Proving $lim_xto 0 dfrac1 + 1/x^21 + 1/x$ does not exist using the sequence definition of a limit
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I'm working through problems in a real analysis textbook. I don't have any solutions. I am using the following definition for a limit:
Definition: Given a function $f : D rightarrow mathbbR$ and a limit point $x_0 in D$, for a number $ell$, we write
$$lim_xto x_0 f(x) = ell $$
provided that whenever $x_n in D - x_0$ that converges to $x_0$,
$$lim_ntoinfty f(x_n) = ell.$$
Problem:
I am trying to show that the limit
$$lim_xto 0 dfrac1 + 1/x^21 + 1/x$$
does not exist.
My attempt:
Let $x_n$ be a sequence in $D - 0$ that converges to $0$. Now we need to show that
$$lim_nto infty dfrac1 + 1/x_n^21 + 1/x_n = ell.$$
Rewrite the limit as follows:
$$ lim_nto inftydfrac1 + 1/x_n^21 + 1/x_n = lim_nto infty dfracx_n^2 + 1x_n^2 + x_n = dfrac10,$$
which is undefined. Thus, we can conclude that our limit does not exist.
real-analysis proof-verification
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up vote
1
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I'm working through problems in a real analysis textbook. I don't have any solutions. I am using the following definition for a limit:
Definition: Given a function $f : D rightarrow mathbbR$ and a limit point $x_0 in D$, for a number $ell$, we write
$$lim_xto x_0 f(x) = ell $$
provided that whenever $x_n in D - x_0$ that converges to $x_0$,
$$lim_ntoinfty f(x_n) = ell.$$
Problem:
I am trying to show that the limit
$$lim_xto 0 dfrac1 + 1/x^21 + 1/x$$
does not exist.
My attempt:
Let $x_n$ be a sequence in $D - 0$ that converges to $0$. Now we need to show that
$$lim_nto infty dfrac1 + 1/x_n^21 + 1/x_n = ell.$$
Rewrite the limit as follows:
$$ lim_nto inftydfrac1 + 1/x_n^21 + 1/x_n = lim_nto infty dfracx_n^2 + 1x_n^2 + x_n = dfrac10,$$
which is undefined. Thus, we can conclude that our limit does not exist.
real-analysis proof-verification
2
That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
â user582578
Aug 8 at 20:20
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm working through problems in a real analysis textbook. I don't have any solutions. I am using the following definition for a limit:
Definition: Given a function $f : D rightarrow mathbbR$ and a limit point $x_0 in D$, for a number $ell$, we write
$$lim_xto x_0 f(x) = ell $$
provided that whenever $x_n in D - x_0$ that converges to $x_0$,
$$lim_ntoinfty f(x_n) = ell.$$
Problem:
I am trying to show that the limit
$$lim_xto 0 dfrac1 + 1/x^21 + 1/x$$
does not exist.
My attempt:
Let $x_n$ be a sequence in $D - 0$ that converges to $0$. Now we need to show that
$$lim_nto infty dfrac1 + 1/x_n^21 + 1/x_n = ell.$$
Rewrite the limit as follows:
$$ lim_nto inftydfrac1 + 1/x_n^21 + 1/x_n = lim_nto infty dfracx_n^2 + 1x_n^2 + x_n = dfrac10,$$
which is undefined. Thus, we can conclude that our limit does not exist.
real-analysis proof-verification
I'm working through problems in a real analysis textbook. I don't have any solutions. I am using the following definition for a limit:
Definition: Given a function $f : D rightarrow mathbbR$ and a limit point $x_0 in D$, for a number $ell$, we write
$$lim_xto x_0 f(x) = ell $$
provided that whenever $x_n in D - x_0$ that converges to $x_0$,
$$lim_ntoinfty f(x_n) = ell.$$
Problem:
I am trying to show that the limit
$$lim_xto 0 dfrac1 + 1/x^21 + 1/x$$
does not exist.
My attempt:
Let $x_n$ be a sequence in $D - 0$ that converges to $0$. Now we need to show that
$$lim_nto infty dfrac1 + 1/x_n^21 + 1/x_n = ell.$$
Rewrite the limit as follows:
$$ lim_nto inftydfrac1 + 1/x_n^21 + 1/x_n = lim_nto infty dfracx_n^2 + 1x_n^2 + x_n = dfrac10,$$
which is undefined. Thus, we can conclude that our limit does not exist.
real-analysis proof-verification
asked Aug 8 at 20:06
Hat
789115
789115
2
That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
â user582578
Aug 8 at 20:20
add a comment |Â
2
That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
â user582578
Aug 8 at 20:20
2
2
That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
â user582578
Aug 8 at 20:20
That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
â user582578
Aug 8 at 20:20
add a comment |Â
2 Answers
2
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oldest
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up vote
2
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As I know, in order to show in this way the limit doesn't exist we
have to address to a specific sequence. Take for example $a_n=frac1n$
which converges to $0$, but
beginalign*
fleft(a_nright) & =fleft(frac1nright)=frac1+frac1left(frac1nright)^21+frac1left(frac1nright)=frac1+n^21+n=fracleft(1+nright)^2-2n1+n=1+n-frac2n1+n=\
& =1+n-frac2left(1+nright)-21+n=-1+n+frac21+ntoinfty
endalign*
1
You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
â user582578
Aug 8 at 20:44
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up vote
0
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The sequence $x_n=1/n$ converges to $0$; on the other hand,
$$
f(1/n)=frac1+n^21+n>fracn^22n=fracn2
$$
Therefore, the sequence $f(1/n)$ doesn't converge.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
As I know, in order to show in this way the limit doesn't exist we
have to address to a specific sequence. Take for example $a_n=frac1n$
which converges to $0$, but
beginalign*
fleft(a_nright) & =fleft(frac1nright)=frac1+frac1left(frac1nright)^21+frac1left(frac1nright)=frac1+n^21+n=fracleft(1+nright)^2-2n1+n=1+n-frac2n1+n=\
& =1+n-frac2left(1+nright)-21+n=-1+n+frac21+ntoinfty
endalign*
1
You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
â user582578
Aug 8 at 20:44
add a comment |Â
up vote
2
down vote
As I know, in order to show in this way the limit doesn't exist we
have to address to a specific sequence. Take for example $a_n=frac1n$
which converges to $0$, but
beginalign*
fleft(a_nright) & =fleft(frac1nright)=frac1+frac1left(frac1nright)^21+frac1left(frac1nright)=frac1+n^21+n=fracleft(1+nright)^2-2n1+n=1+n-frac2n1+n=\
& =1+n-frac2left(1+nright)-21+n=-1+n+frac21+ntoinfty
endalign*
1
You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
â user582578
Aug 8 at 20:44
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As I know, in order to show in this way the limit doesn't exist we
have to address to a specific sequence. Take for example $a_n=frac1n$
which converges to $0$, but
beginalign*
fleft(a_nright) & =fleft(frac1nright)=frac1+frac1left(frac1nright)^21+frac1left(frac1nright)=frac1+n^21+n=fracleft(1+nright)^2-2n1+n=1+n-frac2n1+n=\
& =1+n-frac2left(1+nright)-21+n=-1+n+frac21+ntoinfty
endalign*
As I know, in order to show in this way the limit doesn't exist we
have to address to a specific sequence. Take for example $a_n=frac1n$
which converges to $0$, but
beginalign*
fleft(a_nright) & =fleft(frac1nright)=frac1+frac1left(frac1nright)^21+frac1left(frac1nright)=frac1+n^21+n=fracleft(1+nright)^2-2n1+n=1+n-frac2n1+n=\
& =1+n-frac2left(1+nright)-21+n=-1+n+frac21+ntoinfty
endalign*
answered Aug 8 at 20:35
Jon
500413
500413
1
You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
â user582578
Aug 8 at 20:44
add a comment |Â
1
You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
â user582578
Aug 8 at 20:44
1
1
You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
â user582578
Aug 8 at 20:44
You don't "have to", rather "it is enough". The argument for a general sequence can also be completed.
â user582578
Aug 8 at 20:44
add a comment |Â
up vote
0
down vote
The sequence $x_n=1/n$ converges to $0$; on the other hand,
$$
f(1/n)=frac1+n^21+n>fracn^22n=fracn2
$$
Therefore, the sequence $f(1/n)$ doesn't converge.
add a comment |Â
up vote
0
down vote
The sequence $x_n=1/n$ converges to $0$; on the other hand,
$$
f(1/n)=frac1+n^21+n>fracn^22n=fracn2
$$
Therefore, the sequence $f(1/n)$ doesn't converge.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The sequence $x_n=1/n$ converges to $0$; on the other hand,
$$
f(1/n)=frac1+n^21+n>fracn^22n=fracn2
$$
Therefore, the sequence $f(1/n)$ doesn't converge.
The sequence $x_n=1/n$ converges to $0$; on the other hand,
$$
f(1/n)=frac1+n^21+n>fracn^22n=fracn2
$$
Therefore, the sequence $f(1/n)$ doesn't converge.
answered Aug 8 at 20:50
egreg
165k1180187
165k1180187
add a comment |Â
add a comment |Â
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2
That is correct. You need to judge if your instructors wanted a fuller explanation of the very last equality or that is enough. If that is the case, you can work with arbitrary sequences as you did, but you can also pick a specific one, like $x_n=1/n$ and work with it, since it is enough to show non-existence for one sequence to ensure that the original limit doesn't exist.
â user582578
Aug 8 at 20:20