Proving that the limit of $1/x$ as $x$ approaches negative infinity equals $0$
0
$begingroup$
I am trying to prove that the limit of $1/x$ as $x to -infty$ equals $ 0$.
I get stuck in trying to find a proper epsilon. I know that it is supposed to be $-1/epsilon$ but I don't understand how to manipulate the inequality $epsilon > -1/x$.
calculus limits epsilon-delta
edited Oct 19 '18 at 1:13
HK Lee
14k52360
asked Jul 16 '14 at 3:51
Jonathan DuranJonathan Duran
112
$endgroup$
add a comment |
0
$begingroup$
I am trying to prove that the limit of $1/x$ as $x to -infty$ equals $ 0$.
I get stuck in trying to find a proper epsilon. I know that it is supposed to be $-1/epsilon$ but I don't understand how to manipulate the inequality $epsilon > -1/x$.
calculus limits epsilon-delta
edited Oct 19 '18 at 1:13
HK Lee
14k52360
asked Jul 16 '14 at 3:51
Jonathan DuranJonathan Duran
112
$endgroup$
add a comment |
0
0
0
$begingroup$
I am trying to prove that the limit of $1/x$ as $x to -infty$ equals $ 0$.
I get stuck in trying to find a proper epsilon. I know that it is supposed to be $-1/epsilon$ but I don't understand how to manipulate the inequality $epsilon > -1/x$.
calculus limits epsilon-delta
edited Oct 19 '18 at 1:13
HK Lee
14k52360
asked Jul 16 '14 at 3:51
Jonathan DuranJonathan Duran
112
$endgroup$
I am trying to prove that the limit of $1/x$ as $x to -infty$ equals $ 0$.
I get stuck in trying to find a proper epsilon. I know that it is supposed to be $-1/epsilon$ but I don't understand how to manipulate the inequality $epsilon > -1/x$.
calculus limits epsilon-delta
calculus limits epsilon-delta
edited Oct 19 '18 at 1:13
HK Lee
14k52360
asked Jul 16 '14 at 3:51
Jonathan DuranJonathan Duran
112
edited Oct 19 '18 at 1:13
HK Lee
14k52360
asked Jul 16 '14 at 3:51
Jonathan DuranJonathan Duran
112
edited Oct 19 '18 at 1:13
HK Lee
14k52360
edited Oct 19 '18 at 1:13
HK Lee
14k52360
edited Oct 19 '18 at 1:13
HK Lee
14k52360
14k52360
asked Jul 16 '14 at 3:51
Jonathan DuranJonathan Duran
112
asked Jul 16 '14 at 3:51
Jonathan DuranJonathan Duran
112
asked Jul 16 '14 at 3:51
Jonathan DuranJonathan Duran
112
112
add a comment |
add a comment |
1 Answer
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Fix $epsilon>0$. Choose $x<-N$ where $N$ is chosen so that $N>0$ and $frac{1}{N}<epsilon$. Note that $-x=lvert xrvert$. So $lvert xrvert>N$. Can you finish the argument from here?
answered Jul 16 '14 at 3:56
user71352user71352
11.4k21025
$endgroup$
$begingroup$
wouldn't it be 1/x < epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:25
$begingroup$
I'm sorry did I make a mistake? If you want to exclude the $N$ from the argument then yes you can choose $x<frac{-1}{epsilon}$ then $lvert xrvert=-x>frac{1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:26
$begingroup$
not sure but i can write out my scratch work and you can point out my mistake. x<N s.t |1/x|<e since x is less than 0, |1/x|=-1/x. So now i have -1/x<e which i rewrite as x>-1/e but that can't be write since i need -1/e t0 be bigger than x
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:34
$begingroup$
There appears to be an error in the inequality. Notice that if you want $frac{-1}{x}<epsilon$ then you need $frac{1}{x}>-epsilon$ and finally $x<frac{-1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:36
$begingroup$
glad to have located the error. i think I'm having trouble with the sign of inequalities when i divide with negative unknowns. if you have 1/x < -e … wouldn't you have x > -1/e since you divided by a negative epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:42
|
show 5 more comments
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1 Answer
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1 Answer
1
active
oldest
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oldest
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0
$begingroup$
Fix $epsilon>0$. Choose $x<-N$ where $N$ is chosen so that $N>0$ and $frac{1}{N}<epsilon$. Note that $-x=lvert xrvert$. So $lvert xrvert>N$. Can you finish the argument from here?
answered Jul 16 '14 at 3:56
user71352user71352
11.4k21025
$endgroup$
$begingroup$
wouldn't it be 1/x < epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:25
$begingroup$
I'm sorry did I make a mistake? If you want to exclude the $N$ from the argument then yes you can choose $x<frac{-1}{epsilon}$ then $lvert xrvert=-x>frac{1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:26
$begingroup$
not sure but i can write out my scratch work and you can point out my mistake. x<N s.t |1/x|<e since x is less than 0, |1/x|=-1/x. So now i have -1/x<e which i rewrite as x>-1/e but that can't be write since i need -1/e t0 be bigger than x
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:34
$begingroup$
There appears to be an error in the inequality. Notice that if you want $frac{-1}{x}<epsilon$ then you need $frac{1}{x}>-epsilon$ and finally $x<frac{-1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:36
$begingroup$
glad to have located the error. i think I'm having trouble with the sign of inequalities when i divide with negative unknowns. if you have 1/x < -e … wouldn't you have x > -1/e since you divided by a negative epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:42
|
show 5 more comments
0
$begingroup$
Fix $epsilon>0$. Choose $x<-N$ where $N$ is chosen so that $N>0$ and $frac{1}{N}<epsilon$. Note that $-x=lvert xrvert$. So $lvert xrvert>N$. Can you finish the argument from here?
answered Jul 16 '14 at 3:56
user71352user71352
11.4k21025
$endgroup$
$begingroup$
wouldn't it be 1/x < epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:25
$begingroup$
I'm sorry did I make a mistake? If you want to exclude the $N$ from the argument then yes you can choose $x<frac{-1}{epsilon}$ then $lvert xrvert=-x>frac{1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:26
$begingroup$
not sure but i can write out my scratch work and you can point out my mistake. x<N s.t |1/x|<e since x is less than 0, |1/x|=-1/x. So now i have -1/x<e which i rewrite as x>-1/e but that can't be write since i need -1/e t0 be bigger than x
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:34
$begingroup$
There appears to be an error in the inequality. Notice that if you want $frac{-1}{x}<epsilon$ then you need $frac{1}{x}>-epsilon$ and finally $x<frac{-1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:36
$begingroup$
glad to have located the error. i think I'm having trouble with the sign of inequalities when i divide with negative unknowns. if you have 1/x < -e … wouldn't you have x > -1/e since you divided by a negative epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:42
|
show 5 more comments
0
0
0
$begingroup$
Fix $epsilon>0$. Choose $x<-N$ where $N$ is chosen so that $N>0$ and $frac{1}{N}<epsilon$. Note that $-x=lvert xrvert$. So $lvert xrvert>N$. Can you finish the argument from here?
answered Jul 16 '14 at 3:56
user71352user71352
11.4k21025
$endgroup$
Fix $epsilon>0$. Choose $x<-N$ where $N$ is chosen so that $N>0$ and $frac{1}{N}<epsilon$. Note that $-x=lvert xrvert$. So $lvert xrvert>N$. Can you finish the argument from here?
answered Jul 16 '14 at 3:56
user71352user71352
11.4k21025
answered Jul 16 '14 at 3:56
user71352user71352
11.4k21025
answered Jul 16 '14 at 3:56
user71352user71352
11.4k21025
answered Jul 16 '14 at 3:56
user71352user71352
11.4k21025
11.4k21025
$begingroup$
wouldn't it be 1/x < epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:25
$begingroup$
I'm sorry did I make a mistake? If you want to exclude the $N$ from the argument then yes you can choose $x<frac{-1}{epsilon}$ then $lvert xrvert=-x>frac{1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:26
$begingroup$
not sure but i can write out my scratch work and you can point out my mistake. x<N s.t |1/x|<e since x is less than 0, |1/x|=-1/x. So now i have -1/x<e which i rewrite as x>-1/e but that can't be write since i need -1/e t0 be bigger than x
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:34
$begingroup$
There appears to be an error in the inequality. Notice that if you want $frac{-1}{x}<epsilon$ then you need $frac{1}{x}>-epsilon$ and finally $x<frac{-1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:36
$begingroup$
glad to have located the error. i think I'm having trouble with the sign of inequalities when i divide with negative unknowns. if you have 1/x < -e … wouldn't you have x > -1/e since you divided by a negative epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:42
|
show 5 more comments
$begingroup$
wouldn't it be 1/x < epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:25
$begingroup$
I'm sorry did I make a mistake? If you want to exclude the $N$ from the argument then yes you can choose $x<frac{-1}{epsilon}$ then $lvert xrvert=-x>frac{1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:26
$begingroup$
not sure but i can write out my scratch work and you can point out my mistake. x<N s.t |1/x|<e since x is less than 0, |1/x|=-1/x. So now i have -1/x<e which i rewrite as x>-1/e but that can't be write since i need -1/e t0 be bigger than x
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:34
$begingroup$
There appears to be an error in the inequality. Notice that if you want $frac{-1}{x}<epsilon$ then you need $frac{1}{x}>-epsilon$ and finally $x<frac{-1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:36
$begingroup$
glad to have located the error. i think I'm having trouble with the sign of inequalities when i divide with negative unknowns. if you have 1/x < -e … wouldn't you have x > -1/e since you divided by a negative epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:42
$begingroup$
wouldn't it be 1/x < epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:25
$begingroup$
wouldn't it be 1/x < epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:25
$begingroup$
I'm sorry did I make a mistake? If you want to exclude the $N$ from the argument then yes you can choose $x<frac{-1}{epsilon}$ then $lvert xrvert=-x>frac{1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:26
$begingroup$
I'm sorry did I make a mistake? If you want to exclude the $N$ from the argument then yes you can choose $x<frac{-1}{epsilon}$ then $lvert xrvert=-x>frac{1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:26
$begingroup$
not sure but i can write out my scratch work and you can point out my mistake. x<N s.t |1/x|<e since x is less than 0, |1/x|=-1/x. So now i have -1/x<e which i rewrite as x>-1/e but that can't be write since i need -1/e t0 be bigger than x
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:34
$begingroup$
not sure but i can write out my scratch work and you can point out my mistake. x<N s.t |1/x|<e since x is less than 0, |1/x|=-1/x. So now i have -1/x<e which i rewrite as x>-1/e but that can't be write since i need -1/e t0 be bigger than x
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:34
$begingroup$
There appears to be an error in the inequality. Notice that if you want $frac{-1}{x}<epsilon$ then you need $frac{1}{x}>-epsilon$ and finally $x<frac{-1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:36
$begingroup$
There appears to be an error in the inequality. Notice that if you want $frac{-1}{x}<epsilon$ then you need $frac{1}{x}>-epsilon$ and finally $x<frac{-1}{epsilon}$.
$endgroup$
– user71352
Jul 16 '14 at 4:36
$begingroup$
glad to have located the error. i think I'm having trouble with the sign of inequalities when i divide with negative unknowns. if you have 1/x < -e … wouldn't you have x > -1/e since you divided by a negative epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:42
$begingroup$
glad to have located the error. i think I'm having trouble with the sign of inequalities when i divide with negative unknowns. if you have 1/x < -e … wouldn't you have x > -1/e since you divided by a negative epsilon?
$endgroup$
– Jonathan Duran
Jul 16 '14 at 4:42
|
show 5 more comments
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