Mixing
Computing limits with natural log
$begingroup$
I have the following limit:
$$lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2$$
I know, for sure, that the limit is a finite number. I have tried the following method, but keep getting $infty$:
- $$lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2 = L$$
- $$ ln{L} = lnbigg({lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2}bigg)$$
- $$ Rightarrow ln{L} = ln{bigg(bigg(frac{2}{3}bigg)^nn^2}bigg)$$
- $$ = nln{frac{2}{3}} + 2ln{n}$$
- $$ Rightarrow e^{ln{L}} = e^{nln{frac{2}{3}} + 2ln{n}}$$
- $$ L = bigg(frac{2}{3}bigg)^n+n^2$$
This seems to be $infty$. What am I doing wrong? Please keep in mind that I am looking for the simplest solution; I'm only a second year undergraduate.
calculus limits
asked Jan 26 at 0:28
darylnakdarylnak
184111
$endgroup$
add a comment |
$begingroup$
I have the following limit:
$$lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2$$
I know, for sure, that the limit is a finite number. I have tried the following method, but keep getting $infty$:
- $$lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2 = L$$
- $$ ln{L} = lnbigg({lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2}bigg)$$
- $$ Rightarrow ln{L} = ln{bigg(bigg(frac{2}{3}bigg)^nn^2}bigg)$$
- $$ = nln{frac{2}{3}} + 2ln{n}$$
- $$ Rightarrow e^{ln{L}} = e^{nln{frac{2}{3}} + 2ln{n}}$$
- $$ L = bigg(frac{2}{3}bigg)^n+n^2$$
This seems to be $infty$. What am I doing wrong? Please keep in mind that I am looking for the simplest solution; I'm only a second year undergraduate.
calculus limits
asked Jan 26 at 0:28
darylnakdarylnak
184111
$endgroup$
$begingroup$
At last step, should be $L=(2/3)^n cdot n^2.$ [not $+$ as you have. Also you seem to have dropped lim somewhere, and do you mean $lim_{n to infty}$ at top?
$endgroup$
– coffeemath
Jan 26 at 0:43
add a comment |
$begingroup$
I have the following limit:
$$lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2$$
I know, for sure, that the limit is a finite number. I have tried the following method, but keep getting $infty$:
- $$lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2 = L$$
- $$ ln{L} = lnbigg({lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2}bigg)$$
- $$ Rightarrow ln{L} = ln{bigg(bigg(frac{2}{3}bigg)^nn^2}bigg)$$
- $$ = nln{frac{2}{3}} + 2ln{n}$$
- $$ Rightarrow e^{ln{L}} = e^{nln{frac{2}{3}} + 2ln{n}}$$
- $$ L = bigg(frac{2}{3}bigg)^n+n^2$$
This seems to be $infty$. What am I doing wrong? Please keep in mind that I am looking for the simplest solution; I'm only a second year undergraduate.
calculus limits
asked Jan 26 at 0:28
darylnakdarylnak
184111
$endgroup$
I have the following limit:
$$lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2$$
I know, for sure, that the limit is a finite number. I have tried the following method, but keep getting $infty$:
- $$lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2 = L$$
- $$ ln{L} = lnbigg({lim_{xtoinfty} bigg(frac{2}{3}bigg)^nn^2}bigg)$$
- $$ Rightarrow ln{L} = ln{bigg(bigg(frac{2}{3}bigg)^nn^2}bigg)$$
- $$ = nln{frac{2}{3}} + 2ln{n}$$
- $$ Rightarrow e^{ln{L}} = e^{nln{frac{2}{3}} + 2ln{n}}$$
- $$ L = bigg(frac{2}{3}bigg)^n+n^2$$
This seems to be $infty$. What am I doing wrong? Please keep in mind that I am looking for the simplest solution; I'm only a second year undergraduate.
calculus limits
calculus limits
asked Jan 26 at 0:28
darylnakdarylnak
184111
asked Jan 26 at 0:28
darylnakdarylnak
184111
asked Jan 26 at 0:28
darylnakdarylnak
184111
asked Jan 26 at 0:28
darylnakdarylnak
184111
asked Jan 26 at 0:28
darylnakdarylnak
184111
184111
$begingroup$
At last step, should be $L=(2/3)^n cdot n^2.$ [not $+$ as you have. Also you seem to have dropped lim somewhere, and do you mean $lim_{n to infty}$ at top?
$endgroup$
– coffeemath
Jan 26 at 0:43
add a comment |
$begingroup$
At last step, should be $L=(2/3)^n cdot n^2.$ [not $+$ as you have. Also you seem to have dropped lim somewhere, and do you mean $lim_{n to infty}$ at top?
$endgroup$
– coffeemath
Jan 26 at 0:43
$begingroup$
At last step, should be $L=(2/3)^n cdot n^2.$ [not $+$ as you have. Also you seem to have dropped lim somewhere, and do you mean $lim_{n to infty}$ at top?
$endgroup$
– coffeemath
Jan 26 at 0:43
$begingroup$
At last step, should be $L=(2/3)^n cdot n^2.$ [not $+$ as you have. Also you seem to have dropped lim somewhere, and do you mean $lim_{n to infty}$ at top?
$endgroup$
– coffeemath
Jan 26 at 0:43
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint:
Factor out $n$: $$nln{frac{2}{3}} + 2ln{n}=nBigl(lndfrac23+2dfrac{ln n}nBigr)$$
and remember that $ln x=o(x)$ near $infty$.
answered Jan 26 at 0:42
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Your mistake is going from the next-to-last time to the last line. The right side of the next-to-last line is the exponential of a sum--that is not the same as the sum of exponentials. It is not true that
$$e^{x+y}= e^x+e^y$$
So you cannot reach that last line. If you transform the next-to-last line properly, you just go back to your first line and have gained nothing.
A better approach is to use L'Hopital's rule. Your problem has the form $0cdotinfty$, so transform it to a ratio. Use L'Hopital to find the limit of
$$frac{n^2}{(3/2)^n}$$
Taking the derivative twice of numerator and of denominator gives a constant in the numerator and an exponential that approaches infinity in the denominator. Thus the limit is zero.
I'll leave the details to you.
answered Jan 26 at 0:45
Rory DaultonRory Daulton
29.5k63355
$endgroup$
add a comment |
$begingroup$
Since you are looking for a simple solution, here another approach using
$0<q<1 Rightarrow q^n stackrel{nto infty}{longrightarrow}0$ and- $lim_{ntoinfty}frac{(n+1)^2}{n^2}=1$
Let $a_n = left(frac{2}{3} right)^ncdot n^2$, then there is $N in mathbb{N}$ such that
$$frac{a_{n+1}}{a_n} = frac{2}{3}frac{(n+1)^2}{n^2}stackrel{n>N}{<}frac{3}{4}$$
It follows that for $n > N$ you have
$$a_n =frac{a_n}{a_{n-1}}cdot frac{a_{n-1}}{a_{n-2}} cdots frac{a_{N+1}}{a_{N}}cdot a_N < left(frac{3}{4} right)^{n-N}cdot a_N=Ccdot left(frac{3}{4} right)^{n}stackrel{ntoinfty}{longrightarrow}0$$
answered Jan 26 at 7:02
trancelocationtrancelocation
12.8k1827
$endgroup$
add a comment |
$begingroup$
Option:
$n in mathbb{Z^+}.$
Set $e^{-a}=(2/3)$, where $a>0$, real.
Then
$0 < dfrac{n^2}{e^{an}} < $
$small {dfrac{n^2}{1+an+(1/2!)(an)^2 +(1/3!)(an)^3 +....} }$
$< dfrac{3!n^2}{(an)^3} =left (dfrac{3!}{a^3}right ) dfrac{1}{n}.$
Take the limit.
answered Jan 26 at 7:21
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
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$begingroup$
Hint:
Factor out $n$: $$nln{frac{2}{3}} + 2ln{n}=nBigl(lndfrac23+2dfrac{ln n}nBigr)$$
and remember that $ln x=o(x)$ near $infty$.
answered Jan 26 at 0:42
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Factor out $n$: $$nln{frac{2}{3}} + 2ln{n}=nBigl(lndfrac23+2dfrac{ln n}nBigr)$$
and remember that $ln x=o(x)$ near $infty$.
answered Jan 26 at 0:42
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Factor out $n$: $$nln{frac{2}{3}} + 2ln{n}=nBigl(lndfrac23+2dfrac{ln n}nBigr)$$
and remember that $ln x=o(x)$ near $infty$.
answered Jan 26 at 0:42
BernardBernard
123k741117
$endgroup$
Hint:
Factor out $n$: $$nln{frac{2}{3}} + 2ln{n}=nBigl(lndfrac23+2dfrac{ln n}nBigr)$$
and remember that $ln x=o(x)$ near $infty$.
answered Jan 26 at 0:42
BernardBernard
123k741117
answered Jan 26 at 0:42
BernardBernard
123k741117
answered Jan 26 at 0:42
BernardBernard
123k741117
answered Jan 26 at 0:42
BernardBernard
123k741117
123k741117
add a comment |
add a comment |
$begingroup$
Your mistake is going from the next-to-last time to the last line. The right side of the next-to-last line is the exponential of a sum--that is not the same as the sum of exponentials. It is not true that
$$e^{x+y}= e^x+e^y$$
So you cannot reach that last line. If you transform the next-to-last line properly, you just go back to your first line and have gained nothing.
A better approach is to use L'Hopital's rule. Your problem has the form $0cdotinfty$, so transform it to a ratio. Use L'Hopital to find the limit of
$$frac{n^2}{(3/2)^n}$$
Taking the derivative twice of numerator and of denominator gives a constant in the numerator and an exponential that approaches infinity in the denominator. Thus the limit is zero.
I'll leave the details to you.
answered Jan 26 at 0:45
Rory DaultonRory Daulton
29.5k63355
$endgroup$
add a comment |
$begingroup$
Your mistake is going from the next-to-last time to the last line. The right side of the next-to-last line is the exponential of a sum--that is not the same as the sum of exponentials. It is not true that
$$e^{x+y}= e^x+e^y$$
So you cannot reach that last line. If you transform the next-to-last line properly, you just go back to your first line and have gained nothing.
A better approach is to use L'Hopital's rule. Your problem has the form $0cdotinfty$, so transform it to a ratio. Use L'Hopital to find the limit of
$$frac{n^2}{(3/2)^n}$$
Taking the derivative twice of numerator and of denominator gives a constant in the numerator and an exponential that approaches infinity in the denominator. Thus the limit is zero.
I'll leave the details to you.
answered Jan 26 at 0:45
Rory DaultonRory Daulton
29.5k63355
$endgroup$
add a comment |
$begingroup$
Your mistake is going from the next-to-last time to the last line. The right side of the next-to-last line is the exponential of a sum--that is not the same as the sum of exponentials. It is not true that
$$e^{x+y}= e^x+e^y$$
So you cannot reach that last line. If you transform the next-to-last line properly, you just go back to your first line and have gained nothing.
A better approach is to use L'Hopital's rule. Your problem has the form $0cdotinfty$, so transform it to a ratio. Use L'Hopital to find the limit of
$$frac{n^2}{(3/2)^n}$$
Taking the derivative twice of numerator and of denominator gives a constant in the numerator and an exponential that approaches infinity in the denominator. Thus the limit is zero.
I'll leave the details to you.
answered Jan 26 at 0:45
Rory DaultonRory Daulton
29.5k63355
$endgroup$
Your mistake is going from the next-to-last time to the last line. The right side of the next-to-last line is the exponential of a sum--that is not the same as the sum of exponentials. It is not true that
$$e^{x+y}= e^x+e^y$$
So you cannot reach that last line. If you transform the next-to-last line properly, you just go back to your first line and have gained nothing.
A better approach is to use L'Hopital's rule. Your problem has the form $0cdotinfty$, so transform it to a ratio. Use L'Hopital to find the limit of
$$frac{n^2}{(3/2)^n}$$
Taking the derivative twice of numerator and of denominator gives a constant in the numerator and an exponential that approaches infinity in the denominator. Thus the limit is zero.
I'll leave the details to you.
answered Jan 26 at 0:45
Rory DaultonRory Daulton
29.5k63355
answered Jan 26 at 0:45
Rory DaultonRory Daulton
29.5k63355
answered Jan 26 at 0:45
Rory DaultonRory Daulton
29.5k63355
answered Jan 26 at 0:45
Rory DaultonRory Daulton
29.5k63355
29.5k63355
add a comment |
add a comment |
$begingroup$
Since you are looking for a simple solution, here another approach using
$0<q<1 Rightarrow q^n stackrel{nto infty}{longrightarrow}0$ and- $lim_{ntoinfty}frac{(n+1)^2}{n^2}=1$
Let $a_n = left(frac{2}{3} right)^ncdot n^2$, then there is $N in mathbb{N}$ such that
$$frac{a_{n+1}}{a_n} = frac{2}{3}frac{(n+1)^2}{n^2}stackrel{n>N}{<}frac{3}{4}$$
It follows that for $n > N$ you have
$$a_n =frac{a_n}{a_{n-1}}cdot frac{a_{n-1}}{a_{n-2}} cdots frac{a_{N+1}}{a_{N}}cdot a_N < left(frac{3}{4} right)^{n-N}cdot a_N=Ccdot left(frac{3}{4} right)^{n}stackrel{ntoinfty}{longrightarrow}0$$
answered Jan 26 at 7:02
trancelocationtrancelocation
12.8k1827
$endgroup$
add a comment |
$begingroup$
Since you are looking for a simple solution, here another approach using
$0<q<1 Rightarrow q^n stackrel{nto infty}{longrightarrow}0$ and- $lim_{ntoinfty}frac{(n+1)^2}{n^2}=1$
Let $a_n = left(frac{2}{3} right)^ncdot n^2$, then there is $N in mathbb{N}$ such that
$$frac{a_{n+1}}{a_n} = frac{2}{3}frac{(n+1)^2}{n^2}stackrel{n>N}{<}frac{3}{4}$$
It follows that for $n > N$ you have
$$a_n =frac{a_n}{a_{n-1}}cdot frac{a_{n-1}}{a_{n-2}} cdots frac{a_{N+1}}{a_{N}}cdot a_N < left(frac{3}{4} right)^{n-N}cdot a_N=Ccdot left(frac{3}{4} right)^{n}stackrel{ntoinfty}{longrightarrow}0$$
answered Jan 26 at 7:02
trancelocationtrancelocation
12.8k1827
$endgroup$
add a comment |
$begingroup$
Since you are looking for a simple solution, here another approach using
$0<q<1 Rightarrow q^n stackrel{nto infty}{longrightarrow}0$ and- $lim_{ntoinfty}frac{(n+1)^2}{n^2}=1$
Let $a_n = left(frac{2}{3} right)^ncdot n^2$, then there is $N in mathbb{N}$ such that
$$frac{a_{n+1}}{a_n} = frac{2}{3}frac{(n+1)^2}{n^2}stackrel{n>N}{<}frac{3}{4}$$
It follows that for $n > N$ you have
$$a_n =frac{a_n}{a_{n-1}}cdot frac{a_{n-1}}{a_{n-2}} cdots frac{a_{N+1}}{a_{N}}cdot a_N < left(frac{3}{4} right)^{n-N}cdot a_N=Ccdot left(frac{3}{4} right)^{n}stackrel{ntoinfty}{longrightarrow}0$$
answered Jan 26 at 7:02
trancelocationtrancelocation
12.8k1827
$endgroup$
Since you are looking for a simple solution, here another approach using
$0<q<1 Rightarrow q^n stackrel{nto infty}{longrightarrow}0$ and- $lim_{ntoinfty}frac{(n+1)^2}{n^2}=1$
Let $a_n = left(frac{2}{3} right)^ncdot n^2$, then there is $N in mathbb{N}$ such that
$$frac{a_{n+1}}{a_n} = frac{2}{3}frac{(n+1)^2}{n^2}stackrel{n>N}{<}frac{3}{4}$$
It follows that for $n > N$ you have
$$a_n =frac{a_n}{a_{n-1}}cdot frac{a_{n-1}}{a_{n-2}} cdots frac{a_{N+1}}{a_{N}}cdot a_N < left(frac{3}{4} right)^{n-N}cdot a_N=Ccdot left(frac{3}{4} right)^{n}stackrel{ntoinfty}{longrightarrow}0$$
answered Jan 26 at 7:02
trancelocationtrancelocation
12.8k1827
answered Jan 26 at 7:02
trancelocationtrancelocation
12.8k1827
answered Jan 26 at 7:02
trancelocationtrancelocation
12.8k1827
answered Jan 26 at 7:02
trancelocationtrancelocation
12.8k1827
12.8k1827
add a comment |
add a comment |
$begingroup$
Option:
$n in mathbb{Z^+}.$
Set $e^{-a}=(2/3)$, where $a>0$, real.
Then
$0 < dfrac{n^2}{e^{an}} < $
$small {dfrac{n^2}{1+an+(1/2!)(an)^2 +(1/3!)(an)^3 +....} }$
$< dfrac{3!n^2}{(an)^3} =left (dfrac{3!}{a^3}right ) dfrac{1}{n}.$
Take the limit.
answered Jan 26 at 7:21
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
add a comment |
$begingroup$
Option:
$n in mathbb{Z^+}.$
Set $e^{-a}=(2/3)$, where $a>0$, real.
Then
$0 < dfrac{n^2}{e^{an}} < $
$small {dfrac{n^2}{1+an+(1/2!)(an)^2 +(1/3!)(an)^3 +....} }$
$< dfrac{3!n^2}{(an)^3} =left (dfrac{3!}{a^3}right ) dfrac{1}{n}.$
Take the limit.
answered Jan 26 at 7:21
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
add a comment |
$begingroup$
Option:
$n in mathbb{Z^+}.$
Set $e^{-a}=(2/3)$, where $a>0$, real.
Then
$0 < dfrac{n^2}{e^{an}} < $
$small {dfrac{n^2}{1+an+(1/2!)(an)^2 +(1/3!)(an)^3 +....} }$
$< dfrac{3!n^2}{(an)^3} =left (dfrac{3!}{a^3}right ) dfrac{1}{n}.$
Take the limit.
answered Jan 26 at 7:21
Peter SzilasPeter Szilas
11.6k2822
$endgroup$
Option:
$n in mathbb{Z^+}.$
Set $e^{-a}=(2/3)$, where $a>0$, real.
Then
$0 < dfrac{n^2}{e^{an}} < $
$small {dfrac{n^2}{1+an+(1/2!)(an)^2 +(1/3!)(an)^3 +....} }$
$< dfrac{3!n^2}{(an)^3} =left (dfrac{3!}{a^3}right ) dfrac{1}{n}.$
Take the limit.
answered Jan 26 at 7:21
Peter SzilasPeter Szilas
11.6k2822
answered Jan 26 at 7:21
Peter SzilasPeter Szilas
11.6k2822
answered Jan 26 at 7:21
Peter SzilasPeter Szilas
11.6k2822
answered Jan 26 at 7:21
Peter SzilasPeter Szilas
11.6k2822
11.6k2822
add a comment |
add a comment |
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$begingroup$
At last step, should be $L=(2/3)^n cdot n^2.$ [not $+$ as you have. Also you seem to have dropped lim somewhere, and do you mean $lim_{n to infty}$ at top?
$endgroup$
– coffeemath
Jan 26 at 0:43