How to solve this challenging limit?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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
edited Mar 16 at 14:34
Rócherz
3,0263823
asked Mar 16 at 14:18
TidronicusTidronicus
263
$endgroup$
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
edited Mar 16 at 14:34
Rócherz
3,0263823
asked Mar 16 at 14:18
TidronicusTidronicus
263
$endgroup$
$begingroup$
Please make you question self-contained, without relying on images. You can use MathJax.
$endgroup$
– Carsten S
Mar 16 at 14:26
1
$begingroup$
Thanks Rocherz, great formatting!
$endgroup$
– Tidronicus
Mar 16 at 14:40
$begingroup$
Possible hint: logarithms.
$endgroup$
– Sean Roberson
Mar 16 at 14:59
$begingroup$
That's a question of JEE Advanced I guess, you can look it up on the net solutions are available there.
$endgroup$
– Sahil Silare
Mar 16 at 17:40
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
edited Mar 16 at 14:34
Rócherz
3,0263823
asked Mar 16 at 14:18
TidronicusTidronicus
263
$endgroup$
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
calculus
edited Mar 16 at 14:34
Rócherz
3,0263823
asked Mar 16 at 14:18
TidronicusTidronicus
263
edited Mar 16 at 14:34
Rócherz
3,0263823
asked Mar 16 at 14:18
TidronicusTidronicus
263
edited Mar 16 at 14:34
Rócherz
3,0263823
edited Mar 16 at 14:34
Rócherz
3,0263823
edited Mar 16 at 14:34
Rócherz
3,0263823
3,0263823
asked Mar 16 at 14:18
TidronicusTidronicus
263
asked Mar 16 at 14:18
TidronicusTidronicus
263
asked Mar 16 at 14:18
TidronicusTidronicus
263
263
$begingroup$
Please make you question self-contained, without relying on images. You can use MathJax.
$endgroup$
– Carsten S
Mar 16 at 14:26
1
$begingroup$
Thanks Rocherz, great formatting!
$endgroup$
– Tidronicus
Mar 16 at 14:40
$begingroup$
Possible hint: logarithms.
$endgroup$
– Sean Roberson
Mar 16 at 14:59
$begingroup$
That's a question of JEE Advanced I guess, you can look it up on the net solutions are available there.
$endgroup$
– Sahil Silare
Mar 16 at 17:40
add a comment |
$begingroup$
Please make you question self-contained, without relying on images. You can use MathJax.
$endgroup$
– Carsten S
Mar 16 at 14:26
1
$begingroup$
Thanks Rocherz, great formatting!
$endgroup$
– Tidronicus
Mar 16 at 14:40
$begingroup$
Possible hint: logarithms.
$endgroup$
– Sean Roberson
Mar 16 at 14:59
$begingroup$
That's a question of JEE Advanced I guess, you can look it up on the net solutions are available there.
$endgroup$
– Sahil Silare
Mar 16 at 17:40
$begingroup$
Please make you question self-contained, without relying on images. You can use MathJax.
$endgroup$
– Carsten S
Mar 16 at 14:26
$begingroup$
Please make you question self-contained, without relying on images. You can use MathJax.
$endgroup$
– Carsten S
Mar 16 at 14:26
1
1
$begingroup$
Thanks Rocherz, great formatting!
$endgroup$
– Tidronicus
Mar 16 at 14:40
$begingroup$
Thanks Rocherz, great formatting!
$endgroup$
– Tidronicus
Mar 16 at 14:40
$begingroup$
Possible hint: logarithms.
$endgroup$
– Sean Roberson
Mar 16 at 14:59
$begingroup$
Possible hint: logarithms.
$endgroup$
– Sean Roberson
Mar 16 at 14:59
$begingroup$
That's a question of JEE Advanced I guess, you can look it up on the net solutions are available there.
$endgroup$
– Sahil Silare
Mar 16 at 17:40
$begingroup$
That's a question of JEE Advanced I guess, you can look it up on the net solutions are available there.
$endgroup$
– Sahil Silare
Mar 16 at 17:40
add a comment |
1 Answer
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$begingroup$
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.
answered Mar 16 at 17:19
Fabio LucchiniFabio Lucchini
9,50111426
$endgroup$
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.
Mar 16 at 17:28
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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.
answered Mar 16 at 17:19
Fabio LucchiniFabio Lucchini
9,50111426
$endgroup$
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.
Mar 16 at 17:28
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.
answered Mar 16 at 17:19
Fabio LucchiniFabio Lucchini
9,50111426
$endgroup$
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.
Mar 16 at 17:28
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.
answered Mar 16 at 17:19
Fabio LucchiniFabio Lucchini
9,50111426
$endgroup$
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.
answered Mar 16 at 17:19
Fabio LucchiniFabio Lucchini
9,50111426
edited Mar 16 at 19:07
answered Mar 16 at 17:19
Fabio LucchiniFabio Lucchini
9,50111426
answered Mar 16 at 17:19
Fabio LucchiniFabio Lucchini
9,50111426
answered Mar 16 at 17:19
Fabio LucchiniFabio Lucchini
9,50111426
9,50111426
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.
Mar 16 at 17:28
add a comment |
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.
Mar 16 at 17:28
1
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.
Mar 16 at 17:28
$begingroup$
+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.
Mar 16 at 17:28
add a comment |
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$begingroup$
Please make you question self-contained, without relying on images. You can use MathJax.
$endgroup$
– Carsten S
Mar 16 at 14:26
1
$begingroup$
Thanks Rocherz, great formatting!
$endgroup$
– Tidronicus
Mar 16 at 14:40
$begingroup$
Possible hint: logarithms.
$endgroup$
– Sean Roberson
Mar 16 at 14:59
$begingroup$
That's a question of JEE Advanced I guess, you can look it up on the net solutions are available there.
$endgroup$
– Sahil Silare
Mar 16 at 17:40