Mixing
Solving limits that approach infinity
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Solve limit:
$$lim_ktoinftyfrac4k+32$$
My approach.
$$lim_ktoinftyfrac(4k+3)2 frac1/k1/k$$
$$lim_ktoinftyfrac4+3/k2/k$$
Then, we know that the limit of k as k approaches inf for $$frac3k$$ goes to 0 and the same goes for $$frac2k$$ however for both limits, they can also be viewed as infinitesmall such that the final answer is $$frac40+ = infty$$
Can somebody verify if I did this correctly? I'm confused as to whether the fraction limits approach 0 or infinitesmall because if it's 0 then my limit is undefined
real-analysis limits
asked Jul 26 at 15:25
Donald Devy
63
 |Â
show 1 more comment
up vote
-1
down vote
favorite
Solve limit:
$$lim_ktoinftyfrac4k+32$$
My approach.
$$lim_ktoinftyfrac(4k+3)2 frac1/k1/k$$
$$lim_ktoinftyfrac4+3/k2/k$$
Then, we know that the limit of k as k approaches inf for $$frac3k$$ goes to 0 and the same goes for $$frac2k$$ however for both limits, they can also be viewed as infinitesmall such that the final answer is $$frac40+ = infty$$
Can somebody verify if I did this correctly? I'm confused as to whether the fraction limits approach 0 or infinitesmall because if it's 0 then my limit is undefined
real-analysis limits
asked Jul 26 at 15:25
Donald Devy
63
2
You have limits for $n$, but the arguments are in terms of $k$.
– Benedict W. J. Irwin
Jul 26 at 15:26
Sry I fixed it to be variable k.
– Donald Devy
Jul 26 at 15:28
Isn't it obvious from the beginning that $4k+3toinfty$ as $ktoinfty?$
– saulspatz
Jul 26 at 15:29
True... I actually didn't have to do anything. But for curiosity's sake can you help me with my understanding for the fraction limits?
– Donald Devy
Jul 26 at 15:31
You did this correctly (as in you are using heuristics which can be made rigorous). You should dispense of your worries with the limit begin undefined; the limit must be infinite, since $(4k+3)/2$ becomes arbitrarily large as $k$ does.
– Mike Earnest
Jul 26 at 15:55
 |Â
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Solve limit:
$$lim_ktoinftyfrac4k+32$$
My approach.
$$lim_ktoinftyfrac(4k+3)2 frac1/k1/k$$
$$lim_ktoinftyfrac4+3/k2/k$$
Then, we know that the limit of k as k approaches inf for $$frac3k$$ goes to 0 and the same goes for $$frac2k$$ however for both limits, they can also be viewed as infinitesmall such that the final answer is $$frac40+ = infty$$
Can somebody verify if I did this correctly? I'm confused as to whether the fraction limits approach 0 or infinitesmall because if it's 0 then my limit is undefined
real-analysis limits
asked Jul 26 at 15:25
Donald Devy
63
Solve limit:
$$lim_ktoinftyfrac4k+32$$
My approach.
$$lim_ktoinftyfrac(4k+3)2 frac1/k1/k$$
$$lim_ktoinftyfrac4+3/k2/k$$
Then, we know that the limit of k as k approaches inf for $$frac3k$$ goes to 0 and the same goes for $$frac2k$$ however for both limits, they can also be viewed as infinitesmall such that the final answer is $$frac40+ = infty$$
Can somebody verify if I did this correctly? I'm confused as to whether the fraction limits approach 0 or infinitesmall because if it's 0 then my limit is undefined
real-analysis limits
asked Jul 26 at 15:25
Donald Devy
63
edited Jul 26 at 15:27
asked Jul 26 at 15:25
Donald Devy
63
asked Jul 26 at 15:25
Donald Devy
63
asked Jul 26 at 15:25
Donald Devy
63
63
2
You have limits for $n$, but the arguments are in terms of $k$.
– Benedict W. J. Irwin
Jul 26 at 15:26
Sry I fixed it to be variable k.
– Donald Devy
Jul 26 at 15:28
Isn't it obvious from the beginning that $4k+3toinfty$ as $ktoinfty?$
– saulspatz
Jul 26 at 15:29
True... I actually didn't have to do anything. But for curiosity's sake can you help me with my understanding for the fraction limits?
– Donald Devy
Jul 26 at 15:31
You did this correctly (as in you are using heuristics which can be made rigorous). You should dispense of your worries with the limit begin undefined; the limit must be infinite, since $(4k+3)/2$ becomes arbitrarily large as $k$ does.
– Mike Earnest
Jul 26 at 15:55
 |Â
show 1 more comment
2
You have limits for $n$, but the arguments are in terms of $k$.
– Benedict W. J. Irwin
Jul 26 at 15:26
Sry I fixed it to be variable k.
– Donald Devy
Jul 26 at 15:28
Isn't it obvious from the beginning that $4k+3toinfty$ as $ktoinfty?$
– saulspatz
Jul 26 at 15:29
True... I actually didn't have to do anything. But for curiosity's sake can you help me with my understanding for the fraction limits?
– Donald Devy
Jul 26 at 15:31
You did this correctly (as in you are using heuristics which can be made rigorous). You should dispense of your worries with the limit begin undefined; the limit must be infinite, since $(4k+3)/2$ becomes arbitrarily large as $k$ does.
– Mike Earnest
Jul 26 at 15:55
2
2
You have limits for $n$, but the arguments are in terms of $k$.
– Benedict W. J. Irwin
Jul 26 at 15:26
You have limits for $n$, but the arguments are in terms of $k$.
– Benedict W. J. Irwin
Jul 26 at 15:26
Sry I fixed it to be variable k.
– Donald Devy
Jul 26 at 15:28
Sry I fixed it to be variable k.
– Donald Devy
Jul 26 at 15:28
Isn't it obvious from the beginning that $4k+3toinfty$ as $ktoinfty?$
– saulspatz
Jul 26 at 15:29
Isn't it obvious from the beginning that $4k+3toinfty$ as $ktoinfty?$
– saulspatz
Jul 26 at 15:29
True... I actually didn't have to do anything. But for curiosity's sake can you help me with my understanding for the fraction limits?
– Donald Devy
Jul 26 at 15:31
True... I actually didn't have to do anything. But for curiosity's sake can you help me with my understanding for the fraction limits?
– Donald Devy
Jul 26 at 15:31
You did this correctly (as in you are using heuristics which can be made rigorous). You should dispense of your worries with the limit begin undefined; the limit must be infinite, since $(4k+3)/2$ becomes arbitrarily large as $k$ does.
– Mike Earnest
Jul 26 at 15:55
You did this correctly (as in you are using heuristics which can be made rigorous). You should dispense of your worries with the limit begin undefined; the limit must be infinite, since $(4k+3)/2$ becomes arbitrarily large as $k$ does.
– Mike Earnest
Jul 26 at 15:55
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
1
down vote
Simply note that
$$frac4k+32=2k+frac32ge 2k ge k to infty$$
then conclude by squeeze theorem.
answered Jul 26 at 15:31
gimusi
65k73583
add a comment |Â
up vote
0
down vote
Why so complicated?
$$lim_ktoinftyfrac4k+32=lim_ktoinfty(2k+1.5)=2infty+1.5=infty$$
answered Jul 26 at 15:30
Andrei
7,3702822
3
While right in spirit, treating $infty$ as a number gives me the willies.
– Randall
Jul 26 at 15:34
As long as you don't do things like $infty/infty$ or $inftycdot 0$ or $infty-infty$, you should be fine. It will boil down to the answer form @gimusi
– Andrei
Jul 26 at 15:36
add a comment |Â
up vote
0
down vote
Can't you find $k$ such that $$frac4k+32>1000000 ?$$
And larger numbers ?
If yes, the expression is unbounded.
answered Jul 26 at 16:29
Yves Daoust
110k665203
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Simply note that
$$frac4k+32=2k+frac32ge 2k ge k to infty$$
then conclude by squeeze theorem.
answered Jul 26 at 15:31
gimusi
65k73583
add a comment |Â
up vote
1
down vote
Simply note that
$$frac4k+32=2k+frac32ge 2k ge k to infty$$
then conclude by squeeze theorem.
answered Jul 26 at 15:31
gimusi
65k73583
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Simply note that
$$frac4k+32=2k+frac32ge 2k ge k to infty$$
then conclude by squeeze theorem.
answered Jul 26 at 15:31
gimusi
65k73583
Simply note that
$$frac4k+32=2k+frac32ge 2k ge k to infty$$
then conclude by squeeze theorem.
answered Jul 26 at 15:31
gimusi
65k73583
answered Jul 26 at 15:31
gimusi
65k73583
answered Jul 26 at 15:31
gimusi
65k73583
answered Jul 26 at 15:31
gimusi
65k73583
65k73583
add a comment |Â
add a comment |Â
up vote
0
down vote
Why so complicated?
$$lim_ktoinftyfrac4k+32=lim_ktoinfty(2k+1.5)=2infty+1.5=infty$$
answered Jul 26 at 15:30
Andrei
7,3702822
3
While right in spirit, treating $infty$ as a number gives me the willies.
– Randall
Jul 26 at 15:34
As long as you don't do things like $infty/infty$ or $inftycdot 0$ or $infty-infty$, you should be fine. It will boil down to the answer form @gimusi
– Andrei
Jul 26 at 15:36
add a comment |Â
up vote
0
down vote
Why so complicated?
$$lim_ktoinftyfrac4k+32=lim_ktoinfty(2k+1.5)=2infty+1.5=infty$$
answered Jul 26 at 15:30
Andrei
7,3702822
3
While right in spirit, treating $infty$ as a number gives me the willies.
– Randall
Jul 26 at 15:34
As long as you don't do things like $infty/infty$ or $inftycdot 0$ or $infty-infty$, you should be fine. It will boil down to the answer form @gimusi
– Andrei
Jul 26 at 15:36
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Why so complicated?
$$lim_ktoinftyfrac4k+32=lim_ktoinfty(2k+1.5)=2infty+1.5=infty$$
answered Jul 26 at 15:30
Andrei
7,3702822
Why so complicated?
$$lim_ktoinftyfrac4k+32=lim_ktoinfty(2k+1.5)=2infty+1.5=infty$$
answered Jul 26 at 15:30
Andrei
7,3702822
answered Jul 26 at 15:30
Andrei
7,3702822
answered Jul 26 at 15:30
Andrei
7,3702822
answered Jul 26 at 15:30
Andrei
7,3702822
7,3702822
3
While right in spirit, treating $infty$ as a number gives me the willies.
– Randall
Jul 26 at 15:34
As long as you don't do things like $infty/infty$ or $inftycdot 0$ or $infty-infty$, you should be fine. It will boil down to the answer form @gimusi
– Andrei
Jul 26 at 15:36
add a comment |Â
3
While right in spirit, treating $infty$ as a number gives me the willies.
– Randall
Jul 26 at 15:34
As long as you don't do things like $infty/infty$ or $inftycdot 0$ or $infty-infty$, you should be fine. It will boil down to the answer form @gimusi
– Andrei
Jul 26 at 15:36
3
3
While right in spirit, treating $infty$ as a number gives me the willies.
– Randall
Jul 26 at 15:34
While right in spirit, treating $infty$ as a number gives me the willies.
– Randall
Jul 26 at 15:34
As long as you don't do things like $infty/infty$ or $inftycdot 0$ or $infty-infty$, you should be fine. It will boil down to the answer form @gimusi
– Andrei
Jul 26 at 15:36
As long as you don't do things like $infty/infty$ or $inftycdot 0$ or $infty-infty$, you should be fine. It will boil down to the answer form @gimusi
– Andrei
Jul 26 at 15:36
add a comment |Â
up vote
0
down vote
Can't you find $k$ such that $$frac4k+32>1000000 ?$$
And larger numbers ?
If yes, the expression is unbounded.
answered Jul 26 at 16:29
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
Can't you find $k$ such that $$frac4k+32>1000000 ?$$
And larger numbers ?
If yes, the expression is unbounded.
answered Jul 26 at 16:29
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Can't you find $k$ such that $$frac4k+32>1000000 ?$$
And larger numbers ?
If yes, the expression is unbounded.
answered Jul 26 at 16:29
Yves Daoust
110k665203
Can't you find $k$ such that $$frac4k+32>1000000 ?$$
And larger numbers ?
If yes, the expression is unbounded.
answered Jul 26 at 16:29
Yves Daoust
110k665203
answered Jul 26 at 16:29
Yves Daoust
110k665203
answered Jul 26 at 16:29
Yves Daoust
110k665203
answered Jul 26 at 16:29
Yves Daoust
110k665203
110k665203
add a comment |Â
add a comment |Â
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2
You have limits for $n$, but the arguments are in terms of $k$.
– Benedict W. J. Irwin
Jul 26 at 15:26
Sry I fixed it to be variable k.
– Donald Devy
Jul 26 at 15:28
Isn't it obvious from the beginning that $4k+3toinfty$ as $ktoinfty?$
– saulspatz
Jul 26 at 15:29
True... I actually didn't have to do anything. But for curiosity's sake can you help me with my understanding for the fraction limits?
– Donald Devy
Jul 26 at 15:31
You did this correctly (as in you are using heuristics which can be made rigorous). You should dispense of your worries with the limit begin undefined; the limit must be infinite, since $(4k+3)/2$ becomes arbitrarily large as $k$ does.
– Mike Earnest
Jul 26 at 15:55