Loading [MathJax]/jax/element/mml/optable/MathOperators.js
MCQExams
0:0:2
CBSE
JEE
NTSE
NEET
Practice
Homework
×
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
∵
\Rightarrow -\epsilon <f(x)-l<\epsilon
\Rightarrow l-\epsilon<f(x)<l+\epsilon
\Rightarrow 3-\epsilon<2x+1<3+\epsilon
\Rightarrow -\epsilon<2x-2<\epsilon
\Rightarrow -\dfrac{\epsilon }{2}<x-1<\dfrac{\epsilon }{2}
\Rightarrow |x-1|<\dfrac{\epsilon }{2}
\Rightarrow |x-a| <\dfrac{\epsilon }{2}
\delta = \dfrac{\epsilon}{2} = \dfrac{0.001}{2} = 0.0005
Hence, option A.
Evaluate:
\displaystyle \lim_{h\rightarrow 0}\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right)
Report Question
0%
\displaystyle \frac{1}{12}
0%
-\displaystyle \frac{4}{3}
0%
-\displaystyle \frac{16}{3}
0%
-\displaystyle \frac{1}{48}
Explanation
Let
L =\displaystyle \lim_{h\rightarrow 0}\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right) =\lim_{h\rightarrow 0}\frac{1}{h}\left(\frac{1}{\sqrt[3]{8+h}}-\frac{1}{2}\right)
Let
a= \sqrt[3]{8+h}\Rightarrow a^3 = 8+h
and
b=2\Rightarrow b^3 =8
Also
a^3-b^3=(a-b)(a^2+b^2+ab)=8+h-8 =h\Rightarrow
(1)
\therefore\displaystyle L = \lim_{h\to 0}\frac{1}{h}\left( \frac{1}{a}-\frac{1}{b}\right)=-\lim_{h\to 0}\frac{a-b}{h}\cdot \frac{1}{ab}
\displaystyle \quad = -\lim_{h\to 0}\frac{a^3-b^3}{h}\cdot \frac{1}{ab(a^2+b^2+ab)}
\displaystyle \quad = -\lim_{h\to 0}\frac{h}{h}\cdot \frac{1}{ab(a^2+b^2+ab)}
.....using
(1)
As
{h \rightarrow 0}\Rightarrow a^3=8\Rightarrow a=2
\displaystyle \quad =-\frac{1}{2\times 2(2^2+2^2+2\times 2)}=
=-\dfrac{1}{48}
lf
\displaystyle \lim_{x\rightarrow a^+}f(x)=L
, then for each
\epsilon>0
, there exists
\delta>0
so that
Report Question
0%
0<|x-a|<\delta\Rightarrow |f(x)-L|\geq\epsilon
0%
0<|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon
0%
a < x < a+\delta\Rightarrow f(x) - L <\epsilon
0%
a-\delta < x < a\Rightarrow |f(x)-L| < \epsilon
Explanation
It is fundamental concept that, for limit of a function
f(x)
to exist at any point
a
there exists a real number
\delta>0
, such that
0< |x-a|<\delta
, for which
|f(x)-L| < \epsilon
, where
\epsilon >0
\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos x}{x\log(1+x)}=
Report Question
0%
1
0%
0
0%
-1
0%
\displaystyle \dfrac{1}{2}
Explanation
\displaystyle \lim_{x\rightarrow 0}\dfrac{1-\cos x}{x\log(1+x)}
\displaystyle\lim _{ x\rightarrow 0 } \displaystyle \dfrac { 2\sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ x\log (1+x) }
=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { 2\displaystyle\dfrac { \sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ { \left(\displaystyle\dfrac { x }{ 2 } \right) }^{ 2 } } \times \dfrac { 1 }{ 4 } }{\displaystyle \dfrac { 1 }{ x } \log (1+x) }
=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { \displaystyle\dfrac { 1 }{ 2 } }{ \dfrac{\log{ (1+x) }}{ x } } =\dfrac{1}{2}
\displaystyle \lim_{x\rightarrow 0}\frac{(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 \rightarrow 0 \dfrac{(1-{e}^{x})}{x}) \times (\dfrac{sin x}{x})( lim _{ x \rightarrow 0} \dfrac{1}{1+x} )
Now using,
lim
x \rightarrow 0 \dfrac{({e}^{x} - 1)}{x} = 1
lim
x \rightarrow 0 \dfrac{sin x}{x}
= 1
We get,
(-1)
\times 1 \times lim x \rightarrow 0 \dfrac{1}{1+x}
= -1
lf
f(x)=2x-3,a=2,l=1
and
\epsilon =0.001
then
\delta>0
satisfying
0<|x-a|<\delta, \ \ |f(x)-l|<\epsilon
, is:
Report Question
0%
0.005
0%
0.0005
0%
0.001
0%
0.0001
Explanation
|f(x)-l|<0.001=\epsilon
\Rightarrow |2x-3-1| < 0.001
\Rightarrow -0.001<2x-4<0.001
\Rightarrow -0.0005<x-2<0.0005
\Rightarrow |x-2|<0.0005
\Rightarrow |x-a|<0.0005=\delta
Hence,
\delta = 0.0005 > 0
\displaystyle \lim_{x\rightarrow 0}\frac{e^{x}-e^{\sin x}}{2(x-\sin x)}=
Report Question
0%
-\dfrac{1}{2}
0%
\dfrac{1}{2}
0%
1
0%
\dfrac{3}{2}
Explanation
We simplify the given expression as,
\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}({e}^{x - sin x} - 1)}{2(x - sin x)}
Let
x - sin x = y
As
x \rightarrow 0 \text{ so does y } \rightarrow 0
Hence, the question transforms into
(\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}}{2}) \times (\lim _{y \rightarrow 0}\dfrac{{e}^{y} - 1}{y})
= \dfrac{1}{2} \times 1
= \dfrac{1}{2}
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}
=
Report Question
0%
2
0%
-2
0%
\displaystyle \frac{1}{2}
0%
-\displaystyle \frac{1}{2}
Explanation
\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}
\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { x\dfrac { 2\tan { x } }{ 1-\tan ^{ 2 }{ x } } -2x\tan x }{\left(1-\dfrac { 1-\tan ^{ 2 }{ x } }{ 1+\tan ^{ 2 }{ x } } \right)^{ 2 } } }
\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x\tan ^{ 3 }{ x } }{ 4\tan ^{ 4 }{ x } }}
\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x }{ 4\tan { x } } }
=\displaystyle \dfrac { 1 }{ 2 }
lf
\displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right)
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\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right)
As
{ x\rightarrow 0 }
, denominator tends to 0, so the numerator also tends to 0.
\Rightarrow \lim _{ x\rightarrow 0 } \cos 4x+a\cos 2x+b=0
at
x = 0 \Rightarrow \cos 4x = 1, \cos 2x = 1
\Rightarrow 1+a+b=0
\Rightarrow a+b=-1
Now, put
a = -4
\Rightarrow b = 3
Option C satisfies above equation.
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}
=
Report Question
0%
\displaystyle \frac{1}{2}
0%
\displaystyle \frac{3}{2}
0%
\displaystyle \frac{3}{4}
0%
\displaystyle \frac{1}{4}
Explanation
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}=\lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos x)(1+\cos x+\cos^2x)}{x\sin 2x}
=\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{2\sin^2\frac{x}{2}(1+\cos x+\cos^2x)}{4x\cos x\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\lim_{x\rightarrow 0}\displaystyle \dfrac{\sin\dfrac{x}{2}(1+\cos x+\cos^2x)}{4 \times \dfrac x2\cos \tfrac x2 \times \cos x } =\frac{3}{4}
...... As
\left \{ \lim_{\tfrac x2 \to0} \dfrac {\sin \tfrac x2}{\tfrac x2}=0 \right \}
\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \dfrac{(\dfrac{\pi}{2}-x)\sec x}{cosecx}
=
Report Question
0%
1
0%
0
0%
-1
0%
\displaystyle \dfrac{1}{\pi}
Explanation
\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { (\dfrac { \pi }{ 2 } -x)\sec x }{ \csc { x } }
\displaystyle=\lim _{ h\rightarrow 0 } \dfrac { (\dfrac { \pi }{ 2 } -\dfrac { \pi }{ 2 } +h)\sec { (\dfrac { \pi }{ 2 } -h) } }{ \csc { (\dfrac { \pi }{ 2 } -h } ) }
\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\csc { h } }{ \sec { h } }
\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\cos { h } }{ \sin { h } } =1
Solve :
\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}
Report Question
0%
\displaystyle \frac{3}{4}
0%
\displaystyle \frac{1}{4}
0%
\displaystyle \frac{4}{3}
0%
0
Explanation
Solve :
\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}
=\underset{x\rightarrow0}\lim\dfrac{\sin x}{x}\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}+x)\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}-x)
=1\times\sin\dfrac{\pi}{3}\times\sin\dfrac{\pi}{3}
=1\times\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}
=\dfrac34
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}=
Report Question
0%
\dfrac{10}{3}
0%
\dfrac{3}{10}
0%
\dfrac{6}{5}
0%
\dfrac{5}{6}
Explanation
Using,
1 - cos 2x = 2{sin}^{2} x
,
The expression transforms to,
lim _{ x \rightarrow 0 } \dfrac{ 2{sin}^{2} xsin 5x}{{x}^{2}sin 3x}
Rewriting the expression in a different form,
lim _{ x \rightarrow 0 } \dfrac{2{sin}^{2} x}{{x}^{2}} \times \dfrac{sin 5x}{5x} \times \dfrac{3x}{sin 3x} \times \dfrac{5}{3}
Therefore, the limit to the expression is
\dfrac{10}{3}
Hence, option 'A' is correct.
The value of
\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] }
is:
Report Question
0%
0
0%
1
0%
\displaystyle \frac { \sin { \beta } }{ \beta }
0%
\displaystyle \frac { \sin { 2\beta } }{ 2\beta }
Explanation
\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } \sin { \left( \alpha +\beta \right) } }{ \left( \alpha -\beta \right) \left( \alpha +\beta \right) } \right] }
\displaystyle =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } \times \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha +\beta \right) } }{ \left( \alpha +\beta \right) } \right] }
\displaystyle =\lim _{ \alpha -\beta \rightarrow 0 }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } .\left( \frac { \sin { 2\beta } }{ 2\beta } \right)
.......
[\because \alpha \rightarrow \beta \implies \alpha - \beta \rightarrow 0]
\displaystyle =1.\frac { \sin { 2\beta } }{ 2\beta }
.....
[\because \displaystyle \lim_{x\rightarrow 0} \dfrac{\sin x}{x}=1]
=\dfrac { \sin { 2\beta } }{ 2\beta }
Evaluate:
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}
Report Question
0%
\displaystyle \frac{3}{2}
0%
\displaystyle \frac{2}{3}
0%
\displaystyle \frac{1}{3}
0%
\displaystyle \frac{3}{4}
Explanation
Given,
\lim_{x\rightarrow 0}\dfrac{\sec 4x-\sec 2x}{\sec 3x-\sec x}
\sec x=\dfrac{1}{\cos x}
=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{\cos 4x}-\dfrac{1}{\cos 2x}}{\dfrac{1}{\cos 3x}-\dfrac{1}{\cos x}}
=\lim_{x\rightarrow 0}\dfrac{\cos 2x-\cos 4x}{\cos x-\cos 3x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}
=\lim_{x\rightarrow 0}\dfrac{-2\sin 3x \sin (-x)}{-2\sin (2x)\sin (-x)}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}
=\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin 3x}{3x}\times 3x}{\dfrac{\sin 2x}{2x}\times 2x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}
as
\cos 0=1
and
\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1
=\dfrac{3x}{2x}=\dfrac{3}{2}
\displaystyle \lim_{x \rightarrow\frac{\pi}{2}}\displaystyle \dfrac{cosecx-\cot x}{x}
=
Report Question
0%
1
0%
\displaystyle \frac{2}{\pi}
0%
\displaystyle \frac{1}{2}
0%
\displaystyle \frac{1}{5}
Explanation
\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { \csc { x } -\cot x }{ x }
\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { 1-\cos x }{ x\sin { x } }
\displaystyle =\dfrac{2}{\pi}
\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \frac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}
=
Report Question
0%
\sqrt{2}
0%
\displaystyle \frac{1}{\sqrt{2}}
0%
-\sqrt{2}
0%
\displaystyle \frac{-1}{\sqrt{2}}
Explanation
\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \dfrac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}
\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 4 } } \dfrac { \sec x.\dfrac { \tan (4x-\pi ) }{ (4x-\pi ) } }{ \dfrac { \sin (4x-\pi ) }{ (4x-\pi ) } }
=\sqrt { 2 }
\displaystyle \lim_{x\rightarrow \frac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=
Report Question
0%
\sqrt{3}
0%
\dfrac{1}{\sqrt{3}}
0%
-\sqrt{3}
0%
-\dfrac{1}{\sqrt{3}}
Explanation
\displaystyle \lim_{x\rightarrow \dfrac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=2\sqrt{3}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\dfrac{\sqrt{3}}{2}\sin x-\dfrac{1}{2}\cos x}{6x-\pi }
=\displaystyle \dfrac{2\sqrt{3}}{6}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\cos\dfrac{\pi}{6}\sin x-\sin\dfrac{\pi}{6}\cos x}{x-\dfrac{\pi}{6} }
=\displaystyle \dfrac{1}{\sqrt{3}}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\sin \left(x-\dfrac{\pi}{6}\right)}{\left(x-\dfrac{\pi}{6} \right)}=\dfrac{1}{\sqrt{3}}
\displaystyle \lim_{x\rightarrow 0}\frac{tan x^{0}}{x}=
Report Question
0%
0
0%
1
0%
\displaystyle \frac{\pi}{180}
0%
\pi
Explanation
\displaystyle \lim_{x\rightarrow 0}\dfrac{tan x^{0}}{x}
\displaystyle={ \dfrac { \pi }{ 180 } }\lim _{ x\rightarrow 0 } \dfrac { \tan { (\dfrac { \pi }{ 180 } x) } }{ { (\dfrac { \pi }{ 180 } ) }x }
\displaystyle={ \dfrac { \pi }{ 180 } }
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}
=
Report Question
0%
0
0%
1
0%
(\displaystyle \frac{\pi}{180})^{3}
0%
4.(\displaystyle \frac{\pi}{180})^{3}
Explanation
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}
\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ x } }{ x^{ 3 } }
\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 } }
\displaystyle ={ (\frac { \pi }{ 180 } ) }^{ 3 }\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 }{ (\frac { \pi }{ 180 } ) }^{ 3 } }
\displaystyle =4{ (\frac { \pi }{ 180 } ) }^{ 3 }
Solve:
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\tan x-\tan 3x}{2x^{3}}
Report Question
0%
\dfrac{1}{4}
0%
\dfrac{3}{4}
0%
4
0%
-4
Explanation
\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{3\tan x-\tan 3x}{2x^{3}}
\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{3(x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+......)-(3x+\dfrac{(3x)^3}{3}+\dfrac{2(3x)^5}{15}+....)}{2x^{3}}=\dfrac{1-9}{2}=-4
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}
=
Report Question
0%
\displaystyle \log(\frac{a}{b})
0%
\displaystyle \log(\frac{b}{a})
0%
\log(ab)
0%
{\log_{b}}{a}
Explanation
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}
=\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-1-b^{x}+1}{{x}}
=\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(a^{x}-1)-(b^{x}-1)}{{x}}
=\log a - \log b
......... Using formula
=\log\left(\dfrac{a}{b}\right)
Hence,
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}=\log\left(\dfrac{a}{b}\right)
\displaystyle \lim_{x\rightarrow 0}(\frac{\sin x-x}{x})(\sin\frac{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
sin x = x -\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}....(1)
sin\ x-x =-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\dfrac{x^{7}}{7!}
\dfrac{sin\ x-x}{x} =-\dfrac{x^{2}}{3!}+\dfrac{x^{4}}{5!}-\dfrac{x^{6}}{7!}....=0\ (as\ x\ found\ to\ 0)
sin
(\dfrac{1}{x})
is an oscillatory function means same finite no.
0\times finite\ no.=0
\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}
=
Report Question
0%
2
0%
1
0%
0
0%
3
Explanation
\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}
=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{1-\tan x}{(\frac{\pi}{4}-x)(1+\tan x)}
.................(dividing the num and deno by cosx)
=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\tan\left(\frac{\pi}{4}-x\right) }{(\frac{\pi}{4}-x)} = \lim_{h\to 0}\frac{\tan h}{h} = 1
\displaystyle \lim_{x\rightarrow \displaystyle \frac{\pi }{2}}\displaystyle \frac{1-\sin\theta}{\cos\theta\left(\dfrac{\pi}{2}-{\theta}\right)}=
Report Question
0%
1
0%
-1
0%
-\displaystyle \frac{1}{2}
0%
\displaystyle \frac{1}{2}
Explanation
Put\ \theta = \dfrac{\pi }{2}-\phi
\begin{matrix}Lim\\\phi \rightarrow 0 \end{matrix}\ \dfrac{1-cos\phi}{\dfrac{sin\phi}{\phi}\phi ^{2}}=\dfrac{1}{2}
\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}
=
Report Question
0%
\displaystyle \frac{1}{3}
0%
\displaystyle \frac{1}{4}
0%
\displaystyle \frac{1}{6}
0%
\displaystyle \frac{1}{8}
Explanation
\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}
\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\sin (\dfrac { \pi }{ 2 } -h) }{ (\pi -2(\dfrac { \pi }{ 2 } -h))^{ 2 } }
......(x-->h-pi/2)
\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\cos { h } }{ 4h^{ 2 } }
\displaystyle =\dfrac { 1 }{ 8 }
\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}=
Report Question
0%
1
0%
-1
0%
\dfrac{1}{3}
0%
0
Explanation
\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}
=\displaystyle \lim_{x\rightarrow \infty }\frac{\displaystyle 1+\frac{\sin x}{x}}{\displaystyle 1+ \frac{\cos x}{x}}
=\displaystyle \frac{\displaystyle 1+\frac{\text{Any finite number between -1 and 1}}{\infty}}{\displaystyle 1+ \frac{\text{Any finite number between -1 and 1}}{\infty}}=\frac{1+0}{1+0}=1
\displaystyle f(x)=\left\{ \begin{matrix} \dfrac { 3 }{ { x }^{ 2 } } \sin { 2{ x }^{ 2 } } ,\; x<0 \\ \dfrac { { x }^{ 2 }+2x+C }{ 1-3{ x }^{ 2 } } ,\; x\ge 0,\; x\neq \frac { 1 }{ \sqrt { 3 } } \\ 0,\; x=\frac { 1 }{ \sqrt { 3 } } \end{matrix} \right.
\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=
Report Question
0%
\frac{C}{4}
0%
2C
0%
\frac{C}{3}
0%
C
Let
f(x)=\begin{cases}x^{2}-1,0 < x < 2\\2x+3,2 \leq x < 3\end{cases},
The quadratic equation whose roots are
\displaystyle \lim_{x\rightarrow 2^{-}}f(x)
and
\displaystyle \lim_{x\rightarrow 2^{+}}f(x)
is
Report Question
0%
x^{2}-10x+21=0
0%
x^{2}-6x+9=0
0%
x^{2}-14x+49=0
0%
x^{2}+6x+9=0
Explanation
For
0 < x < 2
,
f(x) = {x}^{2} - 1
\lim_{ x \rightarrow {2}^{-}} f(x) = 3
For,
2 < x < 3
f(x) = 2x + 3
\lim_{ x \rightarrow {2}^{+}} f(x) = 7
Thus the equation with roots, 3 and 7 is
{x}^{2} - 10x + 21 = 0
The value of
\sqrt{e}
e
upto four decimal places is?
Report Question
0%
1.6484
0%
1.5237
0%
1.2589
0%
1.9190
Explanation
(e)^{\frac {1}{2}}=1+\dfrac {1}{2}+\dfrac {(\dfrac {1}{2})^2}{2!}+\dfrac {(\dfrac {1}{2})^3}{3!}+\dfrac {(\dfrac {1}{2})^4}{4!}
=1+0\cdot 5+0\cdot 125+0\cdot 02083+0\cdot 00260
=1\cdot 6484
If
f(x) = \displaystyle \frac {x^2+6x}{\sin x}
, then
\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=
Report Question
0%
2
0%
4
0%
6
0%
8
Explanation
\lim_{x\rightarrow 0 }^{ - }f\left( x \right) =\dfrac { { x }^{ 2 }+6x }{ \sin x } \\= \lim_{ h\rightarrow 0 } f\left( h \right)
=\dfrac { { h }^{ 2 }+6(0-h) }{ \sin(0-h) }= \lim_{h\rightarrow 0 }f\left( h \right) =\dfrac { { h }-6 }{ -\sin(h)/h } = \dfrac { -6 }{ -1 } =6
Hence, the answer is "C".
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\log(1+ax)-\log(1+bx)}{x}
=
Report Question
0%
a+b
0%
a-b
0%
-(a+b)
0%
ab
Explanation
\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\log(1+ax)-\log(1+bx)}{x}
=\displaystyle a.\lim_{x\to 0}\frac{\log(1+ax)}{ax}-b.\lim_{x\to 0}\frac{\log(1+bx)}{bx}=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%
\displaystyle \frac{f(1)+f(3)}{3}
0%
20
0%
10
Explanation
Since
f
is continuous, so it must take all real values between
f\left(1\right)
and
f\left(3\right)
.
But since
f
takes only rational values so
f
must be a constant function.
Hence
f\left(1.5\right)=f\left(2\right)=10
Evaluate:
\displaystyle \underset{x\rightarrow 0}{\lim}\ \ \frac{\sin3x^{2}}{\cos(2x^{2}-x)}
Report Question
0%
0
0%
-1
0%
4
0%
-6
Explanation
Given,
\lim _{x\to \:0}\left(\dfrac{\sin \left(3\right)x^2}{\cos \left(2x^2-x\right)}\right)
=\dfrac{\sin \left(3\right)\cdot \:0^2}{\cos \left(2\cdot \:0^2-0\right)}
=0
\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { 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
\displaystyle \lim_{x\rightarrow { 0^- }} \dfrac { 3\sin({ 2x }^{ 2 }) }{ { x }^{ 2 } } =\lim_{x\rightarrow { 0^- }}\dfrac { 2*3\sin({ 2x }^{ 2 }) }{ { 2x }^{ 2 } } = 6
Hence, answer is option
C
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
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
1
Not Answered
49
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page