Finding set of limit points as $ x rightarrow 0 $
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
If the function $ f $ is defined as
$$
f(x)
=
begincases
|x|^phi - 1 sin(1/x), & x neq 0 \
0, & x = 0
endcases
$$
where $ phi > 0 $, how does one find the set of limit points of $ f(x) $ as $ x rightarrow 0 $, for each $ phi > 0 $?
real-analysis
asked Sep 3 at 14:19
Kendry Johnson
393212
add a comment |Â
up vote
2
down vote
favorite
If the function $ f $ is defined as
$$
f(x)
=
begincases
|x|^phi - 1 sin(1/x), & x neq 0 \
0, & x = 0
endcases
$$
where $ phi > 0 $, how does one find the set of limit points of $ f(x) $ as $ x rightarrow 0 $, for each $ phi > 0 $?
real-analysis
asked Sep 3 at 14:19
Kendry Johnson
393212
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
If the function $ f $ is defined as
$$
f(x)
=
begincases
|x|^phi - 1 sin(1/x), & x neq 0 \
0, & x = 0
endcases
$$
where $ phi > 0 $, how does one find the set of limit points of $ f(x) $ as $ x rightarrow 0 $, for each $ phi > 0 $?
real-analysis
asked Sep 3 at 14:19
Kendry Johnson
393212
If the function $ f $ is defined as
$$
f(x)
=
begincases
|x|^phi - 1 sin(1/x), & x neq 0 \
0, & x = 0
endcases
$$
where $ phi > 0 $, how does one find the set of limit points of $ f(x) $ as $ x rightarrow 0 $, for each $ phi > 0 $?
real-analysis
real-analysis
asked Sep 3 at 14:19
Kendry Johnson
393212
asked Sep 3 at 14:19
Kendry Johnson
393212
asked Sep 3 at 14:19
Kendry Johnson
393212
asked Sep 3 at 14:19
Kendry Johnson
393212
asked Sep 3 at 14:19
Kendry Johnson
393212
393212
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
HINT
We need to consider the following cases
- $phi>1$
$$0le||x|^phi - 1 sin(1/x)|le |x|^phi - 1to 0$$
- $phi=1$
$$|x|^phi - 1 sin(1/x)=sin(1/x)$$
- $0<phi<1$
$$|x|^phi - 1 sin(1/x)=fracsin(1/x)^1-phi$$
answered Sep 3 at 14:26
gimusi
72.6k73888
Just to confirm... in the first case, the only limit point is 0. In the second and third case, there's no limit points?
– Kendry Johnson
Sep 4 at 6:59
1
Yes the limit exists only for the first case, for the second we have limsup=$1$ and liminf=$-1$ and for the third limsup=$infty$ and liminf=$-infty$
– gimusi
Sep 4 at 7:12
add a comment |Â
up vote
-1
down vote
So, $lim_x to 0+ x^phi - 1sin(frac1x) = lim_x to 0+ x^phi-2 xsin(frac1x)$. As $xsin(frac1x) to 1$ as $x to 0$ that part has no importance, limit exists where limit $x^phi - 2$ exists and has same value as that limit, and the latter limit is zero when $phi > 2$, 1 when $phi = 2$ and doesn't exist when $phi < 2$
answered Sep 3 at 14:27
nikola
618314
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
accepted
HINT
We need to consider the following cases
- $phi>1$
$$0le||x|^phi - 1 sin(1/x)|le |x|^phi - 1to 0$$
- $phi=1$
$$|x|^phi - 1 sin(1/x)=sin(1/x)$$
- $0<phi<1$
$$|x|^phi - 1 sin(1/x)=fracsin(1/x)^1-phi$$
answered Sep 3 at 14:26
gimusi
72.6k73888
Just to confirm... in the first case, the only limit point is 0. In the second and third case, there's no limit points?
– Kendry Johnson
Sep 4 at 6:59
1
Yes the limit exists only for the first case, for the second we have limsup=$1$ and liminf=$-1$ and for the third limsup=$infty$ and liminf=$-infty$
– gimusi
Sep 4 at 7:12
add a comment |Â
up vote
2
down vote
accepted
HINT
We need to consider the following cases
- $phi>1$
$$0le||x|^phi - 1 sin(1/x)|le |x|^phi - 1to 0$$
- $phi=1$
$$|x|^phi - 1 sin(1/x)=sin(1/x)$$
- $0<phi<1$
$$|x|^phi - 1 sin(1/x)=fracsin(1/x)^1-phi$$
answered Sep 3 at 14:26
gimusi
72.6k73888
Just to confirm... in the first case, the only limit point is 0. In the second and third case, there's no limit points?
– Kendry Johnson
Sep 4 at 6:59
1
Yes the limit exists only for the first case, for the second we have limsup=$1$ and liminf=$-1$ and for the third limsup=$infty$ and liminf=$-infty$
– gimusi
Sep 4 at 7:12
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
HINT
We need to consider the following cases
- $phi>1$
$$0le||x|^phi - 1 sin(1/x)|le |x|^phi - 1to 0$$
- $phi=1$
$$|x|^phi - 1 sin(1/x)=sin(1/x)$$
- $0<phi<1$
$$|x|^phi - 1 sin(1/x)=fracsin(1/x)^1-phi$$
answered Sep 3 at 14:26
gimusi
72.6k73888
HINT
We need to consider the following cases
- $phi>1$
$$0le||x|^phi - 1 sin(1/x)|le |x|^phi - 1to 0$$
- $phi=1$
$$|x|^phi - 1 sin(1/x)=sin(1/x)$$
- $0<phi<1$
$$|x|^phi - 1 sin(1/x)=fracsin(1/x)^1-phi$$
answered Sep 3 at 14:26
gimusi
72.6k73888
answered Sep 3 at 14:26
gimusi
72.6k73888
answered Sep 3 at 14:26
gimusi
72.6k73888
answered Sep 3 at 14:26
gimusi
72.6k73888
72.6k73888
Just to confirm... in the first case, the only limit point is 0. In the second and third case, there's no limit points?
– Kendry Johnson
Sep 4 at 6:59
1
Yes the limit exists only for the first case, for the second we have limsup=$1$ and liminf=$-1$ and for the third limsup=$infty$ and liminf=$-infty$
– gimusi
Sep 4 at 7:12
add a comment |Â
Just to confirm... in the first case, the only limit point is 0. In the second and third case, there's no limit points?
– Kendry Johnson
Sep 4 at 6:59
1
Yes the limit exists only for the first case, for the second we have limsup=$1$ and liminf=$-1$ and for the third limsup=$infty$ and liminf=$-infty$
– gimusi
Sep 4 at 7:12
Just to confirm... in the first case, the only limit point is 0. In the second and third case, there's no limit points?
– Kendry Johnson
Sep 4 at 6:59
Just to confirm... in the first case, the only limit point is 0. In the second and third case, there's no limit points?
– Kendry Johnson
Sep 4 at 6:59
1
1
Yes the limit exists only for the first case, for the second we have limsup=$1$ and liminf=$-1$ and for the third limsup=$infty$ and liminf=$-infty$
– gimusi
Sep 4 at 7:12
Yes the limit exists only for the first case, for the second we have limsup=$1$ and liminf=$-1$ and for the third limsup=$infty$ and liminf=$-infty$
– gimusi
Sep 4 at 7:12
add a comment |Â
up vote
-1
down vote
So, $lim_x to 0+ x^phi - 1sin(frac1x) = lim_x to 0+ x^phi-2 xsin(frac1x)$. As $xsin(frac1x) to 1$ as $x to 0$ that part has no importance, limit exists where limit $x^phi - 2$ exists and has same value as that limit, and the latter limit is zero when $phi > 2$, 1 when $phi = 2$ and doesn't exist when $phi < 2$
answered Sep 3 at 14:27
nikola
618314
add a comment |Â
up vote
-1
down vote
So, $lim_x to 0+ x^phi - 1sin(frac1x) = lim_x to 0+ x^phi-2 xsin(frac1x)$. As $xsin(frac1x) to 1$ as $x to 0$ that part has no importance, limit exists where limit $x^phi - 2$ exists and has same value as that limit, and the latter limit is zero when $phi > 2$, 1 when $phi = 2$ and doesn't exist when $phi < 2$
answered Sep 3 at 14:27
nikola
618314
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
So, $lim_x to 0+ x^phi - 1sin(frac1x) = lim_x to 0+ x^phi-2 xsin(frac1x)$. As $xsin(frac1x) to 1$ as $x to 0$ that part has no importance, limit exists where limit $x^phi - 2$ exists and has same value as that limit, and the latter limit is zero when $phi > 2$, 1 when $phi = 2$ and doesn't exist when $phi < 2$
answered Sep 3 at 14:27
nikola
618314
So, $lim_x to 0+ x^phi - 1sin(frac1x) = lim_x to 0+ x^phi-2 xsin(frac1x)$. As $xsin(frac1x) to 1$ as $x to 0$ that part has no importance, limit exists where limit $x^phi - 2$ exists and has same value as that limit, and the latter limit is zero when $phi > 2$, 1 when $phi = 2$ and doesn't exist when $phi < 2$
answered Sep 3 at 14:27
nikola
618314
answered Sep 3 at 14:27
nikola
618314
answered Sep 3 at 14:27
nikola
618314
answered Sep 3 at 14:27
nikola
618314
618314
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2903910%2ffinding-set-of-limit-points-as-x-rightarrow-0%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
