An infinity of natural numbers for which a function doesn't have a limit at $infty$
$begingroup$
Let $f,g:[0,infty)rightarrow mathbb{R}$ be two continuous functions with the property that $lim_{x to infty} f(x)=lim_{x to infty} g(x)=infty$. Prove that there is an infinity of natural numbers $k$ for which $h_k:[0,infty)rightarrow mathbb{R},h_k(x)=f(k+{g(x)})$ doesn't have a limit as $x to infty$.(${a}$ denotes the fractional part of the real number $a$)
My attempt : $k+{g(x)}=lfloor{k}rfloor$,so $lim_{x to infty}h_k(x)=f(lim_{x to infty}lfloor{k}rfloor)=infty$ and this is not what I need to prove.Why is my approach wrong and how should I solve this problem?
real-analysis continuity
asked Jan 6 at 20:41
JustAnAmateurJustAnAmateur
1096
$endgroup$
add a comment |
$begingroup$
Let $f,g:[0,infty)rightarrow mathbb{R}$ be two continuous functions with the property that $lim_{x to infty} f(x)=lim_{x to infty} g(x)=infty$. Prove that there is an infinity of natural numbers $k$ for which $h_k:[0,infty)rightarrow mathbb{R},h_k(x)=f(k+{g(x)})$ doesn't have a limit as $x to infty$.(${a}$ denotes the fractional part of the real number $a$)
My attempt : $k+{g(x)}=lfloor{k}rfloor$,so $lim_{x to infty}h_k(x)=f(lim_{x to infty}lfloor{k}rfloor)=infty$ and this is not what I need to prove.Why is my approach wrong and how should I solve this problem?
real-analysis continuity
asked Jan 6 at 20:41
JustAnAmateurJustAnAmateur
1096
$endgroup$
2
$begingroup$
The result isn't true. In particular, if $g$ is always an integer, then ${g}$ is always $0$, so $h_k(x) = f(k),$ for all $x$, so each $h_k$ is constant, so has a limit. Do you have some extra conditions on $g$?
$endgroup$
– user3482749
Jan 6 at 20:45
1
$begingroup$
Maybe the functions need to be continuous?
$endgroup$
– A. Pongrácz
Jan 6 at 20:47
$begingroup$
By the way, your approach is just nonsense. Why would $k+{g(x)}$ be the integer part of $k$? Also, the limit of the floor of $k$ as x (!!!) tends to infinity is the floor of $k$, so there would be a finite limit.
$endgroup$
– A. Pongrácz
Jan 6 at 20:49
add a comment |
$begingroup$
Let $f,g:[0,infty)rightarrow mathbb{R}$ be two continuous functions with the property that $lim_{x to infty} f(x)=lim_{x to infty} g(x)=infty$. Prove that there is an infinity of natural numbers $k$ for which $h_k:[0,infty)rightarrow mathbb{R},h_k(x)=f(k+{g(x)})$ doesn't have a limit as $x to infty$.(${a}$ denotes the fractional part of the real number $a$)
My attempt : $k+{g(x)}=lfloor{k}rfloor$,so $lim_{x to infty}h_k(x)=f(lim_{x to infty}lfloor{k}rfloor)=infty$ and this is not what I need to prove.Why is my approach wrong and how should I solve this problem?
real-analysis continuity
asked Jan 6 at 20:41
JustAnAmateurJustAnAmateur
1096
$endgroup$
Let $f,g:[0,infty)rightarrow mathbb{R}$ be two continuous functions with the property that $lim_{x to infty} f(x)=lim_{x to infty} g(x)=infty$. Prove that there is an infinity of natural numbers $k$ for which $h_k:[0,infty)rightarrow mathbb{R},h_k(x)=f(k+{g(x)})$ doesn't have a limit as $x to infty$.(${a}$ denotes the fractional part of the real number $a$)
My attempt : $k+{g(x)}=lfloor{k}rfloor$,so $lim_{x to infty}h_k(x)=f(lim_{x to infty}lfloor{k}rfloor)=infty$ and this is not what I need to prove.Why is my approach wrong and how should I solve this problem?
real-analysis continuity
real-analysis continuity
asked Jan 6 at 20:41
JustAnAmateurJustAnAmateur
1096
asked Jan 6 at 20:41
JustAnAmateurJustAnAmateur
1096
edited Jan 6 at 21:49
JustAnAmateur
asked Jan 6 at 20:41
JustAnAmateurJustAnAmateur
1096
asked Jan 6 at 20:41
JustAnAmateurJustAnAmateur
1096
asked Jan 6 at 20:41
JustAnAmateurJustAnAmateur
1096
1096
2
$begingroup$
The result isn't true. In particular, if $g$ is always an integer, then ${g}$ is always $0$, so $h_k(x) = f(k),$ for all $x$, so each $h_k$ is constant, so has a limit. Do you have some extra conditions on $g$?
$endgroup$
– user3482749
Jan 6 at 20:45
1
$begingroup$
Maybe the functions need to be continuous?
$endgroup$
– A. Pongrácz
Jan 6 at 20:47
$begingroup$
By the way, your approach is just nonsense. Why would $k+{g(x)}$ be the integer part of $k$? Also, the limit of the floor of $k$ as x (!!!) tends to infinity is the floor of $k$, so there would be a finite limit.
$endgroup$
– A. Pongrácz
Jan 6 at 20:49
add a comment |
2
$begingroup$
The result isn't true. In particular, if $g$ is always an integer, then ${g}$ is always $0$, so $h_k(x) = f(k),$ for all $x$, so each $h_k$ is constant, so has a limit. Do you have some extra conditions on $g$?
$endgroup$
– user3482749
Jan 6 at 20:45
1
$begingroup$
Maybe the functions need to be continuous?
$endgroup$
– A. Pongrácz
Jan 6 at 20:47
$begingroup$
By the way, your approach is just nonsense. Why would $k+{g(x)}$ be the integer part of $k$? Also, the limit of the floor of $k$ as x (!!!) tends to infinity is the floor of $k$, so there would be a finite limit.
$endgroup$
– A. Pongrácz
Jan 6 at 20:49
2
2
$begingroup$
The result isn't true. In particular, if $g$ is always an integer, then ${g}$ is always $0$, so $h_k(x) = f(k),$ for all $x$, so each $h_k$ is constant, so has a limit. Do you have some extra conditions on $g$?
$endgroup$
– user3482749
Jan 6 at 20:45
$begingroup$
The result isn't true. In particular, if $g$ is always an integer, then ${g}$ is always $0$, so $h_k(x) = f(k),$ for all $x$, so each $h_k$ is constant, so has a limit. Do you have some extra conditions on $g$?
$endgroup$
– user3482749
Jan 6 at 20:45
1
1
$begingroup$
Maybe the functions need to be continuous?
$endgroup$
– A. Pongrácz
Jan 6 at 20:47
$begingroup$
Maybe the functions need to be continuous?
$endgroup$
– A. Pongrácz
Jan 6 at 20:47
$begingroup$
By the way, your approach is just nonsense. Why would $k+{g(x)}$ be the integer part of $k$? Also, the limit of the floor of $k$ as x (!!!) tends to infinity is the floor of $k$, so there would be a finite limit.
$endgroup$
– A. Pongrácz
Jan 6 at 20:49
$begingroup$
By the way, your approach is just nonsense. Why would $k+{g(x)}$ be the integer part of $k$? Also, the limit of the floor of $k$ as x (!!!) tends to infinity is the floor of $k$, so there would be a finite limit.
$endgroup$
– A. Pongrácz
Jan 6 at 20:49
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Your argument isn't correct, and we need additional assumptions to deduce the desired conclusion, as is pointed out on the above comments. So, I'll presume that $f,g$ are continuous. Fix $k$ and suppose $h_k$ tends to $L_k$ as $xtoinfty$. For any given $M>0$, since $lim_{xtoinfty}g(x) = infty$, we can find $x>M$ such that $g(x)>g(M)+3$. Then there exists a natural number $N$ such that $g(M)le N<N+1le g(x)$. By intermediate value theorem, $[N,N+1]subset g([M,infty))$ for all $M>0$. This implies $[0,1)subset {g}([M,infty))$ for all $M>0$. Now, it follows that for any $uin [0,1)$, we can find a sequence $x_n to infty$ such that ${g(x_n)} = u$ for all $n$. This implies
$$
L_k=lim_{xtoinfty}h_k(x) = lim_{ntoinfty} h_k(x_n) = f(k+u)
$$ for all $uin [0,1)$. That is, $f(x)= L_k$ on $xin [k,k+1)$.
Now, assume to the contrary that $lim_{xtoinfty}h_k(x)$ does not exist for at most a finite number of $k$'s. Then, there is $k_0$ such that $$
L_k=lim_{xtoinfty}h_k(x)
$$ exists for all $kge k_0$. This means $f(x)=L_k$ on $[k,k+1)$ as we've seen in the above argument. If $f(x) = L_{k+1} $ on $[k+1,k+2)$, then by continuity of $f$, it must be $L_k = L_{k+1}$. This readily shows that $L_{k_0} = L_k$ for all $kge k_0$. This says that $f(x) = L_{k_0}$ for all $xge k_0$, and leads to a contradiction that $lim_{xtoinfty} f(x) = infty$.
answered Jan 6 at 21:28
SongSong
18.5k21651
$endgroup$
$begingroup$
They are indeed continuous,my bad,great solution.
$endgroup$
– JustAnAmateur
Jan 6 at 21:50
add a comment |
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1 Answer
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$begingroup$
Your argument isn't correct, and we need additional assumptions to deduce the desired conclusion, as is pointed out on the above comments. So, I'll presume that $f,g$ are continuous. Fix $k$ and suppose $h_k$ tends to $L_k$ as $xtoinfty$. For any given $M>0$, since $lim_{xtoinfty}g(x) = infty$, we can find $x>M$ such that $g(x)>g(M)+3$. Then there exists a natural number $N$ such that $g(M)le N<N+1le g(x)$. By intermediate value theorem, $[N,N+1]subset g([M,infty))$ for all $M>0$. This implies $[0,1)subset {g}([M,infty))$ for all $M>0$. Now, it follows that for any $uin [0,1)$, we can find a sequence $x_n to infty$ such that ${g(x_n)} = u$ for all $n$. This implies
$$
L_k=lim_{xtoinfty}h_k(x) = lim_{ntoinfty} h_k(x_n) = f(k+u)
$$ for all $uin [0,1)$. That is, $f(x)= L_k$ on $xin [k,k+1)$.
Now, assume to the contrary that $lim_{xtoinfty}h_k(x)$ does not exist for at most a finite number of $k$'s. Then, there is $k_0$ such that $$
L_k=lim_{xtoinfty}h_k(x)
$$ exists for all $kge k_0$. This means $f(x)=L_k$ on $[k,k+1)$ as we've seen in the above argument. If $f(x) = L_{k+1} $ on $[k+1,k+2)$, then by continuity of $f$, it must be $L_k = L_{k+1}$. This readily shows that $L_{k_0} = L_k$ for all $kge k_0$. This says that $f(x) = L_{k_0}$ for all $xge k_0$, and leads to a contradiction that $lim_{xtoinfty} f(x) = infty$.
answered Jan 6 at 21:28
SongSong
18.5k21651
$endgroup$
$begingroup$
They are indeed continuous,my bad,great solution.
$endgroup$
– JustAnAmateur
Jan 6 at 21:50
add a comment |
$begingroup$
Your argument isn't correct, and we need additional assumptions to deduce the desired conclusion, as is pointed out on the above comments. So, I'll presume that $f,g$ are continuous. Fix $k$ and suppose $h_k$ tends to $L_k$ as $xtoinfty$. For any given $M>0$, since $lim_{xtoinfty}g(x) = infty$, we can find $x>M$ such that $g(x)>g(M)+3$. Then there exists a natural number $N$ such that $g(M)le N<N+1le g(x)$. By intermediate value theorem, $[N,N+1]subset g([M,infty))$ for all $M>0$. This implies $[0,1)subset {g}([M,infty))$ for all $M>0$. Now, it follows that for any $uin [0,1)$, we can find a sequence $x_n to infty$ such that ${g(x_n)} = u$ for all $n$. This implies
$$
L_k=lim_{xtoinfty}h_k(x) = lim_{ntoinfty} h_k(x_n) = f(k+u)
$$ for all $uin [0,1)$. That is, $f(x)= L_k$ on $xin [k,k+1)$.
Now, assume to the contrary that $lim_{xtoinfty}h_k(x)$ does not exist for at most a finite number of $k$'s. Then, there is $k_0$ such that $$
L_k=lim_{xtoinfty}h_k(x)
$$ exists for all $kge k_0$. This means $f(x)=L_k$ on $[k,k+1)$ as we've seen in the above argument. If $f(x) = L_{k+1} $ on $[k+1,k+2)$, then by continuity of $f$, it must be $L_k = L_{k+1}$. This readily shows that $L_{k_0} = L_k$ for all $kge k_0$. This says that $f(x) = L_{k_0}$ for all $xge k_0$, and leads to a contradiction that $lim_{xtoinfty} f(x) = infty$.
answered Jan 6 at 21:28
SongSong
18.5k21651
$endgroup$
$begingroup$
They are indeed continuous,my bad,great solution.
$endgroup$
– JustAnAmateur
Jan 6 at 21:50
add a comment |
$begingroup$
Your argument isn't correct, and we need additional assumptions to deduce the desired conclusion, as is pointed out on the above comments. So, I'll presume that $f,g$ are continuous. Fix $k$ and suppose $h_k$ tends to $L_k$ as $xtoinfty$. For any given $M>0$, since $lim_{xtoinfty}g(x) = infty$, we can find $x>M$ such that $g(x)>g(M)+3$. Then there exists a natural number $N$ such that $g(M)le N<N+1le g(x)$. By intermediate value theorem, $[N,N+1]subset g([M,infty))$ for all $M>0$. This implies $[0,1)subset {g}([M,infty))$ for all $M>0$. Now, it follows that for any $uin [0,1)$, we can find a sequence $x_n to infty$ such that ${g(x_n)} = u$ for all $n$. This implies
$$
L_k=lim_{xtoinfty}h_k(x) = lim_{ntoinfty} h_k(x_n) = f(k+u)
$$ for all $uin [0,1)$. That is, $f(x)= L_k$ on $xin [k,k+1)$.
Now, assume to the contrary that $lim_{xtoinfty}h_k(x)$ does not exist for at most a finite number of $k$'s. Then, there is $k_0$ such that $$
L_k=lim_{xtoinfty}h_k(x)
$$ exists for all $kge k_0$. This means $f(x)=L_k$ on $[k,k+1)$ as we've seen in the above argument. If $f(x) = L_{k+1} $ on $[k+1,k+2)$, then by continuity of $f$, it must be $L_k = L_{k+1}$. This readily shows that $L_{k_0} = L_k$ for all $kge k_0$. This says that $f(x) = L_{k_0}$ for all $xge k_0$, and leads to a contradiction that $lim_{xtoinfty} f(x) = infty$.
answered Jan 6 at 21:28
SongSong
18.5k21651
$endgroup$
Your argument isn't correct, and we need additional assumptions to deduce the desired conclusion, as is pointed out on the above comments. So, I'll presume that $f,g$ are continuous. Fix $k$ and suppose $h_k$ tends to $L_k$ as $xtoinfty$. For any given $M>0$, since $lim_{xtoinfty}g(x) = infty$, we can find $x>M$ such that $g(x)>g(M)+3$. Then there exists a natural number $N$ such that $g(M)le N<N+1le g(x)$. By intermediate value theorem, $[N,N+1]subset g([M,infty))$ for all $M>0$. This implies $[0,1)subset {g}([M,infty))$ for all $M>0$. Now, it follows that for any $uin [0,1)$, we can find a sequence $x_n to infty$ such that ${g(x_n)} = u$ for all $n$. This implies
$$
L_k=lim_{xtoinfty}h_k(x) = lim_{ntoinfty} h_k(x_n) = f(k+u)
$$ for all $uin [0,1)$. That is, $f(x)= L_k$ on $xin [k,k+1)$.
Now, assume to the contrary that $lim_{xtoinfty}h_k(x)$ does not exist for at most a finite number of $k$'s. Then, there is $k_0$ such that $$
L_k=lim_{xtoinfty}h_k(x)
$$ exists for all $kge k_0$. This means $f(x)=L_k$ on $[k,k+1)$ as we've seen in the above argument. If $f(x) = L_{k+1} $ on $[k+1,k+2)$, then by continuity of $f$, it must be $L_k = L_{k+1}$. This readily shows that $L_{k_0} = L_k$ for all $kge k_0$. This says that $f(x) = L_{k_0}$ for all $xge k_0$, and leads to a contradiction that $lim_{xtoinfty} f(x) = infty$.
answered Jan 6 at 21:28
SongSong
18.5k21651
edited Jan 6 at 21:34
answered Jan 6 at 21:28
SongSong
18.5k21651
answered Jan 6 at 21:28
SongSong
18.5k21651
answered Jan 6 at 21:28
SongSong
18.5k21651
18.5k21651
$begingroup$
They are indeed continuous,my bad,great solution.
$endgroup$
– JustAnAmateur
Jan 6 at 21:50
add a comment |
$begingroup$
They are indeed continuous,my bad,great solution.
$endgroup$
– JustAnAmateur
Jan 6 at 21:50
$begingroup$
They are indeed continuous,my bad,great solution.
$endgroup$
– JustAnAmateur
Jan 6 at 21:50
$begingroup$
They are indeed continuous,my bad,great solution.
$endgroup$
– JustAnAmateur
Jan 6 at 21:50
add a comment |
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$begingroup$
The result isn't true. In particular, if $g$ is always an integer, then ${g}$ is always $0$, so $h_k(x) = f(k),$ for all $x$, so each $h_k$ is constant, so has a limit. Do you have some extra conditions on $g$?
$endgroup$
– user3482749
Jan 6 at 20:45
1
$begingroup$
Maybe the functions need to be continuous?
$endgroup$
– A. Pongrácz
Jan 6 at 20:47
$begingroup$
By the way, your approach is just nonsense. Why would $k+{g(x)}$ be the integer part of $k$? Also, the limit of the floor of $k$ as x (!!!) tends to infinity is the floor of $k$, so there would be a finite limit.
$endgroup$
– A. Pongrácz
Jan 6 at 20:49