2
x
(
√
2
s
i
n
2
x
+
3
s
i
n
x
+
4
−
√
s
i
n
2
x
+
6
s
i
n
x
+
2
)
is equal to
Report Question
0%
3
4
0%
1
6
0%
1
12
0%
5
12
lim
x
→
0
1
−
cos
x
x
log
(
1
+
x
)
=
Report Question
0%
1
0%
0
0%
−
1
0%
1
2
Evaluate
lim
x
→
0
sin
[
cos
x
]
1
+
[
cos
x
]
(
[
.
]
denotes the greatest integer function)
Report Question
0%
−
1
0%
1
0%
0
0%
d
o
e
s
n
o
t
e
x
i
s
t
lim
x
→
0
x
(
e
sin
x
−
1
)
1
−
cos
x
Report Question
0%
1
2
0%
2
0%
0
0%
1
P
=
lim
x
→
o
+
(
1
+
tan
2
√
x
)
1
/
2
x
=
____
Report Question
0%
P
=
1
/
4
0%
P
=
1
/
2
0%
P
=
2
0%
P
=
1
lim
x
→
0
(
1
1
/
x
+
2
1
/
x
+
3
1
/
x
+
.
.
.
.
.
n
1
/
x
n
)
n
x
,
n
ϵ
N
is equal to
Report Question
0%
n
!
0%
1
0%
1
n
!
0%
0
Explanation
l
i
m
x
→
∞
(
1
1
/
x
+
2
1
/
x
+
3
1
/
x
.
.
.
.
.
.
+
n
1
/
x
n
)
n
x
if we put the limits then we have
1
∞
form. We have for
l
i
m
x
→
∞
[
f
(
x
)
]
9
(
x
)
such that limit has the form
1
∞
, then
l
i
m
x
→
∞
[
f
(
x
)
]
9
(
x
)
=
e
l
i
m
x
→
∞
9
(
x
)
[
f
(
x
)
−
1
]
⟶
(
1
)
Using this we can evalute the limit asked,
l
i
m
x
→
∞
(
1
1
/
x
+
2
1
/
x
+
3
1
/
x
+
.
.
.
.
.
.
n
1
/
x
n
)
n
x
=
e
l
i
m
x
→
∞
n
x
(
1
1
/
x
+
2
1
/
x
+
3
1
/
x
.
.
.
.
.
.
n
1
/
x
−
n
n
)
=
e
l
i
m
x
→
∞
(
1
1
/
x
+
2
1
/
x
+
3
1
/
x
+
.
.
.
.
.
.
n
1
/
x
−
n
(
1
/
x
)
)
if we put the limits we find that it is in
÷
form, so we will employ the L'Hospital's rule
=
e
l
i
m
x
→
∞
(
1
1
/
x
(
−
1
x
2
)
l
n
1
+
2
1
/
x
(
−
1
x
2
)
l
n
2......
+
n
1
/
x
(
−
1
x
2
)
l
n
n
−
0
(
−
1
/
x
2
)
)
=
e
l
i
m
x
→
∞
(
1
1
/
x
l
n
1
+
2
1
/
x
l
n
2......
+
n
1
/
x
l
n
n
)
=
e
l
n
1
+
l
n
2
+
l
n
3......
+
l
n
n
[
∵
\underset { x\rightarrow \infty }{ lim } { n }^{ 1/x }={ n }^{ 0 }=1
]
={ e }^{ ln\left( 1.2.3.4.5.6......n \right) }
[
\because
lna+lnb=lnab
]
={ e }^{ lnn! }
=n!
[
\because
{ e }^{ lnb }=b
]
Answer : Option A
The values of
\displaystyle\lim_{n\rightarrow \infty}\dfrac{\sqrt[4]{n^5+2}-\sqrt[3]{n^2+1}}{\sqrt[5]{n^4+2}-\sqrt[2]{n^3+1}}
is?
Report Question
0%
1
0%
0
0%
-1
0%
\infty
\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { x\tan { 2x } -2x\tan { x } }{ \left( 1-\cos { 2x } \right) ^{ 2 } } }
equal to
Report Question
0%
\dfrac{1}{4}
0%
1
0%
\dfrac{1}{2}
0%
None of these
Let
f(x)=\displaystyle\lim _{ n\rightarrow \infty }{ \dfrac { { 2x }^{ 2n }\sin { \frac { 1 }{ x } +x } }{ 1+{ x }^{ 2x } } }
then find
Report Question
0%
\displaystyle\lim _{ x\rightarrow \infty }{ xf\left( x \right) }
0%
\displaystyle\lim _{ x\rightarrow 1 }{ f\left( x \right) }
0%
\displaystyle\lim _{ x\rightarrow 0 }{ f\left( x \right) }
0%
\displaystyle\lim _{ x\rightarrow -\infty }{ f\left( x \right) }
If
\lim _{x \rightarrow 0}\left(\cos x+a^{3} \sin \left(b^{6} x\right)\right)^{\frac{1}{x}}=e^{512}
then value of
ab^2
is equal to
Report Question
0%
-512
0%
512
0%
8
0%
none of these
if\left( x \right) =\left[ x-3 \right] +\left[ x-4 \right] \quad for\quad x\epsilon R\quad then\quad \underset { x\rightarrow 3 }{ lim } f\left( x \right) =
Report Question
0%
-2
0%
-1
0%
0
0%
2
\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { { \left( \cos { x } \right) }^{ 1/2 }-{ \left( \cos { x } \right) }^{ 1/3 } }{ \sin ^{ 2 }{ x } } }
is $$
Report Question
0%
1/6
0%
-1/12
0%
2/3
0%
1/3
The value of
\begin{matrix} lim \\ x\rightarrow y \end{matrix}\dfrac { { sin }^{ 2 }x-sin^{ 2 }y }{ { x }^{ 2 }-{ y }^{ 2 } }
Report Question
0%
0
0%
1
0%
\dfrac { siny }{ y }
0%
\dfrac { sin2y }{ 2y }
\displaystyle\lim_{x\rightarrow 0}\dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^2}
is equal to?
Report Question
0%
2
0%
-2
0%
1/2
0%
-1/2
\underset { x\rightarrow 0 }{ lim } (\cos x+a\sin b{ x) }^{ \frac { 1 }{ x } }
is equal to
Report Question
0%
e^a
0%
e^{ab}
0%
e^b
0%
e^{a/b}
\displaystyle \lim_{x\rightarrow 0}{\dfrac{x(1+a\cos x)-b\sin x}{x^{3}}}=1
then
Report Question
0%
a=-5/2
0%
a=-3/2,b=-1/2
0%
a=-3/2,b=-5/2
0%
a=-5/2,b=-3/2
If
\displaystyle\lim_{x\rightarrow 0}[1+x ln (1+b^2)]^{1/x}=2b\sin^2\theta, b >
s m f
\theta\in (-\pi, \pi]
, then the value of
\theta
is?
Report Question
0%
\pm \dfrac{\pi}{4}
0%
\pm \dfrac{\pi}{3}
0%
\pm \dfrac{\pi}{6}
0%
\pm \dfrac{\pi}{2}
\lim _ { x \rightarrow \infty } \dfrac { \sin \left( \pi \cos ^ { -2 } x \right) } { x ^ { 2 } }
is equal to:
Report Question
0%
- \pi
0%
\pi
0%
\dfrac { \pi } { 2 }
0%
1
The value of
\displaystyle\lim_{\theta \rightarrow 0^+}\dfrac{\sin\sqrt{\theta}}{\sqrt{\sin\theta}}
is equal to?
Report Question
0%
0
0%
1
0%
-1
0%
4
Let
f
:
\left ( \dfrac{\pi}{2}, \dfrac{\pi}{2} \right )\rightarrow R
,
f(x) = \left \{\begin{matrix} \lim_{n\rightarrow \infty }\dfrac{(tanx)^{2n} + x^2}{sin^2x + (tanx)^{2n}}; & x \neq 0 \\ 1; & x = 0 \end{matrix} \right., n \in N
. Which of the following holds good ?
Report Question
0%
f \left( -\dfrac{\pi^-}{4} \right) = f \left (\dfrac{\pi^+}{4} \right)
0%
f \left( -\dfrac{\pi^-}{4} \right) = f \left (-\dfrac{\pi^+}{4} \right)
0%
f \left( \dfrac{\pi^-}{4} \right) = f \left (\dfrac{\pi^+}{4} \right)
0%
f(0^+) = f(0) = f(0^)
\displaystyle\lim_{x\rightarrow 0}\left(\dfrac{1}{\sin^2x}-\dfrac{1}{\sin h^2x}\right)=?
Report Question
0%
2_{\displaystyle /3}
0%
0
0%
1_{\displaystyle /3}
0%
-2_{\displaystyle /3}
Consider
\lim _ { x \rightarrow 0 } \dfrac { a x + b e ^ { - x } + \sin x + 1 } { a x - b \sin x } = \ell ( \ell
is a finite number)
Report Question
0%
\ell = 0 , \text { if } a = - 2 , b = - 1
0%
\ell = 5 , i f a = 1 , b = - 1
0%
\ell = 1 , \text { if } a = b = - 1
0%
None of these
\displaystyle\lim_{x\rightarrow 0}\dfrac{\sin(\pi \cos^2x)}{x^2}
is equal to?
Report Question
0%
-\pi
0%
\pi
0%
\pi/2
0%
1
If
f(x)
is odd linear polynomial with
f(1)=1
, then
\underset{x \to 0} {\lim} \dfrac{2^{f(\tan x)}-2^{f(\sin x)}}{x^{2}f(\sin x)}
is
Report Question
0%
1
0%
\ln 2
0%
\dfrac{1}{2}\ln 2
0%
\cos 2
0:0:2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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