Proof from definition that $lim_{xto infty} e^{-x^2}e^{2x}=0$
I'm trying to give a proof from definition that $lim_{xto infty} e^{-x^2}e^{2x}=0$ I know $|e^{-x^2}e^{2x}|<|e^{2x}|$. From there trying to get $|e^{2x}|<varepsilon$, but that isn't useful for finding something to have x greater than so clearly that's not the way to go about it. Any help is appreciated.
real-analysis limits
asked Dec 11 '18 at 4:09
Henry Mullen
162
add a comment |
I'm trying to give a proof from definition that $lim_{xto infty} e^{-x^2}e^{2x}=0$ I know $|e^{-x^2}e^{2x}|<|e^{2x}|$. From there trying to get $|e^{2x}|<varepsilon$, but that isn't useful for finding something to have x greater than so clearly that's not the way to go about it. Any help is appreciated.
real-analysis limits
asked Dec 11 '18 at 4:09
Henry Mullen
162
add a comment |
I'm trying to give a proof from definition that $lim_{xto infty} e^{-x^2}e^{2x}=0$ I know $|e^{-x^2}e^{2x}|<|e^{2x}|$. From there trying to get $|e^{2x}|<varepsilon$, but that isn't useful for finding something to have x greater than so clearly that's not the way to go about it. Any help is appreciated.
real-analysis limits
asked Dec 11 '18 at 4:09
Henry Mullen
162
I'm trying to give a proof from definition that $lim_{xto infty} e^{-x^2}e^{2x}=0$ I know $|e^{-x^2}e^{2x}|<|e^{2x}|$. From there trying to get $|e^{2x}|<varepsilon$, but that isn't useful for finding something to have x greater than so clearly that's not the way to go about it. Any help is appreciated.
real-analysis limits
real-analysis limits
asked Dec 11 '18 at 4:09
Henry Mullen
162
asked Dec 11 '18 at 4:09
Henry Mullen
162
asked Dec 11 '18 at 4:09
Henry Mullen
162
asked Dec 11 '18 at 4:09
Henry Mullen
162
asked Dec 11 '18 at 4:09
Henry Mullen
162
162
add a comment |
add a comment |
1 Answer
1
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$e^{-x^2}e^{2x}=e cdot e^{-(x-1)^2}$ and, with power series, $e^{(x-1)^2} ge (x-1)^2.$ Thus
$ 0 le e^{-x^2}e^{2x} le frac{e}{(x-1)^2}$.
Can you proceed ?
answered Dec 11 '18 at 4:16
Fred
44.2k1845
OP is asking $ epsilon - $ $delta$ proof
– deleteprofile
Dec 11 '18 at 4:18
3
I know. Make a $ epsilon - delta$ proof to show that $frac{e}{(x-1)^2} to 0$ as $x to infty$ and you are done !
– Fred
Dec 11 '18 at 4:21
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$e^{-x^2}e^{2x}=e cdot e^{-(x-1)^2}$ and, with power series, $e^{(x-1)^2} ge (x-1)^2.$ Thus
$ 0 le e^{-x^2}e^{2x} le frac{e}{(x-1)^2}$.
Can you proceed ?
answered Dec 11 '18 at 4:16
Fred
44.2k1845
OP is asking $ epsilon - $ $delta$ proof
– deleteprofile
Dec 11 '18 at 4:18
3
I know. Make a $ epsilon - delta$ proof to show that $frac{e}{(x-1)^2} to 0$ as $x to infty$ and you are done !
– Fred
Dec 11 '18 at 4:21
add a comment |
$e^{-x^2}e^{2x}=e cdot e^{-(x-1)^2}$ and, with power series, $e^{(x-1)^2} ge (x-1)^2.$ Thus
$ 0 le e^{-x^2}e^{2x} le frac{e}{(x-1)^2}$.
Can you proceed ?
answered Dec 11 '18 at 4:16
Fred
44.2k1845
OP is asking $ epsilon - $ $delta$ proof
– deleteprofile
Dec 11 '18 at 4:18
3
I know. Make a $ epsilon - delta$ proof to show that $frac{e}{(x-1)^2} to 0$ as $x to infty$ and you are done !
– Fred
Dec 11 '18 at 4:21
add a comment |
$e^{-x^2}e^{2x}=e cdot e^{-(x-1)^2}$ and, with power series, $e^{(x-1)^2} ge (x-1)^2.$ Thus
$ 0 le e^{-x^2}e^{2x} le frac{e}{(x-1)^2}$.
Can you proceed ?
answered Dec 11 '18 at 4:16
Fred
44.2k1845
$e^{-x^2}e^{2x}=e cdot e^{-(x-1)^2}$ and, with power series, $e^{(x-1)^2} ge (x-1)^2.$ Thus
$ 0 le e^{-x^2}e^{2x} le frac{e}{(x-1)^2}$.
Can you proceed ?
answered Dec 11 '18 at 4:16
Fred
44.2k1845
edited Dec 11 '18 at 6:44
answered Dec 11 '18 at 4:16
Fred
44.2k1845
answered Dec 11 '18 at 4:16
Fred
44.2k1845
answered Dec 11 '18 at 4:16
Fred
44.2k1845
44.2k1845
OP is asking $ epsilon - $ $delta$ proof
– deleteprofile
Dec 11 '18 at 4:18
3
I know. Make a $ epsilon - delta$ proof to show that $frac{e}{(x-1)^2} to 0$ as $x to infty$ and you are done !
– Fred
Dec 11 '18 at 4:21
add a comment |
OP is asking $ epsilon - $ $delta$ proof
– deleteprofile
Dec 11 '18 at 4:18
3
I know. Make a $ epsilon - delta$ proof to show that $frac{e}{(x-1)^2} to 0$ as $x to infty$ and you are done !
– Fred
Dec 11 '18 at 4:21
OP is asking $ epsilon - $ $delta$ proof
– deleteprofile
Dec 11 '18 at 4:18
OP is asking $ epsilon - $ $delta$ proof
– deleteprofile
Dec 11 '18 at 4:18
3
3
I know. Make a $ epsilon - delta$ proof to show that $frac{e}{(x-1)^2} to 0$ as $x to infty$ and you are done !
– Fred
Dec 11 '18 at 4:21
I know. Make a $ epsilon - delta$ proof to show that $frac{e}{(x-1)^2} to 0$ as $x to infty$ and you are done !
– Fred
Dec 11 '18 at 4:21
add a comment |
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