Mixing
Limit of $frac{1}{7}e^{-2x^2}(1-4x^2)$ as $xtoinfty$
$begingroup$
I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)
I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$
limits exponential-function
edited Jan 4 at 7:51
Asaf Karagila♦
302k32429760
asked Jan 3 at 15:25
B. CzostekB. Czostek
506
$endgroup$
add a comment |
$begingroup$
I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)
I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$
limits exponential-function
edited Jan 4 at 7:51
Asaf Karagila♦
302k32429760
asked Jan 3 at 15:25
B. CzostekB. Czostek
506
$endgroup$
6
$begingroup$
Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
$endgroup$
– Doug M
Jan 3 at 15:45
$begingroup$
Is $*$ multiplication or convolution in your question?
$endgroup$
– Asaf Karagila♦
Jan 4 at 7:50
add a comment |
$begingroup$
I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)
I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$
limits exponential-function
edited Jan 4 at 7:51
Asaf Karagila♦
302k32429760
asked Jan 3 at 15:25
B. CzostekB. Czostek
506
$endgroup$
I calculated the derivative of $frac{x}{7}*e^{-2x^2}$ and got $frac{1}{7}e^{-2x^2}(1-4x^2)$ (I included it cause if I got that wrong calculating the rest is pointless)
I don't know how to find the limit of this function:
$$frac{1}{7}e^{-2x^2}(1-4x^2)$$ I tried splitting it into two but I still don't know how to handle this$$lim_{x to infty} frac{1}{7}e^{-2x^2}+lim_{x to infty} frac{1}{7}xe^{-2x^2}*(-4x) $$
limits exponential-function
limits exponential-function
edited Jan 4 at 7:51
Asaf Karagila♦
302k32429760
asked Jan 3 at 15:25
B. CzostekB. Czostek
506
edited Jan 4 at 7:51
Asaf Karagila♦
302k32429760
asked Jan 3 at 15:25
B. CzostekB. Czostek
506
edited Jan 4 at 7:51
Asaf Karagila♦
302k32429760
edited Jan 4 at 7:51
Asaf Karagila♦
302k32429760
edited Jan 4 at 7:51
Asaf Karagila♦
302k32429760
302k32429760
asked Jan 3 at 15:25
B. CzostekB. Czostek
506
asked Jan 3 at 15:25
B. CzostekB. Czostek
506
asked Jan 3 at 15:25
B. CzostekB. Czostek
506
506
6
$begingroup$
Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
$endgroup$
– Doug M
Jan 3 at 15:45
$begingroup$
Is $*$ multiplication or convolution in your question?
$endgroup$
– Asaf Karagila♦
Jan 4 at 7:50
add a comment |
6
$begingroup$
Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
$endgroup$
– Doug M
Jan 3 at 15:45
$begingroup$
Is $*$ multiplication or convolution in your question?
$endgroup$
– Asaf Karagila♦
Jan 4 at 7:50
6
6
$begingroup$
Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
$endgroup$
– Doug M
Jan 3 at 15:45
$begingroup$
Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
$endgroup$
– Doug M
Jan 3 at 15:45
$begingroup$
Is $*$ multiplication or convolution in your question?
$endgroup$
– Asaf Karagila♦
Jan 4 at 7:50
$begingroup$
Is $*$ multiplication or convolution in your question?
$endgroup$
– Asaf Karagila♦
Jan 4 at 7:50
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Using L'Hôpital's rule , we get
begin{align}
lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
&=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
end{align}.
answered Jan 3 at 15:32
Thomas ShelbyThomas Shelby
2,235220
$endgroup$
2
$begingroup$
L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
$endgroup$
– Andreas Rejbrand
Jan 4 at 7:01
add a comment |
$begingroup$
Hint: Rewrite the expression as
$$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$
Now, notice the growth of the numerator and denominator. Which grows more quickly?
answered Jan 3 at 16:00
KM101KM101
5,9431523
$endgroup$
add a comment |
$begingroup$
Your derivative is correct.
Then arrange it as:
$$frac{1-4x^2}{7e^{2x^2}}$$
Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$
answered Jan 3 at 16:49
Rhys HughesRhys Hughes
5,2791427
$endgroup$
add a comment |
$begingroup$
Note that for $xge0$,
$$
begin{align}
e^x
&=1+x+frac{x^2}2+dots\
&gefrac{x^2}2tag1
end{align}
$$
Thus,
$$
e^{-2x^2}lefrac1{2x^4}tag2
$$
Applying $(2)$ to the expression for the derivative gives
$$
begin{align}
left|frac17e^{-2x^2}!!left(1-4x^2right)right|
&lefrac{4x^2+1}7frac1{2x^4}\
&=frac2{7x^2}+frac1{14x^4}tag3
end{align}
$$
answered Jan 3 at 20:10
robjohn♦robjohn
266k27304626
$endgroup$
add a comment |
$begingroup$
You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$
I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
$$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.
answered Jan 3 at 15:42
AndrewAndrew
346115
$endgroup$
$begingroup$
Thank you I completely forgot the L'Hopitals Rule
$endgroup$
– B. Czostek
Jan 3 at 16:04
$begingroup$
@B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
$endgroup$
– Ethan Bolker
Jan 4 at 1:34
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using L'Hôpital's rule , we get
begin{align}
lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
&=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
end{align}.
answered Jan 3 at 15:32
Thomas ShelbyThomas Shelby
2,235220
$endgroup$
2
$begingroup$
L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
$endgroup$
– Andreas Rejbrand
Jan 4 at 7:01
add a comment |
$begingroup$
Using L'Hôpital's rule , we get
begin{align}
lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
&=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
end{align}.
answered Jan 3 at 15:32
Thomas ShelbyThomas Shelby
2,235220
$endgroup$
2
$begingroup$
L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
$endgroup$
– Andreas Rejbrand
Jan 4 at 7:01
add a comment |
$begingroup$
Using L'Hôpital's rule , we get
begin{align}
lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
&=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
end{align}.
answered Jan 3 at 15:32
Thomas ShelbyThomas Shelby
2,235220
$endgroup$
Using L'Hôpital's rule , we get
begin{align}
lim_{xto infty}frac{1-4x^2}{7e^{2x^2}}&=lim_{xto infty}frac{-8x}{28cdot xe^{2x^2}}\
&=lim_{xto infty}frac{-8}{28e^{2x^2}}=0
end{align}.
answered Jan 3 at 15:32
Thomas ShelbyThomas Shelby
2,235220
edited Jan 3 at 15:39
answered Jan 3 at 15:32
Thomas ShelbyThomas Shelby
2,235220
answered Jan 3 at 15:32
Thomas ShelbyThomas Shelby
2,235220
answered Jan 3 at 15:32
Thomas ShelbyThomas Shelby
2,235220
2,235220
2
$begingroup$
L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
$endgroup$
– Andreas Rejbrand
Jan 4 at 7:01
add a comment |
2
$begingroup$
L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
$endgroup$
– Andreas Rejbrand
Jan 4 at 7:01
2
2
$begingroup$
L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
$endgroup$
– Andreas Rejbrand
Jan 4 at 7:01
$begingroup$
L'Hôpital feels overkill for this problem (and, indeed, for most similar problems). KM101's solution is much simpler.
$endgroup$
– Andreas Rejbrand
Jan 4 at 7:01
add a comment |
$begingroup$
Hint: Rewrite the expression as
$$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$
Now, notice the growth of the numerator and denominator. Which grows more quickly?
answered Jan 3 at 16:00
KM101KM101
5,9431523
$endgroup$
add a comment |
$begingroup$
Hint: Rewrite the expression as
$$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$
Now, notice the growth of the numerator and denominator. Which grows more quickly?
answered Jan 3 at 16:00
KM101KM101
5,9431523
$endgroup$
add a comment |
$begingroup$
Hint: Rewrite the expression as
$$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$
Now, notice the growth of the numerator and denominator. Which grows more quickly?
answered Jan 3 at 16:00
KM101KM101
5,9431523
$endgroup$
Hint: Rewrite the expression as
$$frac{1}{7}e^{-2x^2}left(1-4x^2right) = frac{1-4x^2}{7e^{2x^2}}$$
Now, notice the growth of the numerator and denominator. Which grows more quickly?
answered Jan 3 at 16:00
KM101KM101
5,9431523
edited Jan 3 at 17:17
answered Jan 3 at 16:00
KM101KM101
5,9431523
answered Jan 3 at 16:00
KM101KM101
5,9431523
answered Jan 3 at 16:00
KM101KM101
5,9431523
5,9431523
add a comment |
add a comment |
$begingroup$
Your derivative is correct.
Then arrange it as:
$$frac{1-4x^2}{7e^{2x^2}}$$
Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$
answered Jan 3 at 16:49
Rhys HughesRhys Hughes
5,2791427
$endgroup$
add a comment |
$begingroup$
Your derivative is correct.
Then arrange it as:
$$frac{1-4x^2}{7e^{2x^2}}$$
Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$
answered Jan 3 at 16:49
Rhys HughesRhys Hughes
5,2791427
$endgroup$
add a comment |
$begingroup$
Your derivative is correct.
Then arrange it as:
$$frac{1-4x^2}{7e^{2x^2}}$$
Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$
answered Jan 3 at 16:49
Rhys HughesRhys Hughes
5,2791427
$endgroup$
Your derivative is correct.
Then arrange it as:
$$frac{1-4x^2}{7e^{2x^2}}$$
Use that, due to exponentiation having a much greater effect than indices, $e^{2x^2}>>4x^2$ for sufficiently large $x$, to show this limit is very clearly $0$
answered Jan 3 at 16:49
Rhys HughesRhys Hughes
5,2791427
answered Jan 3 at 16:49
Rhys HughesRhys Hughes
5,2791427
answered Jan 3 at 16:49
Rhys HughesRhys Hughes
5,2791427
answered Jan 3 at 16:49
Rhys HughesRhys Hughes
5,2791427
5,2791427
add a comment |
add a comment |
$begingroup$
Note that for $xge0$,
$$
begin{align}
e^x
&=1+x+frac{x^2}2+dots\
&gefrac{x^2}2tag1
end{align}
$$
Thus,
$$
e^{-2x^2}lefrac1{2x^4}tag2
$$
Applying $(2)$ to the expression for the derivative gives
$$
begin{align}
left|frac17e^{-2x^2}!!left(1-4x^2right)right|
&lefrac{4x^2+1}7frac1{2x^4}\
&=frac2{7x^2}+frac1{14x^4}tag3
end{align}
$$
answered Jan 3 at 20:10
robjohn♦robjohn
266k27304626
$endgroup$
add a comment |
$begingroup$
Note that for $xge0$,
$$
begin{align}
e^x
&=1+x+frac{x^2}2+dots\
&gefrac{x^2}2tag1
end{align}
$$
Thus,
$$
e^{-2x^2}lefrac1{2x^4}tag2
$$
Applying $(2)$ to the expression for the derivative gives
$$
begin{align}
left|frac17e^{-2x^2}!!left(1-4x^2right)right|
&lefrac{4x^2+1}7frac1{2x^4}\
&=frac2{7x^2}+frac1{14x^4}tag3
end{align}
$$
answered Jan 3 at 20:10
robjohn♦robjohn
266k27304626
$endgroup$
add a comment |
$begingroup$
Note that for $xge0$,
$$
begin{align}
e^x
&=1+x+frac{x^2}2+dots\
&gefrac{x^2}2tag1
end{align}
$$
Thus,
$$
e^{-2x^2}lefrac1{2x^4}tag2
$$
Applying $(2)$ to the expression for the derivative gives
$$
begin{align}
left|frac17e^{-2x^2}!!left(1-4x^2right)right|
&lefrac{4x^2+1}7frac1{2x^4}\
&=frac2{7x^2}+frac1{14x^4}tag3
end{align}
$$
answered Jan 3 at 20:10
robjohn♦robjohn
266k27304626
$endgroup$
Note that for $xge0$,
$$
begin{align}
e^x
&=1+x+frac{x^2}2+dots\
&gefrac{x^2}2tag1
end{align}
$$
Thus,
$$
e^{-2x^2}lefrac1{2x^4}tag2
$$
Applying $(2)$ to the expression for the derivative gives
$$
begin{align}
left|frac17e^{-2x^2}!!left(1-4x^2right)right|
&lefrac{4x^2+1}7frac1{2x^4}\
&=frac2{7x^2}+frac1{14x^4}tag3
end{align}
$$
answered Jan 3 at 20:10
robjohn♦robjohn
266k27304626
answered Jan 3 at 20:10
robjohn♦robjohn
266k27304626
answered Jan 3 at 20:10
robjohn♦robjohn
266k27304626
answered Jan 3 at 20:10
robjohn♦robjohn
266k27304626
266k27304626
add a comment |
add a comment |
$begingroup$
You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$
I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
$$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.
answered Jan 3 at 15:42
AndrewAndrew
346115
$endgroup$
$begingroup$
Thank you I completely forgot the L'Hopitals Rule
$endgroup$
– B. Czostek
Jan 3 at 16:04
$begingroup$
@B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
$endgroup$
– Ethan Bolker
Jan 4 at 1:34
add a comment |
$begingroup$
You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$
I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
$$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.
answered Jan 3 at 15:42
AndrewAndrew
346115
$endgroup$
$begingroup$
Thank you I completely forgot the L'Hopitals Rule
$endgroup$
– B. Czostek
Jan 3 at 16:04
$begingroup$
@B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
$endgroup$
– Ethan Bolker
Jan 4 at 1:34
add a comment |
$begingroup$
You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$
I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
$$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.
answered Jan 3 at 15:42
AndrewAndrew
346115
$endgroup$
You pretty much have the solution. The following limit: $$frac{1}{7}lim_{xto infty}frac{1}{e^{2x^2}}=0.$$
I do not believe that was your problem. The last bit is what you were probably having trouble with. I would try u substitution. Let $u=x^2$ then we have the following:
$$-frac{4}{7} lim_{utoinfty}frac{u}{e^{2u}}=0 $$by L'Hopitals Rule.
answered Jan 3 at 15:42
AndrewAndrew
346115
answered Jan 3 at 15:42
AndrewAndrew
346115
answered Jan 3 at 15:42
AndrewAndrew
346115
answered Jan 3 at 15:42
AndrewAndrew
346115
346115
$begingroup$
Thank you I completely forgot the L'Hopitals Rule
$endgroup$
– B. Czostek
Jan 3 at 16:04
$begingroup$
@B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
$endgroup$
– Ethan Bolker
Jan 4 at 1:34
add a comment |
$begingroup$
Thank you I completely forgot the L'Hopitals Rule
$endgroup$
– B. Czostek
Jan 3 at 16:04
$begingroup$
@B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
$endgroup$
– Ethan Bolker
Jan 4 at 1:34
$begingroup$
Thank you I completely forgot the L'Hopitals Rule
$endgroup$
– B. Czostek
Jan 3 at 16:04
$begingroup$
Thank you I completely forgot the L'Hopitals Rule
$endgroup$
– B. Czostek
Jan 3 at 16:04
$begingroup$
@B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
$endgroup$
– Ethan Bolker
Jan 4 at 1:34
$begingroup$
@B.Czostek I think you'd be better off rarely remembering iL'Hopital. Most limit problems have easier and more informative solutions - as the other answers show for this one.
$endgroup$
– Ethan Bolker
Jan 4 at 1:34
add a comment |
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6
$begingroup$
Exponential growth is more powerful than polynomial growth. $e^{-2x^2}$ goes to $0$ more powerfully than $(1-4x^2)$ goes to $-infty$. Their product goes to $0.$
$endgroup$
– Doug M
Jan 3 at 15:45
$begingroup$
Is $*$ multiplication or convolution in your question?
$endgroup$
– Asaf Karagila♦
Jan 4 at 7:50