ISI MStat PSB 2009 Problem 2 | Linear Difference Equation

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 2 based on Convergence of a sequence. Let's give it a try !!

Problem- ISI MStat PSB 2009 Problem 2

Let $ \{x_{n}: n \geq 0\} $ be a sequence of real numbers such that
$ x_{n+1}=\lambda x_{n}+(1-\lambda) x_{n-1}, n \geq 1,$ for some $ 0<\lambda<1$
(a) Show that $ x_{n}=x_{0}+(x_{1}-x_{0}) \sum_{k=0}^{n-1}(\lambda-1)^{k} $
(b) Hence, or otherwise, show that $ x_{n}$ converges and find the limit.

Prerequisites

Limit

Sequence

Linear Difference Equation

Solution :

(a) We are given that $ x_{n+1}=\lambda x_{n}+(1-\lambda) x_{n-1}, n \geq 1,$ for some $ 0<\lambda<1$

So, $ x_{n+1} - x_{n} = -(1- \lambda)( x_n-x_{n-1}) $ ---- (1)

Again using (1) we have $ ( x_n-x_{n-1})= -(1- \lambda)( x_{n-1}-x_{n-2}) $ .

Now putting this in (1) we have , $ x_{n+1} - x_{n} = {(-(1- \lambda))}^2 ( x_{n-1}-x_{n-2}) $ .

So, proceeding like this we have $ x_{n+1} - x_{n} = {(-(1- \lambda))}^n ( x_{1}-x_{0}) $ for all $ n \geq 1$ and for some $ 0<\lambda<1$---- (2)

So, from (2) we have $ x_{n} - x_{n-1} = {(-(1- \lambda))}^{n-1} ( x_{1}-x_{0}) $ , $ \cdots , (x_2-x_1)=-(\lambda-1)(x_1-x_{0}) $ and $ x_1-x_{0}=x_{1}-x_{0} $

Adding all the above n equation we have $ x_{n}-x_{0}=(x_{1}-x_{0}) \sum_{k=0}^{n-1} {(\lambda-1)}^{k} $

Hence , $ x_{n}=x_{0}+(x_{1}-x_{0}) \sum_{k=0}^{n-1}(\lambda-1)^{k} $ (proved ) .

(b) As we now have an explicit form of $ x_{n}=x_{0}+(x_{1}-x_{0}) \times \frac{1-{( \lambda -1)}^n}{1-(\lambda -1)} $ ----(3)

Hence from (3) we can say $ x_{n} $ is bounded and monotonic ( verify ) so , it's convergent .

Now let's take $ \lim_{n\to\infty} $ both side of (3) we get , $ \lim_{x\to\infty} x_{n} = x_{0}+(x_{1}-x_{0}) \times \frac{1}{2 - \lambda} $ .

Since , $ \lim_{x\to\infty} {( \lambda - 1)}^{n} = 0 $ as , $ -1 < \lambda -1 < 0 $ .

Food For Thought

$\mathrm{m}$ and $\mathrm{k}$ are two natural number and $ a_{1}, a_{2}, \ldots, a_{m}$ and $ b_{1}, b_{2}, \ldots, b_{k}$ are two sets of positive real numbers such that $ a_{1}^{\frac{1}{n}}+a_{2}^{\frac{1}{n}}+\cdots+a_{m}^{\frac{1}{n}} $ = $ b_{1}^{\frac{1}{n}}+\cdots+b_{k}^{\frac{1}{n}} $

for all natural number $ \mathrm{n} .$ Then prove that $ \mathrm{m}=\mathrm{k}$ and $ a_{1} a_{2} \ldots a_{m}=b_{1} b_{2} . . b_{k} $ .

Outstanding Statistics Program with Applications

ISI MStat 2018 PSA Problem 8 | Limit of a Function

This is a beautiful problem from ISI MStat 2018 PSA Problem 8 based on limit of a function. Try yourself and use hints if required.

Limit of a Function - ISI MStat Year 2018 PSA Question 8

The value of $ \lim _{x \rightarrow \infty}(\log x)^{1 / x} $

(A) is e
(B) is 0
(C) is 1
(D) does not exist

Key Concepts

Limit

L'hospital Rule

Check the Answer

Answer: is (C)

ISI MStat 2018 PSA Problem 8

Introduction to Real Analysis by Bertle Sherbert

Try with Hints

What is the form of the limit? Can you convert it to some known limit?

$\infty^0$
$\frac{\infty}{\infty}$
L'Hospital Rule

Let , $ \lim _{x \rightarrow \infty}(\log x)^{1 / x} $ =l (say) then taking log on both sides we get , $ \lim _{x \rightarrow \infty} \frac{ log( logx) }{x} $ =log (l) . Now we will apply L'hospital rule .

Applying L'hospital rule we get , log (l)= $ \lim _{x \rightarrow \infty} \frac{1}{logx x} $= 0 $ \Rightarrow l=e^0 \Rightarrow l=1 $

Hence , option (C) is correct .

Outstanding Statistics Program with Applications

ISI MStat PSA 2019 Problem 17 | Limit of a function

This is a beautiful problem from ISI Mstat 2019 PSA problem 17 based on limit of a function . We provide sequential hints so that you can try this.

Limit of a function

If $ f(a)=2, f'(a)=1 , g(a)=-1$ and $g'(a)=2$ , then what is

$ \lim\limits_{x\to a}\frac{(g(x)f(a) – g(a)f(x))}{(x – a)} $ ?

5
3
- 3
-5

Key Concepts

Algebraic manipulation

Limit form of the Derivative

Check the Answer

Answer: is 5

ISI MStat 2019 PSA Problem 17

Introduction to real analysis Robert G. Bartle, Donald R., Sherbert.

Try with Hints

Try to manipulate $ \frac{(g(x)f(a) – g(a)f(x))}{(x – a)} $ so that you can use the Limit form of the Derivative . Let's give a try .

$ \frac{(g(x)f(a) – g(a)f(x))}{(x – a)} $ =

$ \frac{(g(x)f(a) –g(a)f(a) +g(a)f(a) - g(a)f(x))}{(x – a)} $ =

$ f(a)\frac{g(x)-g(a)}{(x-a)} - g(a)\frac{f(x)-f(a)}{(x-a)} $ .

Now calculate the limit using Limit form of the Derivative.

So, we have $ \lim\limits_{x\to a}\frac{(g(x)f(a) – g(a)f(x)}{(x – a)} $ =

$ \lim\limits_{x\to a} f(a)\frac{g(x)-g(a)}{(x-a)} - \lim\limits_{x\to a} g(a)\frac{f(x)-f(a)}{(x-a)} $ =

$ f(a) g'(a) - g(a)f'(a)= 2.(2)-1.(-1)=5 $.

Outstanding Statistics Program with Applications

Problem on Limit | ISI B.Stat Objective | TOMATO 728

Try this beautiful problem based on Limit, useful for ISI B.Stat Entrance

Problem on Limit | ISI B.Stat TOMATO 728

The limit lim $\int\frac {h}{(h^2 + x^2)}$dx (integration running from $x =-1$to $x = 1$) as$ h \to 0$

equals 0
equals $\pi$
equals $-\pi$
deoes not exist

Key Concepts

Limit

Calculas

trigonometry

Check the Answer

Answer: does not exist

TOMATO, Problem 728

Challenges and Thrills in Pre College Mathematics

Try with Hints

Now, $\int{h}{(h^2 + x^2)}$dx (integration running from $x = -1$ to $x = 1$)
Let, $x$ = h tany

$\Rightarrow dx = h sec^2y dy$
$\Rightarrow $ $x = -1$, $y = -tan^{-1}(1/h)$ and $x = 1$, $y = tan^{-1}(1/h)$

$\Rightarrow \int \frac{h}{(h^2 + x^2)}$dx =$\int \frac{h(hsec^2ydy)}{h^2sec^2y}$ (integration running from$ y = -tan^{-1}(1/h) $ to $y = tan^-1(1/h))$

= y (upper limit =$ tan-1(1/h)$) and lower limit = $-tan^-1(1/h)$

= $2tan^-1(1/h)$

Can you now finish the problem ..........

Now, lim $2tan^-(1/h)$ as$ h \to 0$ doesn‟t exist

ISI MStat 2016 Problem 5 | Order Statistics | PSB Sample

This is a beautiful problem from ISI MStat 2016 Problem 5 (sample) PSB based on order statistics. We provide a detailed solution with the prerequisites mentioned explicitly.

Problem- ISI MStat 2016 Problem 5

Let $ n \geq 2,$ and $ X_{1}, X_{2}, \ldots, X_{n}$ be independent and identically distributed Poisson $ (\lambda) $ random variables for some $ \lambda>0 .$ Let $ X_{(1)} \leq$ $ X_{(2)} \leq \cdots \leq X_{(n)}$ denote the corresponding order statistics.
(a) Show that $ \mathrm{P}\left(X_{(2)}=0\right) \geq 1-n\left(1-e^{-\lambda}\right)^{n-1}$
(b) Evaluate the limit of $ \mathrm{P}\left(X_{(2)}>0\right)$ as the sample size $ n \rightarrow \infty $ .

Prerequisites

Pmf of k-th Order Statistic i.e $ x_{(k)} $
Binomial Expansion
Basic Inequality
squeeze (or sandwich) theorem
Limit

Solution

(a) Given , $ n \geq 2,$ and $ X_{1}, X_{2}, \ldots, X_{n}$ be independent and identically distributed Poisson $ (\lambda) $ random variables for some $ \lambda>0 .$ Let $ X_{(1)} \leq$ $ X_{(2)} \leq \cdots \leq X_{(n)}$ denote the corresponding order statistics.

Let , F(j) be the CDF of $ X_{1}, X_{2}, \ldots, X_{n}$ i.e CDF of Poisson $ (\lambda) $

Then , Pmf of k-th Order Statistic i.e $ x_{(k)} $

$ P(x_{(k)} = j)= F_{k} (j)-F_{k} (j-0) $ , where $ F_{k} (j) =P(x_{(k)} \le j) $ i.e the CDF of k-th Order Statistic

$ F_{k} (j) = \sum_{i=k}^{n} $ ${n \choose i} $ $ {(F(j))}^{i} {(1-F(j))}^{n-i} $

So, $ P(x_{(k)} = j) = \sum_{i=k}^{n} {n \choose i} [ {(F(j))}^{i} {(1-F(j))}^{n-i}-{(F(j-0))}^{i} {(1-F(j-0))}^{n-i}] $

Here we have to find , $ P(x_{(2)} = 0)= \sum_{i=2}^{n} {n \choose i} [{(F(0))}^{i} {(1-F(0))}^{n-i} - 0] $

since , Poisson random variable takes values 0 ,1,2,.... i.e it takes all values < 0 with probabiliy 0 , that's why $ {(F(j-0))}^{i} {(1-F(j-0))}^{n-i} =0$ here for j=0 .

And , $ F(0)=P(x \le 0) = P(X=0)={e}^{- \lambda} \frac{{\lambda}^{0}}{0!} ={e}^{- \lambda} $ , as X follows Poisson $ (\lambda) $ .

So, $ {(F(0))}^{i} {(1-F(0))}^{n-i}= {({e}^{- \lambda})}^{i} {(1-{e}^{- \lambda})}^{n-i} $

Therefore , $ P(x_{(2)} = 0)= \sum_{i=2}^{n} {n \choose i} [{({e}^{- \lambda})}^{i} {(1-{e}^{- \lambda})}^{n-i} ] $

$ = {({e}^{- \lambda}+1-{e}^{- \lambda})}^{n} - {n \choose 0}[{({e}^{- \lambda})}^{0} {(1-{e}^{- \lambda})}^{n-0} ]- {n \choose 1}[{({e}^{- \lambda})}^{1} {(1-{e}^{- \lambda})}^{n-1}] =1-{(1-{e}^{- \lambda})}^{n}- n {e}^{- \lambda} {(1-{e}^{- \lambda})}^{n-1} $

$ =1-{(1-{e}^{- \lambda})}^{n-1}[1-{e}^{- \lambda} +n{e}^{- \lambda}] $

$ =1-{(1-{e}^{- \lambda})}^{n-1}[1+(n-1){e}^{- \lambda}] \ge 1- n{(1-{e}^{- \lambda})}^{n-1} $ .

Since , $ 1+(n-1){e}^{- \lambda} \le n \Longleftrightarrow {e}^{ \lambda} \ge 1 $ for $ n \ge 2$ and $ \lambda >0 $ which is true hence our inequality hold's true (proved)

Hence , $ \mathrm{P}\left(X_{(2)}=0\right) \geq 1-n\left(1-e^{-\lambda}\right)^{n-1}$ (proved )

(b) $ 0 \le P(x_{(2)} >0) =1-P(x_{(2)}= 0) $ $ \le 1-1+n\left(1-e^{-\lambda}\right)^{n-1}$ ( Using inequality in (a) )

So, $ 0 \le P(x_{(2)} >0) =1-P(x_{(2)}= 0) $ $ \le n\left(1-e^{-\lambda}\right)^{n-1}$ -----(1)

As $ 0< 1-{e}^{- \lambda} <1$ for $ \lambda >0 $ i.e it's a fraction so it can be written as $ \frac{1}{a} $ for some $ a>1$ , Hence $ \lim_{n\to\infty} n\left(1-e^{-\lambda}\right)^{n-1} = \lim_{n\to\infty} \frac{n}{a^n} =0 $ (Proof -Use l'hospital rule or think intutively that as n tends to infinity the exponential functions grows more rapidly than any polynomial function ).

Now taking limit $ n \to \infty $ in (1) , we get by squeeze (or sandwich) theorem

$ \lim_{n\to\infty} P(x_{(2)} >0) =0 $

Limit of a function | IIT JAM 2017 | Problem 8

Try this problem from IIT JAM 2017 exam. It deals with evaluating Limit of a function.

Limit of a Function | IIT JAM 2017 | Problem 8

Let $$ f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0 $$
Write $L=\displaystyle\lim_{x \to 0^{-}} f(x)$ and $R=\displaystyle\lim_{x \to 0^{+}} f(x) .$

Then which one of the following is true?

$L$ exists but $R$ does not exist
$L$ does not exist but $R$ exists
Both $L$ and $R$ exist
Neither $L$ nor $R$ exists

Key Concepts

Real Analysis

Function

Limit

Check the Answer

Answer: $L$ exists but $R$ does not exists

IIT JAM 2017 , Problem 8

Try with Hints

Given that, $ f(x)=\frac{x+|x|(1+x)}{x} \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

therefore,

$ f(x)=1+\frac{|x|}{x}(1+x) \sin \left(\frac{1}{x}\right), \quad x \neq 0 $

$ f(x)=\bigg\{\begin{array}{cc}
(2+x) \sin \left(\frac{1}{x}\right), & , x>0 \\
-x \sin \left(\frac{1}{x}\right), & x<0 \\
\end{array} $

Let, $L=\displaystyle\lim_{x \to 0^{-}} f(x)$

and , $R= \displaystyle \lim_{x \rightarrow 0^{+}} f(x) .$

Now,

$L= \displaystyle\lim_{x \to 0^{-}} f(x) $

$\quad = \displaystyle\lim_{x \to 0^{-}} -x \sin \left(\frac{1}{x}\right) $

$ \quad = -\displaystyle\lim_{x \to 0^{-}} x \sin \left(\frac{1}{x}\right) $

Theorem : If $D \subset \mathbb R$ and $f,g : D \to \mathbb R$ . Let $c \in D$. If f is bounded on $N'(c)\cup D$ and $\displaystyle\lim_{x \to c} g(x)=0$, then $\displaystyle\lim_{x \to c}(f.g)(x)=0$.

Now , $ \sin \left(\frac{1}{x}\right) $ is bounded in $\mathbb R - \{0\}$ and $ \displaystyle\lim_{x \to 0^{-}} x=0$ , then $\displaystyle\lim_{x \to 0^{-}} f(x)$ exists and equal to $0$.

But,

$R=\displaystyle\lim_{x\to 0^{+}}f(x)$

$\quad = \displaystyle\lim_{x\to 0^{+}} (2+x) \sin \left(\frac{1}{x}\right),$

$\quad= \displaystyle\lim_{x\to 0^{+}} 2\sin \left(\frac{1}{x}\right) + x \sin \left(\frac{1}{x}\right) $

$\lim_{x \to 0^{+}} \sin \left(\frac{1}{x}\right) $ does not exists [Why?]

Then $L$ exists but $R$ does not.

Definite Integral as Limit of a sum | ISI QMS | QMA 2019

Try this problem from ISI QMS 2019 exam. It requires knowledge Real Analysis and integral calculus and is based on Definite Integral as Limit of a sum.

Definite integral as Limit of a Sum - ISI QMS (QMB Problem 1a)

Find the value of : $ \displaystyle \lim_{n \to \infty}\big[\frac1n+\frac{1}{n+1}+\ldots + \frac{1}{3n}\big]$

Key Concepts

Real Analysis

Definite Integral

Riemann Sum

Check the Answer

Answer: $\ln 3$

ISI QMS 2019 (QMB Problem 1a)

Secrets in Inequalities

Try with Hints

Putting it into standard form :

$\displaystyle\lim_{n \to \infty} \big[\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{3n}\big]$

$= \displaystyle \lim_{n\to \infty}\big[\frac{1}{n+0}+\frac{1}{n+1}+\ldots+\frac{1}{n+2n}\big]$

$= \displaystyle \lim_{n \to \infty} \displaystyle\sum_{r=0}^{2n} \frac{1}{n+r}$

$= \displaystyle \lim_{n \to \infty} \frac1n\displaystyle\sum_{r=0}^{2n} \frac{1}{1+\frac rn}$

$= \displaystyle \lim_{n \to \infty} \frac1n\displaystyle\sum_{r=0}^{2n} f(\frac rn)$ where $f(x)=\frac{1}{1+x}$

An useful result : Let $f:[a,b]\to \mathbb R$ be integrable on $[a,b]$ and $\{P_n\}$ be a sequence of partitions of $[a,b]$ such that the sequence $\{\parallel P_n\parallel\}$ converges to $0$. Then if $\epsilon >0$ be given, there exists a natural number $k$ such that $|S(P_n, f)-\int_a^b f|<\epsilon \quad \forall n\geq k$ where $S(P,f)$ is a Riemann sum of $f$ corresponding to $P$ and any choice of intermediate points.

i.e., If $f$ be integrable on $[a,b]$ and $\{P_n\}$ be a sequence of partitions of $[a,b]$ such that $\lim\limits_{n \to \infty} \parallel P_n \parallel=0$, then $\lim\limits_{n\to \infty} S(P_n,f)=\int_a^b f$

Let $P_n=(0,\frac1n,\frac2n,\ldots,\frac{2n}{n})$ be a sequence of partition on $[0,2]$ dividing it into $2n$ sub-intervals of equal length $\frac1n$.

Also $\lim \parallel P_n\parallel=\lim \frac1n=0$.

Let us choose $\xi_r=\frac rn,\quad r=1,2,3,\ldots,2n$

Then the Riemann sum for $f$ on the interval $[0,2]$ corresponding to the partition $P_n$ and chosen intermediate points $\xi_r$

$S(P_n,f)=\frac1n\displaystyle\sum_{r=1}^{2n} f(\frac rn)$

As $f$ is continuous on $[0,2]$, $f$ is integrable on $[0,2]$.

Now can you use the above result to reach to the solution ?

Using the result

$\lim\limits_{n\to\infty} \frac1n\displaystyle\sum_{r=1}^{2n} f(\frac rn) = \displaystyle\int_0^2 f(x) \mathrm d x$

$= \displaystyle\int_0^2 \frac{ \mathrm d x }{1+x} $

$=\ln(1+x)\bigg|_0^2= \ln 3$ [Ans]

Limit of a Sequence | IIT JAM 2018 | Problem 2

Try this beautiful problem from IIT JAM 2018 which requires knowledge of Real Analysis (Limit of a Sequence).

Limit of a Sequence - IIT JAM 2018 (Problem 2)

Let $a_n=\frac{b_{n+1}}{b_n}$ where $b_1=1, b_2=1$ and $b_{n+2}=b_n+b_{n+1}$ , Then $\lim\limits_{n \to \infty} a_n$ is

$\frac{1-\sqrt5}{2}$
$\frac{1+\sqrt5}{2}$
$\frac{1+\sqrt3}{2}$
$\frac{1-\sqrt3}{2}$

Key Concepts

Real Analysis

Sequence of Reals

Limit of a Sequence

Check the Answer

Answer: $\frac{1+\sqrt5}{2}$

IIT JAM 2018 (Problem 2)

Advanced Calculus by Patrick Fitzpatrick

Try with Hints

Given that, $a_n=\frac{b_{n+1}}{b_n}$

$\Rightarrow \lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} \frac{b_{n+1}}{b_n}= \mathcal{L} $ (say)

Now we know that , $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} a_{n+1} $

$\Rightarrow \mathcal{L}=\lim\limits_{n \to \infty} a_{n+1}$

Can you find an equation on $\mathcal{L}$ from which the value of $\mathcal{L}$ can be obtained.

$\mathcal{L}= \lim\limits_{n \to \infty } a_{n+1}$

$= \lim\limits_{n \to \infty} \frac{b_{n+2}}{b_{n+2}}$

$=\lim\limits_{n\to \infty} \frac{b_{n+1}+b_n}{b_{n+1}}$ [By the given recurrence relation]

$=\lim\limits_{n\to \infty} \left(1+\frac{b_n}{b_{n+1}}\right)$

$=1+\lim\limits_{n \to \infty} \frac{b_n}{b_{n+1}}$

$=1+\frac{1}{\lim\limits_{n\to\infty}\frac{b_{n+1}}{b_n}}$

$=1+\frac{1}{\mathcal{L}}$

Now the value of $\mathcal{L}$ can be easily obtained

i.e., $\mathcal{L}=1+\frac{1}{\mathcal{L}}$

$\Rightarrow \mathcal{L}^2-\mathcal{L}-1=0$

$\Rightarrow \mathcal{L}=\frac{1\pm \sqrt{5}}{2}$

$\Rightarrow \mathcal{L}=\frac{1+\sqrt{5}}{2}$ [Since $a_n>0$] [ANS]

Limit of a Function of two variables

Limit of a two variable Function|IIT JAM 2016 |Problem 6

Find the value of $\lim\limits_{(x,y) \to (2,-2)} \frac{\sqrt{x-y}-2}{x-y-4}$ is

$0$
$\frac14$
$\frac13$
$\frac12$

Key Concepts

Function of Several Variables

Limit of a function of several variables

Check the Answer

Answer: $\textbf{(B)} \frac14$

IIT JAM 2016 (Problem 6)

Function of several variables: Fleming, Wendell H

Try with Hints

So, this problem has double limit. But we will not consider this first i.e., first try to solve the above limit without seeing $(x,y) \to (2,-2)$ , just like we used to do in our school days!!!

Here also we cannot put $x,y$ directly, as it will become undefined.

Now,

$\lim\limits_{(x,y) \to (2,-2)} \frac{(x-y)-4}{(x-y-4)(\sqrt{x-y}+2)}$ [Multiplied by $\sqrt{x-y}+2$ on numerator and denominator ]

Now again, try to simplify the above limit !!!

After simplifying,

We have,

$\lim\limits_{(x,y) \to (2,-2)} \frac{1}{\sqrt{x-y}+2}$ [ Since $(x,y) \to (2,-2) \Rightarrow x-y-4 \ne 0$]

Now in this part we will use the limits, i.e., $(x,y) \to (2,-2)$

$\Rightarrow \frac{1}{\sqrt{2+2}+2}=\frac14$[ANS]

Calculating the limit of the function I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2016, problem 21

[et_pb_section fb_built="1" _builder_version="3.22.4" custom_margin="-1px|||||"][et_pb_row _builder_version="4.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#0c71c3" background_color="#0c71c3" transform_scale="5%|5%" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

What are we learning ?

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Competency in Focus: Calculating the limit of the function This problem from I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2016 is based on simple manipulations and limit of a function .[/et_pb_text][et_pb_text _builder_version="4.1" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#ffffff" background_color="#0c71c3" width="100%" custom_margin="48px||48px||false|false" custom_padding="20px|20px|20px|20px|false|false" border_radii="on|5px|5px|5px|5px" border_color_all="#0c71c3"]

First look at the knowledge graph.

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Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" hover_enabled="0" border_radii="on|100px|100px|100px|100px" box_shadow_style="preset3" box_shadow_color="#ffffff"]Let $f:R \to R$ be a non-zero function such that $\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3}$ exists for all y >0. Let $g(y)= \lim_{x \rightarrow \infty} \frac{f(xy)}{x^3}$ .If g(1)=1 ,then for all y>0 ? [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1"]

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2016. Obective Problem no. 21.

[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.1" open="off"]Limit of a function [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]7 out of 10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.22.4" open="off"]Elementary Number Theory by David M. Burton

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

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So , what we have to find in general the value of g(y) in terms of y given that $g(y)=\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3}$ provided $g(y)=\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3}$ exists .[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1" hover_enabled="0"]

See what can we do is that g(y) can be written as follows , $g(y)=\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3}=
(y^3)\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3y^3} $ [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1" hover_enabled="0"] Again from the given condition we have $g(1)=\lim_{x \rightarrow \infty} f(x) /x^3=\lim_{xy \rightarrow \infty} \frac{f(xy)}{x^3y^3}=1$.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.1" hover_enabled="0"]Therefore , $g(y)=(y^3)\lim_{x \rightarrow \infty} \frac{f(xy)}{x^3y^3} =y^3 $ by previous argument . [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Problem- ISI MStat PSB 2009 Problem 2

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Limit of a Function - ISI MStat Year 2018 PSA Question 8

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Problem- ISI MStat 2016 Problem 5

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Definite integral as Limit of a Sum - ISI QMS (QMB Problem 1a)

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Limit of a Sequence - IIT JAM 2018 (Problem 2)

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