IISER Mathematics - Limits
Exam Duration: 45 Mins Total Questions : 30
\(\underset{x\rightarrow 0}{lim}{tan 2x-x\over3x-sinx}\)is equal to
- (a)
2
- (b)
1/2
- (c)
\(-1\over2\)
- (d)
\(1\over4\)
\(\underset{x \rightarrow \infty}{lim}[{2n\over2n^2-1}.cos{n+1\over2n-1}-{n\over1-2n}.{n(-1)^n\over n^2+1}]\) is equal to
- (a)
1
- (b)
-1
- (c)
0
- (d)
2
if \(\underset { x\rightarrow 0 }{ lim } \frac { log(3+x)-log(3-x) }{ x } =k,\)hen the value of k is
- (a)
0
- (b)
\(-1\over3\)
- (c)
\(2\over3\)
- (d)
\(-2\over3\)
if \(f(x)=\begin{cases}x+2, x\le -1 \\ cx^2, x>-1\end{cases},\)then find c, if \(\underset { x\rightarrow -1 }{ lim } f(x)\) exists.
- (a)
-1
- (b)
1
- (c)
0
- (d)
2
\(\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \cot { ^{ -1 } } ({ r }^{ 2 }+3/4) } } \) is
- (a)
0
- (b)
tan-1 1
- (c)
tan-1 2
- (d)
none of these
If \(\lim _{ x\rightarrow \infty }{ \left( 1+\frac { a }{ x } +\frac { b }{ x^{ 2 } } \right) ^{ 2x } } ={ e }^{ 2 }\) , then
- (a)
a=1,b=2
- (b)
a=2,b=1
- (c)
a=1,b∈R
- (d)
none of these
\(\lim _{ x\rightarrow 1+0 }{ \frac { \int _{ 1 }^{ x }{ \left| t-1 \right| dt } }{ sin\quad (x-1) } } \) is equal to
- (a)
0
- (b)
1
- (c)
-1
- (d)
none of these
If x is a real number in [0,1] then the value of \(f(x)=\lim _{ m\rightarrow \infty }{ \lim _{ n\rightarrow \infty }{ \{ 1+{ cos }^{ 2m }(n!\pi x) } \} } \) is given by
- (a)
2 or 1according as x is rational or irrational
- (b)
1 or 2 according as x is rational or irrational
- (c)
1for all x
- (d)
2 or 1for all x
Let \(f(x)=\lim _{ n\rightarrow \infty }{ \frac { { x }^{ 2n }-1 }{ { x }^{ 2n }+1 } } \) then
- (a)
f(x)=1 for |x|>1
- (b)
f(x)=-1 for |x|<1
- (c)
f(x) is not defined for any value of x
- (d)
f(x)=1 for |x|=1
If a>0,b<0,then \(\lim _{ x\rightarrow 0+ }{ \frac { \sqrt { (1-cos\quad 2\quad ax) } }{ sin\quad bx } } \) is equal to
- (a)
\(\frac { a\sqrt { 2 } }{ b } \)
- (b)
\(-\frac { a\sqrt { 2 } }{ b } \)
- (c)
\(\frac { |a|\sqrt { 2 } }{ |b| } \)
- (d)
\(-\frac { |a|\sqrt { 2 } }{ |b| } \)
If \(\lim _{ x\rightarrow { 0 }^{ + } }{ f(x)= } \) finite where \(f\left( x \right) =\frac { \sin { x+{ ae }^{ x }+{ be }^{ -x }+cIn\quad (1+x) } }{ { x }^{ 3 } } \\ \) and a,b,c are real numbers For the same value of a,b,c as obtain above, then the value of \(\lim _{ x\rightarrow { 0 }^{ + } }{ { x }^{ 2 } } f(x)\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
Evaluate of the following limits.
\(\lim _{ x\rightarrow 2 }{ \left[ \frac { { x }^{ 3 }-2x^{ 2 } }{ { x }^{ 2 }-5x+6 } \right] } \)=
- (a)
1
- (b)
-1
- (c)
4
- (d)
-4
Evaluate of the following limits.
\(\lim _{ x\rightarrow 1 }{ \left[ \frac { x-2 }{ { x }^{ 2 }-x } -\frac { 1 }{ { x }^{ 3 }-3x^{ 2 }+2x } \right] } \)=
- (a)
0
- (b)
1
- (c)
2
- (d)
-2
The value of \(\lim _{ x\rightarrow 1 }{ \frac { x^{ 1/4 }-x^{ 1/5 } }{ { x }^{ 3 }-1 } } \) is
- (a)
\(\frac{1}{20}\)
- (b)
\(\frac{1}{40}\)
- (c)
\(\frac{1}{60}\)
- (d)
\(\frac{3}{20}\)
\(\lim _{ x\rightarrow 1 }{ \frac { x+{ x }^{ 2 }+...+{ x }^{ n }-n }{ x-1 } } \) is
- (a)
n
- (b)
\(\frac { n+1 }{ 2 } \)
- (c)
\(\frac { n(n+1) }{ 2 } \)
- (d)
\(\frac { n(n-1) }{ 2 } \)
\(\lim _{ x\rightarrow 0 }{ \frac { \sqrt { 2+x } -\sqrt { 2 } }{ x } } \) is equal to
- (a)
\(2\sqrt { 2 } \)
- (b)
\(\frac { -1 }{ 2\sqrt { 2 } } \)
- (c)
\(\frac { 1 }{ 2\sqrt { 2 } } \)
- (d)
-\(2\sqrt { 2 } \)
\(\lim _{ x\rightarrow \frac { \pi }{ 4 } }{ \frac { cosx-sinx }{ \left( \frac { \pi }{ 4 } -x \right) (cosx+sinx) } } \) is
- (a)
0
- (b)
1
- (c)
-1
- (d)
2
\(\lim _{ x\rightarrow 0 }{ \left( \frac { { 1 }^{ x }+{ 2 }^{ x }+{ 3 }^{ x }+....+{ n }^{ x } }{ n } \right) ^{ \frac { a }{ x } } } \)
- (a)
\((n!)^{ \frac { a }{ n } }\)
- (b)
\((n!)^{ \frac { n }{ a } }\)
- (c)
(n!)an
- (d)
None of these
Compute the dervivative of 6x100-x55+x
- (a)
x99-x54+1
- (b)
100x99-55x4+1
- (c)
600x99-55x54+1
- (d)
None of these
Find the derivative of f(x)=\(\frac { x+1 }{ x } \) .
- (a)
\(\frac{1}{x}\)
- (b)
\(\frac { 1 }{ { x }^{ 2 } } \)
- (c)
0
- (d)
-\(\frac { 1 }{ { x }^{ 2 } } \)
If \(f(x)=1-\sqrt { x } +(1+\sqrt { x } )^{ 2 },f'(1)\) is equal to
- (a)
\(\cfrac { 1 }{ 2 } \)
- (b)
1
- (c)
\(\cfrac { 3 }{ 2 } \)
- (d)
2
If x2+2x+2y2=1, then \(\cfrac { dy }{ dx } \) at the point where y=1 is equal to
- (a)
1
- (b)
0
- (c)
-1
- (d)
-2
Match the following
Column I | Column II |
\(\underset { x\rightarrow 4 }{ lim } \cfrac { { x }^{ 2 }-16 }{ \sqrt { 3x-8 } -\sqrt { x } } =\) | (p) 8/5 |
(ii) \(\underset { x-\sqrt { 2 } }{ lim } \cfrac { { x }^{ 4 }-4 }{ { x }^{ 2 }+3\sqrt { 2x } -8 } =\) | (q) 16 |
(iii) \(\underset { x\rightarrow 1 }{ lim } \cfrac { { x }^{ 7 }-2{ x }^{ 5 }+1 }{ { x }^{ 3 }-{ 3x }^{ 2 }+2 } =\) | (r)1 |
- (a)
(i)\(\rightarrow \)(p),(ii) \(\rightarrow \)(q), (iii)\(\rightarrow \)(r)
- (b)
(i)\(\rightarrow \)(r),(ii)\(\rightarrow \)(q),(iii)\(\rightarrow \)(p)
- (c)
(i)\(\rightarrow \) (r), (ii)\(\rightarrow \) (p), (iii)\(\rightarrow \)(q)
- (d)
(i) \(\rightarrow \)(q), (ii) \(\rightarrow \)(p), (iii) \(\rightarrow \)(r)
\(\underset { x\rightarrow 0 }{ lim } \cfrac { { x }^{ m }-1 }{ { x }^{ 2 }-1 } \) is
- (a)
1
- (b)
\(\cfrac { m }{ n } \)
- (c)
\(-\cfrac { m }{ n } \)
- (d)
\(\cfrac { { m }^{ 2 } }{ { n }^{ 2 } } \)
Statements -I If a1, a2.., an > 0, then
\(\overset { lim }{ x\rightarrow \infty } \left( \frac { { a }_{ 1 }^{ \frac { 1 }{ x } }+{ a }_{ 2 }^{ \frac { 1 }{ x } }+...+{ a }_{ n }^{ \frac { 1 }{ x } } }{ n } \right) ={ a }_{ 1 }{ a }_{ 2 }....{ a }_{ n }\)
Statements -II : \(\overset { lim }{ x\rightarrow \infty } f\left( x \right) =1\), \(\overset { lim }{ x\rightarrow \infty } g\left( x \right) =\infty \), then \(\overset { lim }{ x\rightarrow a } f\left( x \right) ^{ g\left( x \right) }={ e }^{ \overset { lim }{ x\rightarrow a } g(x)f(x)-1 }\)
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement -I.
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement -I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement-I is false and Statement-II is true.
\(\underset { \theta \rightarrow 0 }{ lim } \cfrac { 1-cos4\theta }{ 1-cos6\theta } \) is
- (a)
\(\cfrac { 4 }{ 9 } \)
- (b)
\(\cfrac { 1 }{ 2 } \)
- (c)
\(\cfrac {- 1 }{ 2 } \)
- (d)
-1
Statement : \(\overset { lim }{ x\rightarrow 0 } \frac { \sqrt { 1-cos2x } }{ x } \) does not exist
Statemnet II : |sin x| = { \(\begin{matrix} sinx,0<x<\pi /2 \\ -\pi /2<x<0 \end{matrix}\)
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement -I.
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement -I.
- (c)
If Statement-I is true but Statement-II is false
- (d)
If Statement-I is false and Statement-II is true.
Statement-I: \(\lim _{ x\longrightarrow 0 }{ \frac { { e }^{ 1/x }-1 }{ { e }^{ x }+1 } } \) does not exist.
Statement-II: L.H.L. = 1and R.H.L. = -1
- (a)
Ifboth Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement -I.
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement -I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement-I is false and Statement-II is true.
Let u = f(x) and v = g(x). Then,
Statement-I: (uv)' = u'v + uv' is a Leibnitz rule or product rule.
Statement-II: \({ \left( \frac { u }{ v } \right) }^{ ' }=\frac { uv'-u'v }{ { v }^{ 2 } } \) is a Leibnitz rule or quotient rule
- (a)
Ifboth Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement -I.
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement -I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement-I is false and Statement-II is true.
If y =\(\frac { sin(x+9) }{ cosx } ,\quad then\quad \frac { dy }{ dx } \)at x = 0 is
- (a)
cos 9
- (b)
sin 9
- (c)
0
- (d)
1