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## TIFR 2014 Problem 14 Solution – Cardinality of Product of Subgroups TIFR 2014 Problem 14 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
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## Problem:True/False?

Let (G) be a group and (H,K) be two subgroups of (G). If both (H) and (K) has 12 elements, then which of the following numbers cannot be the cardinality of the set (HK={hk|h\in H , k\in K})

A. 72

B. 60

C. 48

D. 36

## Discussion:

We have (|H|=|K|=12).

We know that (|HK|=\frac{|H||K|}{|H\cap K|}).(…*)

Or, in other words (|HK||H\cap K|=|H||K|).

So, at-least we expect to have (|HK|) divides (|H||K|=12^2=144).

Here, (72,48,36) all divide (144) but (60) does not divide (144) therefore (|HK|) can not be (60).

Now, the question still remains whether there exists subgroups which give rise to (|HK|=72,48,36). The answer is yes they do exist. And this is in fact given by the formula (*) above. All we need to do is take two subgroups which have only (\frac{144}{72},\frac{144}{48},\frac{144}{36}) elements common respectively.

For example take (H=D_{2.6}) and (K={1,s}\times\mathbb{Z/6Z}) where (s) is the reflection (element of order 2) and we then get example of (|HK|=72). Here we considered (D_{12}) as (D_{12}\times{\bar{0}}). The intersection is ({1,s}\times{\bar{0}}) which has cardinality 2.

Take (H=A_4) and (K={(1),(12)(34),(13)(24),(14)(23)}\times \mathbb{Z/3Z}). Then we get example of (|HK|=36). Here we considered (A_4) as (A_4\times{\bar{0}}). The intersection is ({(1),(12)(34),(13)(24),(14)(23)}\times {\bar{0}}) which has cardinality 4.

For the same (H) taking (K={(1),(123),(132)}\times \mathbb{Z/4Z}) we get (|HK|=48).

## Helpdesk

• What is this topic: Abstract Algebra
• What are some of the associated concept: Finite Order,Order of Subgroup
• Book Suggestions: Contemporary Abstract Algebra by Joseph A. Gallian
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## TIFR 2014 Problem 13 Solution – Supremum of Function TIFR 2014 Problem 13 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

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## Problem:

Let (S) be the set of all tuples ((x,y)) with (x,y) non-negative real numbers satisfying (x+y=2n) ,for a fixed (n\in\mathbb{N}). Then the supremum value of (x^2y^2(x^2+y^2)) on the set (S) is:

A. (3n^6)

B. (2n^6)

C. (4n^6)

D. (n^6)

## Discussion:

Write the expression in terms of (x) only by substituting (y=2n-x).

Let (f(x)=x^2(2n-x)^2(x^2+(2n-x)^2)). Here, (x\in[0,2n]).

Note that for (x=0) or (x=2n) the function (f(x)=0). Also, (f) is positive everywhere else on the interval.

So we want to find (sup{f(x)|x\in (0,2n)}). Note that it exists because the interval is compact and (f) is continuous.

One can straightaway take derivative and compute, or one can do the following:

Take log. Note that now we are only working on the open interval ((0,2n)).

(log(f(x))=2logx+2log(2n-x)+log(x^2+(2n-x)^2))

Now take derivative.

(\frac{f'(x)}{f(x)}=\frac{2}{x}+\frac{-2}{2n-x}+\frac{2x-2(2n-x)}{x^2+(2n-x)^2})

(=\frac{4n-4x}{x(2n-x)}+\frac{4n-4x}{x^2+(2n-x)^2})

(=4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]).

Now, (f'(x)=0) if and only if (4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]=0).

Note that ([\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]>0) for all (x\in (0,2n)).

Therefore, (f'(x)=0) if and only if (x=n). If now, (x) is slightly bigger than (n) then (\frac{f'(x)}{f(x)}<0) and since (f(x)>0) we have (f'(x)<0) in that case. And if (x) is slightly smaller than (n) then (f'(x)>0).

This proves that indeed the point (x=n) is a point of maxima.

Therefore, the supremum value is (f(n)=2n^6). So the correct answer is option B.

## Chatuspathi

• What is this topic: Real Analysis
• What are some of the associated concept: Supremum Property, Maxima-Minima
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert
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## TIFR 2014 Problem 12 Solution – Mapping Properties TIFR 2014 Problem 12 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

There exists a map (f:\mathbb{Z} \to \mathbb{Q}) such that (f)

A. is bijective and increasing

B. is onto and decreasing

C. is bijective and satisfies (f(n)\ge 0) if (n\le 0)

D. has uncountable image.

## Discussion:

Option D is out of the question right away. Because the co-domain (\mathbb{Q}) is countable, and the image set is a subset of the co-domain set, any subset of countable set is countable, so image set has to be countable.

Let’s assume (f) is onto and decreasing. Then (…\ge f(-1) \ge f(0) \ge f(1) \ge f(2) \ge … ).

At this point, we don’t know for sure whether (f(0)=f(1)) or not. But, for the map to be onto, we must have strict inequality somewhere in the above chain of inequalities.

Let’s say (f(a)>f(a+1)). Then we ask what is the pre-image of the rational number (\frac{f(a)+f(a+1)}{2})?

Let (f(n)=\frac{f(a)+f(a+1)}{2})

Since, (f(n)=\frac{f(a)+f(a+1)}{2}>f(a)) and (f) is decreasing, (a>n).

Also, since (f(n)=\frac{f(a)+f(a+1)}{2}<f(a+1)) and (f) is decreasing, (n>a+1).

The two inequalities (a>n) and (n>a+1) together give a contradiction.

Therefore, option B is false.

What did we use here, only onto-ness and that f is decreasing. If instead our function (f) was bijective and increasing then we will get a similar kind of contradiction:

Suppose (f) is bijective and increasing. Then (f(0)<f(1)) (One-to-one ness guarantees the strict inequality)

We have (\frac{f(0)+f(1)}{2}\in\mathbb{Q}).

Therefore there exists (m\in\mathbb{Z}) such that ( f(m)= \frac{f(0)+f(1)}{2} ).

We then have (f(0)<f(m)<f(1)) which means (0<m<1), a contradiction because there is no natural number in between 0 and 1.

So option A is false.

We know that (\mathbb{Q}) does have a bijection with (\mathbb{Z}), in other words (\mathbb{Q}) is countable.

Now, both the set (A={n\in \mathbb{Z}| n\le 0 }) and (B={q\in \mathbb{Q}| q\ge 0}) are countably infinite, therefore has a bijection between them. To see this, let (f_1:A\to \mathbb{Z}) and (f_2:B\to \mathbb{Z}) be bijections. Then (f_2^{-1}o f_1:A \to B) is a bijection.

Similarly (C={n\in \mathbb{Z}| n> 0 }) has a bijection with (D={q\in \mathbb{Q}| q< 0}).

Now, we have (h:A\to B) and (g:C\to D) two bijections.

Then define (f(n)=h(n)) for (n\in A) and (f(n)=g(n)) for (n\in C) to get a bijection from (\mathbb{Z}) to (\mathbb{Q}). This is true because (A \cup B= \mathbb{Z}) and (C \cup D= \mathbb{Q}).

And this (f) satisfies the condition of option C: (f) is bijective and satisfies (f(n)\ge 0) if (n\le 0)

Therefore, option C is true.

## Chatuspathi

• What is this topic: Real Analysis
• What are some of the associated concept: Bijective Mapping
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert
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## TIFR 2014 Problem 11 Solution – Nilpotent Matrix Eigenvalues TIFR 2014 Problem 11 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Linear Algebra by Gilbert Strang. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (A) be an (nxn) matrix with real entries such that (A^k=0) (0-matrix) for some (k\in\mathbb{N}).

Then

A. A has to be the 0-matrix.

B. trace(A) could be non-zero.

C. A is diagonalizable.

D. 0 is the only eigenvalue of A

## Discussion:

Let (v) be an eigenvector of (A) with eigenvalue (\lambda).

Then (v \neq 0) and (Av=\lambda v).

Again, (A^2 v=A(Av)=A(\lambda v)=\lambda Av= (\lambda)^2v).

We continue to apply A, applying it k times gives: (A^k v=(\lambda)^k v).

By given information, the left hand side of the above equality is 0.

So (\lambda^k v=0) and remember (v \neq 0).

So (\lambda =0).

Therefore (0) is the only eigenvalue for (A).

So D is true.

We analyse the question a little bit further, to check it satisfies no other options above.

We know (trace(A)=) sum of eigenvalues of A= (\sum 0 =0)

So option B is false.

Take (A=\begin{bmatrix} 0 & 1 // 0 & 0 \end{bmatrix} ).

Then (A^2 =0). But (A) is not the zero matrix.

Also, if (A) were diagonalizable then the corresponding diagonal matrix would be the zero matrix. Which would then imply that (A) is the zero matrix, which in this case it is not. (See  TIFR 2013 Probmem 8 Solution-Diagonalizable Nilpotent Matrix ) So this disproves options A and C.

## Helpdesk

• What is this topic: Linear Algebra
• What are some of the associated concept: Eigenvectors,Characteristic Polynomial
• Book Suggestions: Introduction to Linear Algebra by Gilbert Strang
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## Theory of Equation (TIFR 2014 problem 10)

Question:

Let (C \subset \mathbb{ZxZ} ) be the set of integer pairs ((a,b)) for which the three complex roots (r_1,r_2,r_3) of the polynomial (p(x)=x^3-2x^2+ax-b) satisfy (r_1^3+r_2^3+r_3^3=0). Then the cardinality of (C) is

A. (\infty)

B. 0

C. 1

D. (1<|C|<\infty)

Discussion:

We have (r_1+r_2+r_3=-(-2)=2)

(r_1r_2+r_2r_3+r_3r_1=a) and

(r_1r_2r_3=-(-b)=b)

Also, of the top of our head, we can think of one identity involving the quantities (r_1^3+r_2^3+r_3^3=0) and the three mentioned just above.

Let’s apply that:

(r_1^3+r_2^3+r_3^3-3r_1r_2r_3=(r_1+r_2+r_3)(r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1))) …(…(1))

Also, note that (r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1) )

So (r_1^2+r_2^2+r_3^2=(2)^2-2(a)=4-2a ).

So by equation ((1)),

(0-3b=2(4-2a-a))

or, (6a-3b=8).

Now, we notice the strangest thing about the above equation: (3|(6a-3b)) but (3) does not divide (8).

Why did this contradiction occur? We didn’t start off by saying “assume that … something …”. Well, even though we pretend to not assume anything, we did assume that this equation has a solution for some (a,b). So, the

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## TIFR 2014 Problem 9 Solution – Eigenvalues of Rotation TIFR 2014 Problem 9 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Linear Algebra by Gilbert Strang. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (A(\theta)=\begin{pmatrix} cos(\theta) &  sin(\theta) \ -sin(\theta) & cos(\theta) \end{pmatrix} )

where (\theta \in (0,2\pi) ). Mark the correct statement below:

A. (A(\theta)) has eigenvectors in (\mathbb{R}^2) for all (\theta \in (0,2\pi))

B.  (A(\theta)) does not have eigenvectors in (\mathbb{R}^2) for any (\theta \in (0,2\pi))

C.  (A(\theta)) has eigenvectors in (\mathbb{R}^2) for exactly one values of  (\theta \in (0,2\pi))

D. (A(\theta)) has eigenvectors in (\mathbb{R}^2) for exactly 2 values of (\theta \in (0,2\pi))

## Discussion:

If we try to write the linear operator on (\mathbb{R}^2) which is the clockwise rotation by angle (\theta) then we see that with respect to standard basis, the matrix is (A(\theta)). This is quite easy to see, for ((1,0) \to (cos(\theta),-sin(\theta)) ), and (…) (draw the picture if you are not convinced about this).

What does having an eigenvector mean in this context? Well, as always, it means that (Av=\lambda v) for some (v \neq 0) in (\mathbb{R}^2). But geometrically, this means that upon applying the transformation (in this case rotation) we get the resulting vector in the spanning space of (v), that is the resulting vector must be in the line passing through (v) and the origin.

Geometrically, after rotation, the vector is in the same line is possible only when either the angle of rotation is (0) or the angle of rotation is ( \pi). That is either we did not move at all or we basically reflected about origin.

(\theta \neq 0), so we are left with (\theta=\pi). Therefore, we know option C is true.

If you do not trust your geometric sense at all, then you would want to look at the characteristic polynomial.

We have (det(A(\theta))=cos^2(\theta)+sin^2(\theta)=1) and (trace(A(\theta))=2cos(\theta)).

Therefore, the characteristic polynomial is (x^2-2cos(\theta)x+1). This has a real root if the discriminant is positive. That is if (4cos^2(\theta)-4 \ge 0) i.e, (cos^2(\theta) \ge 1). We know that the maximum value of cos is 1 so (cos(\theta)=1) or (cos(\theta)=-1). Since (\theta \ne 0) we are forced to conclude (\theta = \pi) and we have the answer once more.

## Helpdesk

• What is this topic: Linear Algebra
• What are some of the associated concept: Eigenvectors,Characteristic Polynomial
• Book Suggestions: Linear Algebra done Right by Sheldon Axler
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## TIFR 2014 Problem 8 Solution -Checking one-one/onto-ness TIFR 2014 Problem 8 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:True/False

Let (f:\mathbb{R} \to \mathbb{R}) be a continuous function such that (|f(x)-f(y)| \ge |x-y| ), for all (x,y \in \mathbb{R}). Then

A. f is both one-one and onto.

B. f is one-one and may be onto.

C. f is onto but may not be one-one.

D. f is neither one-one nor onto.

## Discussion:

Let (f(x)=f(y)) for some (x,y\in \mathbb{R}). Then from the given inequality, we get (0 \ge |x-y| ) which is saying, (|x-y|=0 ) (since modulus can take non-negative values only) and that implies (x=y). So (f) is one-one.

From the inequality, we can see that (f) increases in a steady rate, we want to see whether it is onto or not.

We have (|f(x)-f(0)| \ge |x-0| = |x| ).

If now, (|f(x)| \le M ) then we will end up having (|f(x)-f(0)| \le |f(x)|+|f(0)| \le M+|f(0)| ) which is a contradiction.

So, (|f(x)|) is not bounded.

Already, (f) is continuous and one-one, so f must be increasing or decreasing (strictly). And since (|f|) is not bounded above, we use intermediate value theorem to conclude that (f) must be onto.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuous function, One-One Function
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert
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## TIFR 2014 Problem 7 Solution -Limit of Integration TIFR 2014 Problem 7 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Real and Complex Analysis by Walter Rudin. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem

Let (f_n(x); n\ge 1) be a sequence of continuous nonnegative functions on ([0,1]) such that

( \lim_{n\to\infty} \int_{0}^{1}f_n(x)dx = 0 )

Which of the following statements is always correct?

A. (f_n \to 0) uniformly of ([0,1])

B. (f_n) may not converge uniformly but converges to (0) point-wise.

C. (f_n) will converge point-wise and the limit may be non-zero.

D. (f_n) is not guaranteed to have a point-wise limit.

## Discussion:

We start by a very well known example: (g_n(x)=x^n).

(  \int_{0}^{1}g_n(x)dx= \frac{1}{n+1}\to 0 \int_{0}^{1}f_n(x)dx ) as (n\to \infty).

We know (g_n) does not converge uniformly on ([0,1]) because the limit is (1) at (x=1) and (0) everywhere else so we have a non-continuous limit.

So straight-away A,B are false. Question is now whether at all the sequence has to have a point-wise limit or not.

For this, we take our hint from (g_n) and construct (f_n(x)= \sqrt{n}x^n ).

Then ( \int_{0}^{1}f_n(x)dx= \sqrt{n}\frac{1}{n+1}\to 0 ) as ( n\to \infty).

But look at (f_n(1)= \sqrt{n} ). Therefore, (f_n) does not converge at the point (x=1).

So option (D) i.e., (f_n) is not guaranteed to have a point-wise limit. is true.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Point wise limit, convergence, Uniformly Continuous
• Book Suggestions: Real and Complex Analysis by Walter Rudin
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## TIFR 2014 problem 6 Solution – Inequality in Integration TIFR 2014 Problem 6 Solution is a part of TIFR entrance preparation series offered by Cheenta. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects. Generally, the exams scheduled in December.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem

Let (f:[0,1] \to \mathbb{R} ) be a continuous function. Which of the following statement is always true?

A. ( \int_{0}^{1} f^2(x)dx = (\int_{0}^{1}f(x)dx)^2 )

B.  ( \int_{0}^{1} f^2(x)dx \le (\int_{0}^{1}|f(x)|dx)^2 )

C.  ( \int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2 )

D.  ( \int_{0}^{1} f^2(x)dx < (\int_{0}^{1}f(x)dx)^2 )

## Discussion:

We first recall the Cauchy-Schwartz inequality for an inner product space (V) and two vectors (a,b\in V)

( <a,b> \le ||a||||b|| ).

Here, we also remember the fact that (C[0,1]) (the set of all continuous real (/complex) valued functions on [0,1] ) forms an inner product space with respect to the inner product

( <f,g>=  \int_{0}^{1} f(x)g(x)dx ) (We are only taking real valued functions so we neglect the conjugation…)

We want to apply this inequality to suitable functions so that we get some inequality from the options above.

Let’s try (|f|) and (1) (the constant function).

We get: ( \int_{0}^{1} |f(x)|1dx \le (\int_{0}^{1} |f(x)|^2dx)^{1/2} (\int_{0}^{1} 1^2dx)^{1/2} )

Squaring, we get

( \int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2 ).

So option C ( \int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2 ) is correct.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Inner Product Space, Cauchy – Schwartz inequality,continuous real (/complex) valued functions
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert
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## TIFR 2014 Problem 5 Solution -Convergence of sequence TIFR 2014 Problem 5 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (a_n= (n+1)^{100}e^{-\sqrt{n}}) for (n \ge 1). Then the sequence (a_n) is

A. unbounded

B. bounded but not convergent

C. bounded and converges to 1

D. bounded and converges to 0

## Discussion:

There can be various ways to do this. One way would be to check limit of

(f(x)=(x+1)^{100}e^{-\sqrt{x}}) as (x\to \infty) using L’Hospital rule.

One other way would be to use some inequalities involving (e^{-\sqrt{n}}) and ((n+1)^{100}).

We use the first approach.

( \lim_{x\to\infty}f(x)) is in (\infty /\infty) form.

We apply L’Hospital rule.

The derivative of (e^{\sqrt x} ) is (e^{\sqrt x}\frac{1}{2\sqrt x })

The derivative of ((x+1)^{100}) is (100(x+1)^{99}).

So ( \lim_{x\to\infty}f(x) = lim \frac{200(x+1)^{99}x^{1/2} }{e^{\sqrt x} } )

This is again in the (\infty / \infty ) form. So we continue this process.

Note that:

• At each step, the power of x in the numerator decreases by 1/2,
• and the $e^{\sqrt x }$ remains as it is in the denominator.

Thus, after finitely many steps, we are left with an x with a power of 0 in the numerator. That is after finitely many steps, the numerator will become constant, while the denominator is still (e^{\sqrt x } ).

Thus the limit becomes 0.

So we conclude that the given sequence (a_n) converges to 0. Also, since any convergent sequence is bounded, this sequence is bounded. So option D is correct.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: L’Hospital rule, Convergent Sequence, Bounded Sequence
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert