show that the limit does not exist: $lim_{xto0}frac{1}{x^2}$
$begingroup$
Show that the following limit does not exist:
$$lim_{xto 0}frac{1}{x^2}$$
$(x>0)$
The $delta$ - $varepsilon$ definition can be used to prove a given limit exists for some function at a particular point. My question is, can we prove the non-existence of a limit using the $delta$ - $varepsilon$ definition? (A little hint on how, if yes.)
Besides that, what (other) methods can we use?
[SIMILAR POST: Can we prove that there is no limit at $x=0$ for $f(x)=1/x$ using epsilon-delta definition? ]
real-analysis limits
asked Dec 20 '18 at 9:51
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
Show that the following limit does not exist:
$$lim_{xto 0}frac{1}{x^2}$$
$(x>0)$
The $delta$ - $varepsilon$ definition can be used to prove a given limit exists for some function at a particular point. My question is, can we prove the non-existence of a limit using the $delta$ - $varepsilon$ definition? (A little hint on how, if yes.)
Besides that, what (other) methods can we use?
[SIMILAR POST: Can we prove that there is no limit at $x=0$ for $f(x)=1/x$ using epsilon-delta definition? ]
real-analysis limits
asked Dec 20 '18 at 9:51
Za IraZa Ira
161115
$endgroup$
2
$begingroup$
It does exist though.
$endgroup$
– Rebellos
Dec 20 '18 at 9:59
1
$begingroup$
The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
$endgroup$
– Daniele Tampieri
Dec 20 '18 at 10:00
1
$begingroup$
Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:00
add a comment |
$begingroup$
Show that the following limit does not exist:
$$lim_{xto 0}frac{1}{x^2}$$
$(x>0)$
The $delta$ - $varepsilon$ definition can be used to prove a given limit exists for some function at a particular point. My question is, can we prove the non-existence of a limit using the $delta$ - $varepsilon$ definition? (A little hint on how, if yes.)
Besides that, what (other) methods can we use?
[SIMILAR POST: Can we prove that there is no limit at $x=0$ for $f(x)=1/x$ using epsilon-delta definition? ]
real-analysis limits
asked Dec 20 '18 at 9:51
Za IraZa Ira
161115
$endgroup$
Show that the following limit does not exist:
$$lim_{xto 0}frac{1}{x^2}$$
$(x>0)$
The $delta$ - $varepsilon$ definition can be used to prove a given limit exists for some function at a particular point. My question is, can we prove the non-existence of a limit using the $delta$ - $varepsilon$ definition? (A little hint on how, if yes.)
Besides that, what (other) methods can we use?
[SIMILAR POST: Can we prove that there is no limit at $x=0$ for $f(x)=1/x$ using epsilon-delta definition? ]
real-analysis limits
real-analysis limits
asked Dec 20 '18 at 9:51
Za IraZa Ira
161115
asked Dec 20 '18 at 9:51
Za IraZa Ira
161115
edited Dec 20 '18 at 10:14
Za Ira
asked Dec 20 '18 at 9:51
Za IraZa Ira
161115
asked Dec 20 '18 at 9:51
Za IraZa Ira
161115
asked Dec 20 '18 at 9:51
Za IraZa Ira
161115
161115
2
$begingroup$
It does exist though.
$endgroup$
– Rebellos
Dec 20 '18 at 9:59
1
$begingroup$
The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
$endgroup$
– Daniele Tampieri
Dec 20 '18 at 10:00
1
$begingroup$
Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:00
add a comment |
2
$begingroup$
It does exist though.
$endgroup$
– Rebellos
Dec 20 '18 at 9:59
1
$begingroup$
The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
$endgroup$
– Daniele Tampieri
Dec 20 '18 at 10:00
1
$begingroup$
Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:00
2
2
$begingroup$
It does exist though.
$endgroup$
– Rebellos
Dec 20 '18 at 9:59
$begingroup$
It does exist though.
$endgroup$
– Rebellos
Dec 20 '18 at 9:59
1
1
$begingroup$
The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
$endgroup$
– Daniele Tampieri
Dec 20 '18 at 10:00
$begingroup$
The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
$endgroup$
– Daniele Tampieri
Dec 20 '18 at 10:00
1
1
$begingroup$
Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:00
$begingroup$
Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:00
add a comment |
1 Answer
1
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oldest
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$begingroup$
What is asserted is that there is no finite limit. Prove by contradiction. Suppose $lim_{xto 0+}frac 1 {x^{2}}=L$ exists and is finite. Then we can find $delta >0$ such that $|frac 1 {x^{2}}-L|<1$ whenever $0<x<delta$. If $n$ is any sufficiently large integer then $0<frac 1 n<delta$ so, taking $x=frac 1 n$ we get $| n^{2}-L|<1$. This gives $n^{2} <L+1$ for any sufficiently large integer $n$ which is obviously a contradiction.
answered Dec 20 '18 at 10:00
Kavi Rama MurthyKavi Rama Murthy
72.8k53170
$endgroup$
add a comment |
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$begingroup$
What is asserted is that there is no finite limit. Prove by contradiction. Suppose $lim_{xto 0+}frac 1 {x^{2}}=L$ exists and is finite. Then we can find $delta >0$ such that $|frac 1 {x^{2}}-L|<1$ whenever $0<x<delta$. If $n$ is any sufficiently large integer then $0<frac 1 n<delta$ so, taking $x=frac 1 n$ we get $| n^{2}-L|<1$. This gives $n^{2} <L+1$ for any sufficiently large integer $n$ which is obviously a contradiction.
answered Dec 20 '18 at 10:00
Kavi Rama MurthyKavi Rama Murthy
72.8k53170
$endgroup$
add a comment |
$begingroup$
What is asserted is that there is no finite limit. Prove by contradiction. Suppose $lim_{xto 0+}frac 1 {x^{2}}=L$ exists and is finite. Then we can find $delta >0$ such that $|frac 1 {x^{2}}-L|<1$ whenever $0<x<delta$. If $n$ is any sufficiently large integer then $0<frac 1 n<delta$ so, taking $x=frac 1 n$ we get $| n^{2}-L|<1$. This gives $n^{2} <L+1$ for any sufficiently large integer $n$ which is obviously a contradiction.
answered Dec 20 '18 at 10:00
Kavi Rama MurthyKavi Rama Murthy
72.8k53170
$endgroup$
add a comment |
$begingroup$
What is asserted is that there is no finite limit. Prove by contradiction. Suppose $lim_{xto 0+}frac 1 {x^{2}}=L$ exists and is finite. Then we can find $delta >0$ such that $|frac 1 {x^{2}}-L|<1$ whenever $0<x<delta$. If $n$ is any sufficiently large integer then $0<frac 1 n<delta$ so, taking $x=frac 1 n$ we get $| n^{2}-L|<1$. This gives $n^{2} <L+1$ for any sufficiently large integer $n$ which is obviously a contradiction.
answered Dec 20 '18 at 10:00
Kavi Rama MurthyKavi Rama Murthy
72.8k53170
$endgroup$
What is asserted is that there is no finite limit. Prove by contradiction. Suppose $lim_{xto 0+}frac 1 {x^{2}}=L$ exists and is finite. Then we can find $delta >0$ such that $|frac 1 {x^{2}}-L|<1$ whenever $0<x<delta$. If $n$ is any sufficiently large integer then $0<frac 1 n<delta$ so, taking $x=frac 1 n$ we get $| n^{2}-L|<1$. This gives $n^{2} <L+1$ for any sufficiently large integer $n$ which is obviously a contradiction.
answered Dec 20 '18 at 10:00
Kavi Rama MurthyKavi Rama Murthy
72.8k53170
answered Dec 20 '18 at 10:00
Kavi Rama MurthyKavi Rama Murthy
72.8k53170
answered Dec 20 '18 at 10:00
Kavi Rama MurthyKavi Rama Murthy
72.8k53170
answered Dec 20 '18 at 10:00
Kavi Rama MurthyKavi Rama Murthy
72.8k53170
72.8k53170
add a comment |
add a comment |
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2
$begingroup$
It does exist though.
$endgroup$
– Rebellos
Dec 20 '18 at 9:59
1
$begingroup$
The limit exists but is $+infty$: to see this consider simply $limsup$ and $liminf$: the same is true even if you consider the whole $mathbb{R}$ instead of the positive axis. Perhaps the standard terminology is a bit hazy since in such cases it is said that the function diverges: however this does not mean tha the limit does not exists, but simply that the limit is $infty$.
$endgroup$
– Daniele Tampieri
Dec 20 '18 at 10:00
1
$begingroup$
Show that for every $L>0$ and $varepsilon=1;exists:delta>0$, $|f(x)-L|>varepsilon$ whenever $|x|<delta$.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 10:00