Trjtdtk

How do researchers send unsolicited emails asking for feedback on their works?

How to determine the greatest d orbital splitting?

Have any astronauts/cosmonauts died in space?

How to find the largest number(s) in a list of elements, possibly non-unique?

Does the Shadow Magic sorcerer's Eyes of the Dark feature work on all Darkness spells or just his/her own?

Smaller neighborhood around the identity of Lie Group

Have the tides ever turned twice on any open problem?

What is the tangent at a sharp point on a curve?

Why doesn't Gödel's incompleteness theorem apply to false statements?

Do people actually use the word "kaputt" in conversation?

How to understand 「僕は誰より彼女が好きなんだ。」

Started in 1987 vs. Starting in 1987

PTIJ: Where did Achashverosh's years wander off to?

Do I need to convey a moral for each of my blog post?

Why doesn't the fusion process of the sun speed up?

What do the positive and negative (+/-) transmit and receive pins mean on Ethernet cables?

Will my managed file get deleted?

How to read string as hex number in bash?

Why is participating in the European Parliamentary elections used as a threat?

Animating wave motion in water

Error in master's thesis, I do not know what to do

What will the Frenchman say?

PTIJ: Which Dr. Seuss books should one obtain?

Friend wants my recommendation but I don't want to give it to him

How to solve this challenging limit?

Mixed question based on the concept of limits, application of derivatives.Limit of the sequence $n^n/n!$, is this sequence bounded, convergent and eventually monotonic?Evaluating: $int 3xsinleft(frac x4right) , dx$.can you multiply a limit that doesn't exist times a limit that is equal to zero?Compute this integralFinding the limit of lim$_n rightarrow inftyleft( dfracn^32^n right)$Prove inequality $arccos left( fracsin 1-sin x1-x right) leq sqrtfrac1+x+x^23$What's wrong in this evaluation of the definite integral of $sin^2 (x/3)$?For each $a in mathbbR$ evaluate $ limlimits_n to inftybeginpmatrix 1&fracan\frac-an&1endpmatrix^n.$Demonstrate that $(x+y)ln left(fracx+y2right) leq xln x +yln y$Deriving an expression for price elasticity of demand

4

2

$begingroup$

I believe more than one answer may be correct. I thought about this for a while and I feel like there is a simplification that makes it look a whole lot easier but I can't figure out what it is.

I noticed that every term within the limit is multiplied so I figured that some terms might divide out but dividing $(x+n)$ by $(x^2+n^2)$ doesn't really seem to lead to any useful simplifications.

I thought about multiplying out everything on top and bottom of the inner expression but that again just seems to go nowhere and gets very messy and complicated without any reductions or simplifications.

Since they ask to take the derivative in one of the answer choices, surely there's some simplifications to be made to make this problem a lot easier.

$mathbf 46.$ Let $displaystyle f(x) = lim_ntoinfty left( frac displaystyle n^n (x+n) left(x+frac n2right) ldots left(x+frac nnright) displaystyle n! (x^2+n^2) left(x^2+fracn^24right) ldots left(x^2+fracn^2n^2right)right)^dfrac xn$, for all $x>0$. Then

(A) $f left(dfrac12right) geq f(1)$

(B) $f left(dfrac13right) leq f left(dfrac23right)$

(C) $f'(2) leq 0$

(D) $dfrac f'(3)f(3) geq dfrac f'(2)f(2)$

calculus

share|cite|improve this question

edited 2 days ago

Rócherz

2,9863821

asked 2 days ago

TidronicusTidronicus

263

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please make you question self-contained, without relying on images. You can use MathJax.
  $endgroup$
  – Carsten S
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Rocherz, great formatting!
  $endgroup$
  – Tidronicus
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible hint: logarithms.
  $endgroup$
  – Sean Roberson
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's a question of JEE Advanced I guess, you can look it up on the net solutions are available there.
  $endgroup$
  – Sahil Silare
  2 days ago
  

  
  
  
  

add a comment | 

4

2

$begingroup$

I believe more than one answer may be correct. I thought about this for a while and I feel like there is a simplification that makes it look a whole lot easier but I can't figure out what it is.

I noticed that every term within the limit is multiplied so I figured that some terms might divide out but dividing $(x+n)$ by $(x^2+n^2)$ doesn't really seem to lead to any useful simplifications.

I thought about multiplying out everything on top and bottom of the inner expression but that again just seems to go nowhere and gets very messy and complicated without any reductions or simplifications.

Since they ask to take the derivative in one of the answer choices, surely there's some simplifications to be made to make this problem a lot easier.

$mathbf 46.$ Let $displaystyle f(x) = lim_ntoinfty left( frac displaystyle n^n (x+n) left(x+frac n2right) ldots left(x+frac nnright) displaystyle n! (x^2+n^2) left(x^2+fracn^24right) ldots left(x^2+fracn^2n^2right)right)^dfrac xn$, for all $x>0$. Then

(A) $f left(dfrac12right) geq f(1)$

(B) $f left(dfrac13right) leq f left(dfrac23right)$

(C) $f'(2) leq 0$

(D) $dfrac f'(3)f(3) geq dfrac f'(2)f(2)$

calculus

share|cite|improve this question

edited 2 days ago

Rócherz

2,9863821

asked 2 days ago

TidronicusTidronicus

263

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please make you question self-contained, without relying on images. You can use MathJax.
  $endgroup$
  – Carsten S
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Rocherz, great formatting!
  $endgroup$
  – Tidronicus
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible hint: logarithms.
  $endgroup$
  – Sean Roberson
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's a question of JEE Advanced I guess, you can look it up on the net solutions are available there.
  $endgroup$
  – Sahil Silare
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

I believe more than one answer may be correct. I thought about this for a while and I feel like there is a simplification that makes it look a whole lot easier but I can't figure out what it is.

I noticed that every term within the limit is multiplied so I figured that some terms might divide out but dividing $(x+n)$ by $(x^2+n^2)$ doesn't really seem to lead to any useful simplifications.

I thought about multiplying out everything on top and bottom of the inner expression but that again just seems to go nowhere and gets very messy and complicated without any reductions or simplifications.

Since they ask to take the derivative in one of the answer choices, surely there's some simplifications to be made to make this problem a lot easier.

$mathbf 46.$ Let $displaystyle f(x) = lim_ntoinfty left( frac displaystyle n^n (x+n) left(x+frac n2right) ldots left(x+frac nnright) displaystyle n! (x^2+n^2) left(x^2+fracn^24right) ldots left(x^2+fracn^2n^2right)right)^dfrac xn$, for all $x>0$. Then

(A) $f left(dfrac12right) geq f(1)$

(B) $f left(dfrac13right) leq f left(dfrac23right)$

(C) $f'(2) leq 0$

(D) $dfrac f'(3)f(3) geq dfrac f'(2)f(2)$

calculus

share|cite|improve this question

edited 2 days ago

Rócherz

2,9863821

asked 2 days ago

TidronicusTidronicus

263

$endgroup$

share|cite|improve this question

edited 2 days ago

Rócherz

2,9863821

asked 2 days ago

TidronicusTidronicus

263

share|cite|improve this question

edited 2 days ago

Rócherz

2,9863821

asked 2 days ago

TidronicusTidronicus

263

edited 2 days ago

Rócherz

2,9863821

edited 2 days ago

Rócherz

2,9863821

asked 2 days ago

TidronicusTidronicus

263

asked 2 days ago

TidronicusTidronicus

263

1 Answer
1

active

oldest

votes

6

$begingroup$

beginalign
log f(x)
&=logleft(fracn^nn!prod_k=1^nfracx+frac nkx^2+fracn^2k^2right)^frac xn\
&=frac xnlogleft(fracn!n^2nn!n^2nprod_k=1^nfracfrackxn+1left(frackxnright)^2+1right)\
&=frac xnsum_k=1^nlogfracfrackxn+1left(frackxnright)^2+1\
&xrightarrowntoinftyint_0^xlogfract+1t^2+1mathrm dt
endalign
(Edit) Consequently, $f(x)>0$ for every $x>0$,
$$fracf'(x)f(x)=logfracx+1x^2+1
begincases
>0&0<x<1\
=0&x=1\
<0&x>1
endcases$$
Thus $f$ is increasing for $0<x<1$, hence (A) is false, (B) is true, (C) is true and (D) is equivalent to
$$logfrac410gelogfrac35$$
which is false.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Fabio LucchiniFabio Lucchini

9,28611426

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. In particular (assuming I've made no mistakes), $(ln f)^prime=lnfracx+1x^2+1$ is positive on $[0,,1)$ but negative for $x<1$, so (A) is false, (B) is true, (C) is true and (D) is equivalent to the statement $lnfrac410gelnfrac35$, which is false.
  $endgroup$
  – J.G.
  2 days ago
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150431%2fhow-to-solve-this-challenging-limit%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

beginalign
log f(x)
&=logleft(fracn^nn!prod_k=1^nfracx+frac nkx^2+fracn^2k^2right)^frac xn\
&=frac xnlogleft(fracn!n^2nn!n^2nprod_k=1^nfracfrackxn+1left(frackxnright)^2+1right)\
&=frac xnsum_k=1^nlogfracfrackxn+1left(frackxnright)^2+1\
&xrightarrowntoinftyint_0^xlogfract+1t^2+1mathrm dt
endalign
(Edit) Consequently, $f(x)>0$ for every $x>0$,
$$fracf'(x)f(x)=logfracx+1x^2+1
begincases
>0&0<x<1\
=0&x=1\
<0&x>1
endcases$$
Thus $f$ is increasing for $0<x<1$, hence (A) is false, (B) is true, (C) is true and (D) is equivalent to
$$logfrac410gelogfrac35$$
which is false.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Fabio LucchiniFabio Lucchini

9,28611426

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. In particular (assuming I've made no mistakes), $(ln f)^prime=lnfracx+1x^2+1$ is positive on $[0,,1)$ but negative for $x<1$, so (A) is false, (B) is true, (C) is true and (D) is equivalent to the statement $lnfrac410gelnfrac35$, which is false.
  $endgroup$
  – J.G.
  2 days ago
  

  
  
  
  

add a comment | 

6

$begingroup$

beginalign
log f(x)
&=logleft(fracn^nn!prod_k=1^nfracx+frac nkx^2+fracn^2k^2right)^frac xn\
&=frac xnlogleft(fracn!n^2nn!n^2nprod_k=1^nfracfrackxn+1left(frackxnright)^2+1right)\
&=frac xnsum_k=1^nlogfracfrackxn+1left(frackxnright)^2+1\
&xrightarrowntoinftyint_0^xlogfract+1t^2+1mathrm dt
endalign
(Edit) Consequently, $f(x)>0$ for every $x>0$,
$$fracf'(x)f(x)=logfracx+1x^2+1
begincases
>0&0<x<1\
=0&x=1\
<0&x>1
endcases$$
Thus $f$ is increasing for $0<x<1$, hence (A) is false, (B) is true, (C) is true and (D) is equivalent to
$$logfrac410gelogfrac35$$
which is false.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Fabio LucchiniFabio Lucchini

9,28611426

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. In particular (assuming I've made no mistakes), $(ln f)^prime=lnfracx+1x^2+1$ is positive on $[0,,1)$ but negative for $x<1$, so (A) is false, (B) is true, (C) is true and (D) is equivalent to the statement $lnfrac410gelnfrac35$, which is false.
  $endgroup$
  – J.G.
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

beginalign
log f(x)
&=logleft(fracn^nn!prod_k=1^nfracx+frac nkx^2+fracn^2k^2right)^frac xn\
&=frac xnlogleft(fracn!n^2nn!n^2nprod_k=1^nfracfrackxn+1left(frackxnright)^2+1right)\
&=frac xnsum_k=1^nlogfracfrackxn+1left(frackxnright)^2+1\
&xrightarrowntoinftyint_0^xlogfract+1t^2+1mathrm dt
endalign
(Edit) Consequently, $f(x)>0$ for every $x>0$,
$$fracf'(x)f(x)=logfracx+1x^2+1
begincases
>0&0<x<1\
=0&x=1\
<0&x>1
endcases$$
Thus $f$ is increasing for $0<x<1$, hence (A) is false, (B) is true, (C) is true and (D) is equivalent to
$$logfrac410gelogfrac35$$
which is false.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Fabio LucchiniFabio Lucchini

9,28611426

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Fabio LucchiniFabio Lucchini

9,28611426

answered 2 days ago

Fabio LucchiniFabio Lucchini

9,28611426

answered 2 days ago

Fabio LucchiniFabio Lucchini

9,28611426