How to show that the limit exists for a complex function? [closed]
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When $lvert zrvert ne 1$, it is necessary to show the existence of the limit of the function
$$f(z) = limlimits_ntoinftyfracz^n-1z^n+1$$
complex-analysis limits complex-numbers
asked Sep 9 at 6:12
Alexander Milogorodsky
63
closed as off-topic by Did, Lord Shark the Unknown, Deepesh Meena, Jendrik Stelzner, user91500 Sep 9 at 10:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Deepesh Meena, user91500
add a comment |Â
up vote
0
down vote
favorite
When $lvert zrvert ne 1$, it is necessary to show the existence of the limit of the function
$$f(z) = limlimits_ntoinftyfracz^n-1z^n+1$$
complex-analysis limits complex-numbers
asked Sep 9 at 6:12
Alexander Milogorodsky
63
closed as off-topic by Did, Lord Shark the Unknown, Deepesh Meena, Jendrik Stelzner, user91500 Sep 9 at 10:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Deepesh Meena, user91500
First, you should know $lim_ntoinftyz^n=0$ when $|z|<1$.
– Riemann
Sep 9 at 6:22
2
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
When $lvert zrvert ne 1$, it is necessary to show the existence of the limit of the function
$$f(z) = limlimits_ntoinftyfracz^n-1z^n+1$$
complex-analysis limits complex-numbers
asked Sep 9 at 6:12
Alexander Milogorodsky
63
When $lvert zrvert ne 1$, it is necessary to show the existence of the limit of the function
$$f(z) = limlimits_ntoinftyfracz^n-1z^n+1$$
complex-analysis limits complex-numbers
complex-analysis limits complex-numbers
asked Sep 9 at 6:12
Alexander Milogorodsky
63
asked Sep 9 at 6:12
Alexander Milogorodsky
63
edited Sep 9 at 6:14
asked Sep 9 at 6:12
Alexander Milogorodsky
63
asked Sep 9 at 6:12
Alexander Milogorodsky
63
asked Sep 9 at 6:12
Alexander Milogorodsky
63
63
closed as off-topic by Did, Lord Shark the Unknown, Deepesh Meena, Jendrik Stelzner, user91500 Sep 9 at 10:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Deepesh Meena, user91500
closed as off-topic by Did, Lord Shark the Unknown, Deepesh Meena, Jendrik Stelzner, user91500 Sep 9 at 10:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Deepesh Meena, user91500
First, you should know $lim_ntoinftyz^n=0$ when $|z|<1$.
– Riemann
Sep 9 at 6:22
2
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:28
add a comment |Â
First, you should know $lim_ntoinftyz^n=0$ when $|z|<1$.
– Riemann
Sep 9 at 6:22
2
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:28
First, you should know $lim_ntoinftyz^n=0$ when $|z|<1$.
– Riemann
Sep 9 at 6:22
First, you should know $lim_ntoinftyz^n=0$ when $|z|<1$.
– Riemann
Sep 9 at 6:22
2
2
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:28
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:28
add a comment |Â
2 Answers
2
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oldest
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0
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accepted
First, when $|z|<1$,$$lim_ntoinftyz^n=0.$$
So case one: when $|z|<1$,
$$limlimits_ntoinftyfracz^n-1z^n+1=fraclimlimits_ntoinftyz^n-1limlimits_ntoinftyz^n+1=frac0-10+1=-1.$$
Case two： when $|z|>1$, then $left|frac1zright|<1$,$limlimits_ntoinftyfrac1z^n=0$.
$$limlimits_ntoinftyfracz^n-1z^n+1=limlimits_ntoinftyfrac1-frac1z^n1+frac1z^n=frac1-limlimits_ntoinftyfrac1z^n1+limlimits_ntoinftyfrac1z^n=frac1-01+0=1.$$
answered Sep 9 at 6:43
Riemann
3,0881321
When $|z|=1$, the limit $lim_nto inftyz^n$ does not exist except $z=1$.
– Riemann
Sep 9 at 9:30
add a comment |Â
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Consider that
$$f(z) = lim_ntoinftyfracz^n-1z^n+1=1-2lim_ntoinftyfrac1z^n+1$$
if $|z|>1$ then $|z|^ntoinfty$ and
$$|frac1z^n+1|leqfrac1to0$$
then limit of $f$ is $colorblue1$.
If $|z|<1$ then $|z|^nto0$ and
$$fracz^nzleqleft|fracz^nz^n+1right|leqfracz^n1-$$
shows this limit is $0$, Therfore when $|z|<1$,
$$f(z) = lim_ntoinftyfracz^n-1z^n+1=2lim_ntoinftyfracz^nz^n+1-1$$
and the limit is $colorblue-1$.
answered Sep 9 at 6:29
Nosrati
22.8k61951
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:31
yes. of course.
– Nosrati
Sep 9 at 6:31
$|a|-|b|leq|apm b|leq|a|+|b|$
– Nosrati
Sep 9 at 6:35
You case two is something mistake. $frac1zleq|frac1z^n+1|leqfrac11-$ only implies $|frac1z^n+1|to 1.$ But this does not implies $frac1z^n+1to 1.$
– Riemann
Sep 9 at 6:48
@Nosrati it looks like squeeze theorem, but something vague in the second case indeed.
– Alexander Milogorodsky
Sep 9 at 6:54
 |Â
show 5 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
First, when $|z|<1$,$$lim_ntoinftyz^n=0.$$
So case one: when $|z|<1$,
$$limlimits_ntoinftyfracz^n-1z^n+1=fraclimlimits_ntoinftyz^n-1limlimits_ntoinftyz^n+1=frac0-10+1=-1.$$
Case two： when $|z|>1$, then $left|frac1zright|<1$,$limlimits_ntoinftyfrac1z^n=0$.
$$limlimits_ntoinftyfracz^n-1z^n+1=limlimits_ntoinftyfrac1-frac1z^n1+frac1z^n=frac1-limlimits_ntoinftyfrac1z^n1+limlimits_ntoinftyfrac1z^n=frac1-01+0=1.$$
answered Sep 9 at 6:43
Riemann
3,0881321
When $|z|=1$, the limit $lim_nto inftyz^n$ does not exist except $z=1$.
– Riemann
Sep 9 at 9:30
add a comment |Â
up vote
0
down vote
accepted
First, when $|z|<1$,$$lim_ntoinftyz^n=0.$$
So case one: when $|z|<1$,
$$limlimits_ntoinftyfracz^n-1z^n+1=fraclimlimits_ntoinftyz^n-1limlimits_ntoinftyz^n+1=frac0-10+1=-1.$$
Case two： when $|z|>1$, then $left|frac1zright|<1$,$limlimits_ntoinftyfrac1z^n=0$.
$$limlimits_ntoinftyfracz^n-1z^n+1=limlimits_ntoinftyfrac1-frac1z^n1+frac1z^n=frac1-limlimits_ntoinftyfrac1z^n1+limlimits_ntoinftyfrac1z^n=frac1-01+0=1.$$
answered Sep 9 at 6:43
Riemann
3,0881321
When $|z|=1$, the limit $lim_nto inftyz^n$ does not exist except $z=1$.
– Riemann
Sep 9 at 9:30
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
First, when $|z|<1$,$$lim_ntoinftyz^n=0.$$
So case one: when $|z|<1$,
$$limlimits_ntoinftyfracz^n-1z^n+1=fraclimlimits_ntoinftyz^n-1limlimits_ntoinftyz^n+1=frac0-10+1=-1.$$
Case two： when $|z|>1$, then $left|frac1zright|<1$,$limlimits_ntoinftyfrac1z^n=0$.
$$limlimits_ntoinftyfracz^n-1z^n+1=limlimits_ntoinftyfrac1-frac1z^n1+frac1z^n=frac1-limlimits_ntoinftyfrac1z^n1+limlimits_ntoinftyfrac1z^n=frac1-01+0=1.$$
answered Sep 9 at 6:43
Riemann
3,0881321
First, when $|z|<1$,$$lim_ntoinftyz^n=0.$$
So case one: when $|z|<1$,
$$limlimits_ntoinftyfracz^n-1z^n+1=fraclimlimits_ntoinftyz^n-1limlimits_ntoinftyz^n+1=frac0-10+1=-1.$$
Case two： when $|z|>1$, then $left|frac1zright|<1$,$limlimits_ntoinftyfrac1z^n=0$.
$$limlimits_ntoinftyfracz^n-1z^n+1=limlimits_ntoinftyfrac1-frac1z^n1+frac1z^n=frac1-limlimits_ntoinftyfrac1z^n1+limlimits_ntoinftyfrac1z^n=frac1-01+0=1.$$
answered Sep 9 at 6:43
Riemann
3,0881321
answered Sep 9 at 6:43
Riemann
3,0881321
answered Sep 9 at 6:43
Riemann
3,0881321
answered Sep 9 at 6:43
Riemann
3,0881321
3,0881321
When $|z|=1$, the limit $lim_nto inftyz^n$ does not exist except $z=1$.
– Riemann
Sep 9 at 9:30
add a comment |Â
When $|z|=1$, the limit $lim_nto inftyz^n$ does not exist except $z=1$.
– Riemann
Sep 9 at 9:30
When $|z|=1$, the limit $lim_nto inftyz^n$ does not exist except $z=1$.
– Riemann
Sep 9 at 9:30
When $|z|=1$, the limit $lim_nto inftyz^n$ does not exist except $z=1$.
– Riemann
Sep 9 at 9:30
add a comment |Â
up vote
0
down vote
Consider that
$$f(z) = lim_ntoinftyfracz^n-1z^n+1=1-2lim_ntoinftyfrac1z^n+1$$
if $|z|>1$ then $|z|^ntoinfty$ and
$$|frac1z^n+1|leqfrac1to0$$
then limit of $f$ is $colorblue1$.
If $|z|<1$ then $|z|^nto0$ and
$$fracz^nzleqleft|fracz^nz^n+1right|leqfracz^n1-$$
shows this limit is $0$, Therfore when $|z|<1$,
$$f(z) = lim_ntoinftyfracz^n-1z^n+1=2lim_ntoinftyfracz^nz^n+1-1$$
and the limit is $colorblue-1$.
answered Sep 9 at 6:29
Nosrati
22.8k61951
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:31
yes. of course.
– Nosrati
Sep 9 at 6:31
$|a|-|b|leq|apm b|leq|a|+|b|$
– Nosrati
Sep 9 at 6:35
You case two is something mistake. $frac1zleq|frac1z^n+1|leqfrac11-$ only implies $|frac1z^n+1|to 1.$ But this does not implies $frac1z^n+1to 1.$
– Riemann
Sep 9 at 6:48
@Nosrati it looks like squeeze theorem, but something vague in the second case indeed.
– Alexander Milogorodsky
Sep 9 at 6:54
 |Â
show 5 more comments
up vote
0
down vote
Consider that
$$f(z) = lim_ntoinftyfracz^n-1z^n+1=1-2lim_ntoinftyfrac1z^n+1$$
if $|z|>1$ then $|z|^ntoinfty$ and
$$|frac1z^n+1|leqfrac1to0$$
then limit of $f$ is $colorblue1$.
If $|z|<1$ then $|z|^nto0$ and
$$fracz^nzleqleft|fracz^nz^n+1right|leqfracz^n1-$$
shows this limit is $0$, Therfore when $|z|<1$,
$$f(z) = lim_ntoinftyfracz^n-1z^n+1=2lim_ntoinftyfracz^nz^n+1-1$$
and the limit is $colorblue-1$.
answered Sep 9 at 6:29
Nosrati
22.8k61951
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:31
yes. of course.
– Nosrati
Sep 9 at 6:31
$|a|-|b|leq|apm b|leq|a|+|b|$
– Nosrati
Sep 9 at 6:35
You case two is something mistake. $frac1zleq|frac1z^n+1|leqfrac11-$ only implies $|frac1z^n+1|to 1.$ But this does not implies $frac1z^n+1to 1.$
– Riemann
Sep 9 at 6:48
@Nosrati it looks like squeeze theorem, but something vague in the second case indeed.
– Alexander Milogorodsky
Sep 9 at 6:54
 |Â
show 5 more comments
up vote
0
down vote
up vote
0
down vote
Consider that
$$f(z) = lim_ntoinftyfracz^n-1z^n+1=1-2lim_ntoinftyfrac1z^n+1$$
if $|z|>1$ then $|z|^ntoinfty$ and
$$|frac1z^n+1|leqfrac1to0$$
then limit of $f$ is $colorblue1$.
If $|z|<1$ then $|z|^nto0$ and
$$fracz^nzleqleft|fracz^nz^n+1right|leqfracz^n1-$$
shows this limit is $0$, Therfore when $|z|<1$,
$$f(z) = lim_ntoinftyfracz^n-1z^n+1=2lim_ntoinftyfracz^nz^n+1-1$$
and the limit is $colorblue-1$.
answered Sep 9 at 6:29
Nosrati
22.8k61951
Consider that
$$f(z) = lim_ntoinftyfracz^n-1z^n+1=1-2lim_ntoinftyfrac1z^n+1$$
if $|z|>1$ then $|z|^ntoinfty$ and
$$|frac1z^n+1|leqfrac1to0$$
then limit of $f$ is $colorblue1$.
If $|z|<1$ then $|z|^nto0$ and
$$fracz^nzleqleft|fracz^nz^n+1right|leqfracz^n1-$$
shows this limit is $0$, Therfore when $|z|<1$,
$$f(z) = lim_ntoinftyfracz^n-1z^n+1=2lim_ntoinftyfracz^nz^n+1-1$$
and the limit is $colorblue-1$.
answered Sep 9 at 6:29
Nosrati
22.8k61951
edited Sep 9 at 7:02
answered Sep 9 at 6:29
Nosrati
22.8k61951
answered Sep 9 at 6:29
Nosrati
22.8k61951
answered Sep 9 at 6:29
Nosrati
22.8k61951
22.8k61951
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:31
yes. of course.
– Nosrati
Sep 9 at 6:31
$|a|-|b|leq|apm b|leq|a|+|b|$
– Nosrati
Sep 9 at 6:35
You case two is something mistake. $frac1zleq|frac1z^n+1|leqfrac11-$ only implies $|frac1z^n+1|to 1.$ But this does not implies $frac1z^n+1to 1.$
– Riemann
Sep 9 at 6:48
@Nosrati it looks like squeeze theorem, but something vague in the second case indeed.
– Alexander Milogorodsky
Sep 9 at 6:54
 |Â
show 5 more comments
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:31
yes. of course.
– Nosrati
Sep 9 at 6:31
$|a|-|b|leq|apm b|leq|a|+|b|$
– Nosrati
Sep 9 at 6:35
You case two is something mistake. $frac1zleq|frac1z^n+1|leqfrac11-$ only implies $|frac1z^n+1|to 1.$ But this does not implies $frac1z^n+1to 1.$
– Riemann
Sep 9 at 6:48
@Nosrati it looks like squeeze theorem, but something vague in the second case indeed.
– Alexander Milogorodsky
Sep 9 at 6:54
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:31
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:31
yes. of course.
– Nosrati
Sep 9 at 6:31
yes. of course.
– Nosrati
Sep 9 at 6:31
$|a|-|b|leq|apm b|leq|a|+|b|$
– Nosrati
Sep 9 at 6:35
$|a|-|b|leq|apm b|leq|a|+|b|$
– Nosrati
Sep 9 at 6:35
You case two is something mistake. $frac1zleq|frac1z^n+1|leqfrac11-$ only implies $|frac1z^n+1|to 1.$ But this does not implies $frac1z^n+1to 1.$
– Riemann
Sep 9 at 6:48
You case two is something mistake. $frac1zleq|frac1z^n+1|leqfrac11-$ only implies $|frac1z^n+1|to 1.$ But this does not implies $frac1z^n+1to 1.$
– Riemann
Sep 9 at 6:48
@Nosrati it looks like squeeze theorem, but something vague in the second case indeed.
– Alexander Milogorodsky
Sep 9 at 6:54
@Nosrati it looks like squeeze theorem, but something vague in the second case indeed.
– Alexander Milogorodsky
Sep 9 at 6:54
 |Â
show 5 more comments
First, you should know $lim_ntoinftyz^n=0$ when $|z|<1$.
– Riemann
Sep 9 at 6:22
2
When $|z|<1$, the limit is $-1$, and when $|z|>1$ the limit is $1$.
– Riemann
Sep 9 at 6:28