If $lim f(x) = 0,$ then $lim 1/f(x) = infty.$
$begingroup$
Suppose that $f:DtoBbb R$, where $D$ is a subset of $Bbb R$ and $a$ is an accumulation point of $D$, $lim_{xto a+}f(x)=0$, and $f(x)ne0$ for any $x$ in $D$ in some neighborhood of $a$.
I understand from the following proof that if $lim f(x) = 0,$ then $lim 1/|f(x)| = infty.$ : proof for |f(x)|,
however, does it also apply for $lim 1/f(x)= infty$ ? or $lim 1/f(x)= -infty$ If not then why?
limits
asked Dec 5 '18 at 11:18
Yuki1112Yuki1112
174
$endgroup$
|
show 2 more comments
$begingroup$
Suppose that $f:DtoBbb R$, where $D$ is a subset of $Bbb R$ and $a$ is an accumulation point of $D$, $lim_{xto a+}f(x)=0$, and $f(x)ne0$ for any $x$ in $D$ in some neighborhood of $a$.
I understand from the following proof that if $lim f(x) = 0,$ then $lim 1/|f(x)| = infty.$ : proof for |f(x)|,
however, does it also apply for $lim 1/f(x)= infty$ ? or $lim 1/f(x)= -infty$ If not then why?
limits
asked Dec 5 '18 at 11:18
Yuki1112Yuki1112
174
$endgroup$
3
$begingroup$
What if $f$ only takes negative values?
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:21
2
$begingroup$
the problem is whether you consider $infty$ to be either $pm infty$ or only $+infty$
$endgroup$
– TheD0ubleT
Dec 5 '18 at 11:22
$begingroup$
What if $f$ takes both positive and negative values ($f$ would not be continuous)?
$endgroup$
– Michael Burr
Dec 5 '18 at 11:23
$begingroup$
@MichaelBurr how can f take both of those, conjure to 0 and then not be continuous? when it's $lim_{xto a+}$?
$endgroup$
– Yuki1112
Dec 5 '18 at 11:25
$begingroup$
@TheD0ubleT either, sorry, I edited it to be more clear on the matter
$endgroup$
– Yuki1112
Dec 5 '18 at 11:31
|
show 2 more comments
$begingroup$
Suppose that $f:DtoBbb R$, where $D$ is a subset of $Bbb R$ and $a$ is an accumulation point of $D$, $lim_{xto a+}f(x)=0$, and $f(x)ne0$ for any $x$ in $D$ in some neighborhood of $a$.
I understand from the following proof that if $lim f(x) = 0,$ then $lim 1/|f(x)| = infty.$ : proof for |f(x)|,
however, does it also apply for $lim 1/f(x)= infty$ ? or $lim 1/f(x)= -infty$ If not then why?
limits
asked Dec 5 '18 at 11:18
Yuki1112Yuki1112
174
$endgroup$
Suppose that $f:DtoBbb R$, where $D$ is a subset of $Bbb R$ and $a$ is an accumulation point of $D$, $lim_{xto a+}f(x)=0$, and $f(x)ne0$ for any $x$ in $D$ in some neighborhood of $a$.
I understand from the following proof that if $lim f(x) = 0,$ then $lim 1/|f(x)| = infty.$ : proof for |f(x)|,
however, does it also apply for $lim 1/f(x)= infty$ ? or $lim 1/f(x)= -infty$ If not then why?
limits
limits
asked Dec 5 '18 at 11:18
Yuki1112Yuki1112
174
asked Dec 5 '18 at 11:18
Yuki1112Yuki1112
174
edited Dec 5 '18 at 11:23
Yuki1112
asked Dec 5 '18 at 11:18
Yuki1112Yuki1112
174
asked Dec 5 '18 at 11:18
Yuki1112Yuki1112
174
asked Dec 5 '18 at 11:18
Yuki1112Yuki1112
174
174
3
$begingroup$
What if $f$ only takes negative values?
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:21
2
$begingroup$
the problem is whether you consider $infty$ to be either $pm infty$ or only $+infty$
$endgroup$
– TheD0ubleT
Dec 5 '18 at 11:22
$begingroup$
What if $f$ takes both positive and negative values ($f$ would not be continuous)?
$endgroup$
– Michael Burr
Dec 5 '18 at 11:23
$begingroup$
@MichaelBurr how can f take both of those, conjure to 0 and then not be continuous? when it's $lim_{xto a+}$?
$endgroup$
– Yuki1112
Dec 5 '18 at 11:25
$begingroup$
@TheD0ubleT either, sorry, I edited it to be more clear on the matter
$endgroup$
– Yuki1112
Dec 5 '18 at 11:31
|
show 2 more comments
3
$begingroup$
What if $f$ only takes negative values?
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:21
2
$begingroup$
the problem is whether you consider $infty$ to be either $pm infty$ or only $+infty$
$endgroup$
– TheD0ubleT
Dec 5 '18 at 11:22
$begingroup$
What if $f$ takes both positive and negative values ($f$ would not be continuous)?
$endgroup$
– Michael Burr
Dec 5 '18 at 11:23
$begingroup$
@MichaelBurr how can f take both of those, conjure to 0 and then not be continuous? when it's $lim_{xto a+}$?
$endgroup$
– Yuki1112
Dec 5 '18 at 11:25
$begingroup$
@TheD0ubleT either, sorry, I edited it to be more clear on the matter
$endgroup$
– Yuki1112
Dec 5 '18 at 11:31
3
3
$begingroup$
What if $f$ only takes negative values?
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:21
$begingroup$
What if $f$ only takes negative values?
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:21
2
2
$begingroup$
the problem is whether you consider $infty$ to be either $pm infty$ or only $+infty$
$endgroup$
– TheD0ubleT
Dec 5 '18 at 11:22
$begingroup$
the problem is whether you consider $infty$ to be either $pm infty$ or only $+infty$
$endgroup$
– TheD0ubleT
Dec 5 '18 at 11:22
$begingroup$
What if $f$ takes both positive and negative values ($f$ would not be continuous)?
$endgroup$
– Michael Burr
Dec 5 '18 at 11:23
$begingroup$
What if $f$ takes both positive and negative values ($f$ would not be continuous)?
$endgroup$
– Michael Burr
Dec 5 '18 at 11:23
$begingroup$
@MichaelBurr how can f take both of those, conjure to 0 and then not be continuous? when it's $lim_{xto a+}$?
$endgroup$
– Yuki1112
Dec 5 '18 at 11:25
$begingroup$
@MichaelBurr how can f take both of those, conjure to 0 and then not be continuous? when it's $lim_{xto a+}$?
$endgroup$
– Yuki1112
Dec 5 '18 at 11:25
$begingroup$
@TheD0ubleT either, sorry, I edited it to be more clear on the matter
$endgroup$
– Yuki1112
Dec 5 '18 at 11:31
$begingroup$
@TheD0ubleT either, sorry, I edited it to be more clear on the matter
$endgroup$
– Yuki1112
Dec 5 '18 at 11:31
|
show 2 more comments
0
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3
$begingroup$
What if $f$ only takes negative values?
$endgroup$
– José Carlos Santos
Dec 5 '18 at 11:21
2
$begingroup$
the problem is whether you consider $infty$ to be either $pm infty$ or only $+infty$
$endgroup$
– TheD0ubleT
Dec 5 '18 at 11:22
$begingroup$
What if $f$ takes both positive and negative values ($f$ would not be continuous)?
$endgroup$
– Michael Burr
Dec 5 '18 at 11:23
$begingroup$
@MichaelBurr how can f take both of those, conjure to 0 and then not be continuous? when it's $lim_{xto a+}$?
$endgroup$
– Yuki1112
Dec 5 '18 at 11:25
$begingroup$
@TheD0ubleT either, sorry, I edited it to be more clear on the matter
$endgroup$
– Yuki1112
Dec 5 '18 at 11:31