Is “$limlimits_n to inftyf(x_0+frac1n)=l$” another way of expressing the right-sided limit?Limit $mathop lim limits_n to infty fracnleft( a_1…a_n right)^frac1na_1 + … + a_n$Evaluating $lim limits_x to inftyleft(fracf(x_0 + frac 1x)f(x_0 - frac 1x)right)^x$How can I prove $lim limits_n to inftyleft(1+frac1a_nright)^a_n=e$ without involving function limit?why is $limlimits_ktoinftyfrac1+kk^k=0?$limit $ lim limits_n to infty left(fracz^1/sqrt n + z^-1/sqrt n2right)^n $Prove $limlimits_x to +infty fracf(x)x = limlimits_x to +infty f'(x)$ if both limits existDoes, $mathop lim limits_x to +infty f'(x) = + infty Leftrightarrow mathop lim limits_x to +infty fracf(x)x = + infty $?Formal limit definition in $limlimits_xto+infty fracx^2+29x=+infty$How can I prove that $limlimits_x to x_0 left|fracf(x)g(x)right| = infty$ when working with punctured neighborhoods?Calculate the limit $lim limits_ x to infty left(fracx^2+1x-1right)$
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Is “$limlimits_n to inftyf(x_0+frac1n)=l$” another way of expressing the right-sided limit?
Limit $mathop lim limits_n to infty fracnleft( a_1…a_n right)^frac1na_1 + … + a_n$Evaluating $lim limits_x to inftyleft(fracf(x_0 + frac 1x)f(x_0 - frac 1x)right)^x$How can I prove $lim limits_n to inftyleft(1+frac1a_nright)^a_n=e$ without involving function limit?why is $limlimits_ktoinftyfrac1+kk^k=0?$limit $ lim limits_n to infty left(fracz^1/sqrt n + z^-1/sqrt n2right)^n $Prove $limlimits_x to +infty fracf(x)x = limlimits_x to +infty f'(x)$ if both limits existDoes, $mathop lim limits_x to +infty f'(x) = + infty Leftrightarrow mathop lim limits_x to +infty fracf(x)x = + infty $?Formal limit definition in $limlimits_xto+infty fracx^2+29x=+infty$How can I prove that $limlimits_x to x_0 left|fracf(x)g(x)right| = infty$ when working with punctured neighborhoods?Calculate the limit $lim limits_ x to infty left(fracx^2+1x-1right)$
$begingroup$
Let $f:mathbbR to mathbbR$. Can we say that $limlimits_n to inftyf(x_0+frac1n)=l$ is another way of expressing the right-sided limit at $x_0$?
I tried to use the definition of the limit,but I am stuck.Intuitively, it seems true, but I don't know how to prove it.
real-analysis limits definition
edited Apr 19 at 12:53
user21820
40.4k544163
asked Apr 19 at 10:24
Math GuyMath Guy
1657
$endgroup$
add a comment |
$begingroup$
Let $f:mathbbR to mathbbR$. Can we say that $limlimits_n to inftyf(x_0+frac1n)=l$ is another way of expressing the right-sided limit at $x_0$?
I tried to use the definition of the limit,but I am stuck.Intuitively, it seems true, but I don't know how to prove it.
real-analysis limits definition
edited Apr 19 at 12:53
user21820
40.4k544163
asked Apr 19 at 10:24
Math GuyMath Guy
1657
$endgroup$
add a comment |
$begingroup$
Let $f:mathbbR to mathbbR$. Can we say that $limlimits_n to inftyf(x_0+frac1n)=l$ is another way of expressing the right-sided limit at $x_0$?
I tried to use the definition of the limit,but I am stuck.Intuitively, it seems true, but I don't know how to prove it.
real-analysis limits definition
edited Apr 19 at 12:53
user21820
40.4k544163
asked Apr 19 at 10:24
Math GuyMath Guy
1657
$endgroup$
Let $f:mathbbR to mathbbR$. Can we say that $limlimits_n to inftyf(x_0+frac1n)=l$ is another way of expressing the right-sided limit at $x_0$?
I tried to use the definition of the limit,but I am stuck.Intuitively, it seems true, but I don't know how to prove it.
real-analysis limits definition
real-analysis limits definition
edited Apr 19 at 12:53
user21820
40.4k544163
asked Apr 19 at 10:24
Math GuyMath Guy
1657
edited Apr 19 at 12:53
user21820
40.4k544163
asked Apr 19 at 10:24
Math GuyMath Guy
1657
edited Apr 19 at 12:53
user21820
40.4k544163
edited Apr 19 at 12:53
user21820
40.4k544163
edited Apr 19 at 12:53
user21820
40.4k544163
40.4k544163
asked Apr 19 at 10:24
Math GuyMath Guy
1657
asked Apr 19 at 10:24
Math GuyMath Guy
1657
asked Apr 19 at 10:24
Math GuyMath Guy
1657
1657
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, it is not true. Take, for instance,$$f(x)=begincasessinleft(fracpi xright)&text if xneq0\0&text if x=0.endcases$$Then the limit $lim_xto0^+f(x)$ doesn't exist, in spite of the fact that $lim_ntoinftyfleft(frac1nright)=0$.
answered Apr 19 at 10:27
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
2
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
Apr 19 at 10:35
3
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
Apr 19 at 10:39
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
Apr 19 at 11:42
2
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
Apr 19 at 11:47
add a comment |
$begingroup$
No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.
answered Apr 19 at 10:27
Kavi Rama MurthyKavi Rama Murthy
78.2k53571
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
No, it is not true. Take, for instance,$$f(x)=begincasessinleft(fracpi xright)&text if xneq0\0&text if x=0.endcases$$Then the limit $lim_xto0^+f(x)$ doesn't exist, in spite of the fact that $lim_ntoinftyfleft(frac1nright)=0$.
answered Apr 19 at 10:27
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
2
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
Apr 19 at 10:35
3
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
Apr 19 at 10:39
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
Apr 19 at 11:42
2
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
Apr 19 at 11:47
add a comment |
$begingroup$
No, it is not true. Take, for instance,$$f(x)=begincasessinleft(fracpi xright)&text if xneq0\0&text if x=0.endcases$$Then the limit $lim_xto0^+f(x)$ doesn't exist, in spite of the fact that $lim_ntoinftyfleft(frac1nright)=0$.
answered Apr 19 at 10:27
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
2
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
Apr 19 at 10:35
3
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
Apr 19 at 10:39
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
Apr 19 at 11:42
2
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
Apr 19 at 11:47
add a comment |
$begingroup$
No, it is not true. Take, for instance,$$f(x)=begincasessinleft(fracpi xright)&text if xneq0\0&text if x=0.endcases$$Then the limit $lim_xto0^+f(x)$ doesn't exist, in spite of the fact that $lim_ntoinftyfleft(frac1nright)=0$.
answered Apr 19 at 10:27
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
No, it is not true. Take, for instance,$$f(x)=begincasessinleft(fracpi xright)&text if xneq0\0&text if x=0.endcases$$Then the limit $lim_xto0^+f(x)$ doesn't exist, in spite of the fact that $lim_ntoinftyfleft(frac1nright)=0$.
answered Apr 19 at 10:27
José Carlos SantosJosé Carlos Santos
178k24139251
edited Apr 19 at 20:51
answered Apr 19 at 10:27
José Carlos SantosJosé Carlos Santos
178k24139251
answered Apr 19 at 10:27
José Carlos SantosJosé Carlos Santos
178k24139251
answered Apr 19 at 10:27
José Carlos SantosJosé Carlos Santos
178k24139251
178k24139251
2
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
Apr 19 at 10:35
3
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
Apr 19 at 10:39
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
Apr 19 at 11:42
2
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
Apr 19 at 11:47
add a comment |
2
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
Apr 19 at 10:35
3
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
Apr 19 at 10:39
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
Apr 19 at 11:42
2
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
Apr 19 at 11:47
2
2
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
Apr 19 at 10:35
$begingroup$
Who said that $n$ was an integer ? ;-)
$endgroup$
– Yves Daoust
Apr 19 at 10:35
3
3
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
Apr 19 at 10:39
$begingroup$
LOL. This is why I prefer the notation for limits that we use in my country: the limit of a sequence $(a_n)_ninmathbb N$ is denoted by $lim_ninmathbb Na_n$, instead of $lim_ntoinftya_n$. So, there is no ambiguity.
$endgroup$
– José Carlos Santos
Apr 19 at 10:39
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
Apr 19 at 11:42
$begingroup$
@YvesDaoust it is an integer. In my country, integers are commonly denoted by $n$, this is why I forgot that this may not be the case everywhere.
$endgroup$
– Math Guy
Apr 19 at 11:42
2
2
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
Apr 19 at 11:47
$begingroup$
@MathGuy: in my country, "n" stands for "not-necessarily-natural" and is short for "nnn" :-)
$endgroup$
– Yves Daoust
Apr 19 at 11:47
add a comment |
$begingroup$
No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.
answered Apr 19 at 10:27
Kavi Rama MurthyKavi Rama Murthy
78.2k53571
$endgroup$
add a comment |
$begingroup$
No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.
answered Apr 19 at 10:27
Kavi Rama MurthyKavi Rama Murthy
78.2k53571
$endgroup$
add a comment |
$begingroup$
No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.
answered Apr 19 at 10:27
Kavi Rama MurthyKavi Rama Murthy
78.2k53571
$endgroup$
No. Right hand limit of $f$ at $x_0$ is $l$ if $f(x_0+r_n) to l$ for every sequence $(r_n)$ decreasing to $0$. A particular sequence will not do. For example, take $f(x)=0$ for $x$ rational and $1$ for $x$ irrational. Take $x_0=0$.
answered Apr 19 at 10:27
Kavi Rama MurthyKavi Rama Murthy
78.2k53571
answered Apr 19 at 10:27
Kavi Rama MurthyKavi Rama Murthy
78.2k53571
answered Apr 19 at 10:27
Kavi Rama MurthyKavi Rama Murthy
78.2k53571
answered Apr 19 at 10:27
Kavi Rama MurthyKavi Rama Murthy
78.2k53571
78.2k53571
add a comment |
add a comment |
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