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Limit of a function $g_k(x)$ when the limit $g_k(y)/g_k(z)$ is known.
Clash Royale CLAN TAG#URR8PPP
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Let $g_k(x)$ be a decreasing function respect to $x$ defined in $mathbbR$ which satisfies the condition
$$
fracg_k(y)g_k(z) to infty text when k to infty quad textand whenever , y<z.
$$
Can we deduce anything about the limit $g_k(x) to infty$ as $k to infty$?
Maybe that this limit is almost always $0$ or $infty$?
Or maybe if we add some additional conditions we can say something? I’m a bit unsure about this.
real-analysis analysis limits
asked Jul 22 at 21:13
Kplusn
909
add a comment |Â
up vote
0
down vote
favorite
Let $g_k(x)$ be a decreasing function respect to $x$ defined in $mathbbR$ which satisfies the condition
$$
fracg_k(y)g_k(z) to infty text when k to infty quad textand whenever , y<z.
$$
Can we deduce anything about the limit $g_k(x) to infty$ as $k to infty$?
Maybe that this limit is almost always $0$ or $infty$?
Or maybe if we add some additional conditions we can say something? I’m a bit unsure about this.
real-analysis analysis limits
asked Jul 22 at 21:13
Kplusn
909
What is the purpose of $k$ in $g_k(x)$?
– Shrey Joshi
Jul 22 at 21:24
@Kplusn if $g_k(x)=exp(-kx)$, then $g_k(0)to 1$ as $ktoinfty$ (and $0$ or $+infty$ for $xneq 0$).
– user254433
Jul 22 at 21:57
@ShreyJoshi What do you mean? It’s just an index that denotes that the function also depends on $k$.
– Kplusn
Jul 22 at 21:58
@user254433 Yes, you are correct. But I’m not interested in particular examples.
– Kplusn
Jul 22 at 21:59
Oh so $g_k(x)$ is like $g(k, x)$.
– Shrey Joshi
Jul 22 at 22:09
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $g_k(x)$ be a decreasing function respect to $x$ defined in $mathbbR$ which satisfies the condition
$$
fracg_k(y)g_k(z) to infty text when k to infty quad textand whenever , y<z.
$$
Can we deduce anything about the limit $g_k(x) to infty$ as $k to infty$?
Maybe that this limit is almost always $0$ or $infty$?
Or maybe if we add some additional conditions we can say something? I’m a bit unsure about this.
real-analysis analysis limits
asked Jul 22 at 21:13
Kplusn
909
Let $g_k(x)$ be a decreasing function respect to $x$ defined in $mathbbR$ which satisfies the condition
$$
fracg_k(y)g_k(z) to infty text when k to infty quad textand whenever , y<z.
$$
Can we deduce anything about the limit $g_k(x) to infty$ as $k to infty$?
Maybe that this limit is almost always $0$ or $infty$?
Or maybe if we add some additional conditions we can say something? I’m a bit unsure about this.
real-analysis analysis limits
asked Jul 22 at 21:13
Kplusn
909
edited Jul 22 at 22:01
asked Jul 22 at 21:13
Kplusn
909
asked Jul 22 at 21:13
Kplusn
909
asked Jul 22 at 21:13
Kplusn
909
909
What is the purpose of $k$ in $g_k(x)$?
– Shrey Joshi
Jul 22 at 21:24
@Kplusn if $g_k(x)=exp(-kx)$, then $g_k(0)to 1$ as $ktoinfty$ (and $0$ or $+infty$ for $xneq 0$).
– user254433
Jul 22 at 21:57
@ShreyJoshi What do you mean? It’s just an index that denotes that the function also depends on $k$.
– Kplusn
Jul 22 at 21:58
@user254433 Yes, you are correct. But I’m not interested in particular examples.
– Kplusn
Jul 22 at 21:59
Oh so $g_k(x)$ is like $g(k, x)$.
– Shrey Joshi
Jul 22 at 22:09
add a comment |Â
What is the purpose of $k$ in $g_k(x)$?
– Shrey Joshi
Jul 22 at 21:24
@Kplusn if $g_k(x)=exp(-kx)$, then $g_k(0)to 1$ as $ktoinfty$ (and $0$ or $+infty$ for $xneq 0$).
– user254433
Jul 22 at 21:57
@ShreyJoshi What do you mean? It’s just an index that denotes that the function also depends on $k$.
– Kplusn
Jul 22 at 21:58
@user254433 Yes, you are correct. But I’m not interested in particular examples.
– Kplusn
Jul 22 at 21:59
Oh so $g_k(x)$ is like $g(k, x)$.
– Shrey Joshi
Jul 22 at 22:09
What is the purpose of $k$ in $g_k(x)$?
– Shrey Joshi
Jul 22 at 21:24
What is the purpose of $k$ in $g_k(x)$?
– Shrey Joshi
Jul 22 at 21:24
@Kplusn if $g_k(x)=exp(-kx)$, then $g_k(0)to 1$ as $ktoinfty$ (and $0$ or $+infty$ for $xneq 0$).
– user254433
Jul 22 at 21:57
@Kplusn if $g_k(x)=exp(-kx)$, then $g_k(0)to 1$ as $ktoinfty$ (and $0$ or $+infty$ for $xneq 0$).
– user254433
Jul 22 at 21:57
@ShreyJoshi What do you mean? It’s just an index that denotes that the function also depends on $k$.
– Kplusn
Jul 22 at 21:58
@ShreyJoshi What do you mean? It’s just an index that denotes that the function also depends on $k$.
– Kplusn
Jul 22 at 21:58
@user254433 Yes, you are correct. But I’m not interested in particular examples.
– Kplusn
Jul 22 at 21:59
@user254433 Yes, you are correct. But I’m not interested in particular examples.
– Kplusn
Jul 22 at 21:59
Oh so $g_k(x)$ is like $g(k, x)$.
– Shrey Joshi
Jul 22 at 22:09
Oh so $g_k(x)$ is like $g(k, x)$.
– Shrey Joshi
Jul 22 at 22:09
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
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accepted
Let $x_0$ be such that $lim_ktoinftyg_k(x_0) = A$ and $A$ is not $0$ nor $infty$. Then by your condition for any other $x$ we have either $g_k(x)/g_k(x_0) toinfty$ or $g_k(x)/g_k(x_0)to 0$ as $ktoinfty$. But in both cases we have
$$
lim_ktoinfty fracg_k(x)g_k(x_0) = frac1A lim_ktoinftyg_k(x).
$$
We can make a conclusion now.
answered Jul 24 at 23:23
dsw
963
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $x_0$ be such that $lim_ktoinftyg_k(x_0) = A$ and $A$ is not $0$ nor $infty$. Then by your condition for any other $x$ we have either $g_k(x)/g_k(x_0) toinfty$ or $g_k(x)/g_k(x_0)to 0$ as $ktoinfty$. But in both cases we have
$$
lim_ktoinfty fracg_k(x)g_k(x_0) = frac1A lim_ktoinftyg_k(x).
$$
We can make a conclusion now.
answered Jul 24 at 23:23
dsw
963
add a comment |Â
up vote
3
down vote
accepted
Let $x_0$ be such that $lim_ktoinftyg_k(x_0) = A$ and $A$ is not $0$ nor $infty$. Then by your condition for any other $x$ we have either $g_k(x)/g_k(x_0) toinfty$ or $g_k(x)/g_k(x_0)to 0$ as $ktoinfty$. But in both cases we have
$$
lim_ktoinfty fracg_k(x)g_k(x_0) = frac1A lim_ktoinftyg_k(x).
$$
We can make a conclusion now.
answered Jul 24 at 23:23
dsw
963
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $x_0$ be such that $lim_ktoinftyg_k(x_0) = A$ and $A$ is not $0$ nor $infty$. Then by your condition for any other $x$ we have either $g_k(x)/g_k(x_0) toinfty$ or $g_k(x)/g_k(x_0)to 0$ as $ktoinfty$. But in both cases we have
$$
lim_ktoinfty fracg_k(x)g_k(x_0) = frac1A lim_ktoinftyg_k(x).
$$
We can make a conclusion now.
answered Jul 24 at 23:23
dsw
963
Let $x_0$ be such that $lim_ktoinftyg_k(x_0) = A$ and $A$ is not $0$ nor $infty$. Then by your condition for any other $x$ we have either $g_k(x)/g_k(x_0) toinfty$ or $g_k(x)/g_k(x_0)to 0$ as $ktoinfty$. But in both cases we have
$$
lim_ktoinfty fracg_k(x)g_k(x_0) = frac1A lim_ktoinftyg_k(x).
$$
We can make a conclusion now.
answered Jul 24 at 23:23
dsw
963
answered Jul 24 at 23:23
dsw
963
answered Jul 24 at 23:23
dsw
963
answered Jul 24 at 23:23
dsw
963
963
add a comment |Â
add a comment |Â
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What is the purpose of $k$ in $g_k(x)$?
– Shrey Joshi
Jul 22 at 21:24
@Kplusn if $g_k(x)=exp(-kx)$, then $g_k(0)to 1$ as $ktoinfty$ (and $0$ or $+infty$ for $xneq 0$).
– user254433
Jul 22 at 21:57
@ShreyJoshi What do you mean? It’s just an index that denotes that the function also depends on $k$.
– Kplusn
Jul 22 at 21:58
@user254433 Yes, you are correct. But I’m not interested in particular examples.
– Kplusn
Jul 22 at 21:59
Oh so $g_k(x)$ is like $g(k, x)$.
– Shrey Joshi
Jul 22 at 22:09