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How to solve this challenging limit?

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 11 hours ago

Rócherz

2,9762821

asked 11 hours ago

TidronicusTidronicus

263

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Rocherz, great formatting!
  $endgroup$
  – Tidronicus
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible hint: logarithms.
  $endgroup$
  – Sean Roberson
  11 hours 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
  8 hours 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 11 hours ago

Rócherz

2,9762821

asked 11 hours ago

TidronicusTidronicus

263

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Rocherz, great formatting!
  $endgroup$
  – Tidronicus
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible hint: logarithms.
  $endgroup$
  – Sean Roberson
  11 hours 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
  8 hours 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 11 hours ago

Rócherz

2,9762821

asked 11 hours ago

TidronicusTidronicus

263

$endgroup$

share|cite|improve this question

edited 11 hours ago

Rócherz

2,9762821

asked 11 hours ago

TidronicusTidronicus

263

share|cite|improve this question

edited 11 hours ago

Rócherz

2,9762821

asked 11 hours ago

TidronicusTidronicus

263

edited 11 hours ago

Rócherz

2,9762821

edited 11 hours ago

Rócherz

2,9762821

asked 11 hours ago

TidronicusTidronicus

263

asked 11 hours ago

TidronicusTidronicus

263

1 Answer
1

active

oldest

votes

5

$begingroup$

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 7 hours ago

answered 8 hours ago

Fabio LucchiniFabio Lucchini

9,01311426

$endgroup$

• 
  
  
  

  
  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.
  8 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

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 7 hours ago

answered 8 hours ago

Fabio LucchiniFabio Lucchini

9,01311426

$endgroup$

• 
  
  
  

  
  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.
  8 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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 7 hours ago

answered 8 hours ago

Fabio LucchiniFabio Lucchini

9,01311426

$endgroup$

• 
  
  
  

  
  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.
  8 hours 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 7 hours ago

answered 8 hours ago

Fabio LucchiniFabio Lucchini

9,01311426

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Fabio LucchiniFabio Lucchini

9,01311426

answered 8 hours ago

Fabio LucchiniFabio Lucchini

9,01311426

answered 8 hours ago

Fabio LucchiniFabio Lucchini

9,01311426