Why don't I get the correct limit of a sequence, regardless of how I arrange it (the sequence), while following the rules for solving limits?Finding the upper and lower limit of the following sequence.limit rules - How to apply limit rules when there are some nonexistent limitsHow to compute the limit of $a_n$ from the following expression?How to evaluate the following limits?How can I get the result of this limitsHow to prove that a limit exists for $lim_limitsnto inftyfracn+2n = 1$Updating the Definition of Limit of a Sequence to take in consideration sequences that are not defined for some finite number of natural numbersFind the formula for the nth term of the sequence and find its limitsHow to solve the sequence limit
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Why don't I get the correct limit of a sequence, regardless of how I arrange it (the sequence), while following the rules for solving limits?
Finding the upper and lower limit of the following sequence.limit rules - How to apply limit rules when there are some nonexistent limitsHow to compute the limit of $a_n$ from the following expression?How to evaluate the following limits?How can I get the result of this limitsHow to prove that a limit exists for $lim_limitsnto inftyfracn+2n = 1$Updating the Definition of Limit of a Sequence to take in consideration sequences that are not defined for some finite number of natural numbersFind the formula for the nth term of the sequence and find its limitsHow to solve the sequence limit
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
The fact that, when solving limits of sequences ($n in mathbbN$), it strikes me as very bizarre that, even though I follow all the (elementary) rules for solving limits (of sequences), I can get different results for the same example! Consider the following:
$$
lim_n to infty left( frac3n + 12n - 1 right) = fraclim_n to infty(3n + 1)lim_n to infty (2n - 1) = fraclim_n to infty(3n) + lim_n to infty(1)lim_n to infty (2n) - lim_n to infty(1) = fracinfty + 1infty - 1 Longrightarrow textThis sequence does not have a limit.
$$
I followed all of the rules for solving limits, yet this conclusion is wrong. The correct approach is the following.
$$
lim_n to infty left( frac3n + 12n - 1 right) = lim_n to infty left( fracn(3 + frac1n)n(2 - frac1n) right) = lim_n to infty left( frac3 + frac1n2 - frac1n right) =fraclim_n to infty(3 + frac1n)lim_n to infty (2 - frac1n) = fraclim_n to infty(3) + lim_n to infty(frac1n)lim_n to infty (2) - lim_n to infty(frac1n) = frac3 + 02 - 0 = frac32.
$$
My question is: why do I get the correct answer only if I arrange the sequence a certain way? I don't understand that at all. Thank you for your answers in advance and God, Jesus Christ, bless and keep you always.
sequences-and-series limits
asked 8 hours ago
Gregor PerčičGregor Perčič
6691 gold badge5 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
The fact that, when solving limits of sequences ($n in mathbbN$), it strikes me as very bizarre that, even though I follow all the (elementary) rules for solving limits (of sequences), I can get different results for the same example! Consider the following:
$$
lim_n to infty left( frac3n + 12n - 1 right) = fraclim_n to infty(3n + 1)lim_n to infty (2n - 1) = fraclim_n to infty(3n) + lim_n to infty(1)lim_n to infty (2n) - lim_n to infty(1) = fracinfty + 1infty - 1 Longrightarrow textThis sequence does not have a limit.
$$
I followed all of the rules for solving limits, yet this conclusion is wrong. The correct approach is the following.
$$
lim_n to infty left( frac3n + 12n - 1 right) = lim_n to infty left( fracn(3 + frac1n)n(2 - frac1n) right) = lim_n to infty left( frac3 + frac1n2 - frac1n right) =fraclim_n to infty(3 + frac1n)lim_n to infty (2 - frac1n) = fraclim_n to infty(3) + lim_n to infty(frac1n)lim_n to infty (2) - lim_n to infty(frac1n) = frac3 + 02 - 0 = frac32.
$$
My question is: why do I get the correct answer only if I arrange the sequence a certain way? I don't understand that at all. Thank you for your answers in advance and God, Jesus Christ, bless and keep you always.
sequences-and-series limits
asked 8 hours ago
Gregor PerčičGregor Perčič
6691 gold badge5 silver badges12 bronze badges
$endgroup$
6
$begingroup$
The error is in the first part: you are assuming that if you get $fracinfty+1infty-1$, then the limit doesn't exist. This is wrong. You cannot conclude anything about the existence of the limit if you get a so-called "indeterminate form," such as $fracinftyinfty$ or $frac00$.
$endgroup$
– symplectomorphic
8 hours ago
$begingroup$
@symplectomorphic Could you elaborate on that, please?
$endgroup$
– Gregor Perčič
8 hours ago
1
$begingroup$
Take the limit of x/x as x approaches zero. Obviously the answer is 1, but taking the limit of the top and bottom separately gives 0/0. This is an indeterminate form, not a limit that doesn't exist.
$endgroup$
– Gabe
8 hours ago
$begingroup$
If a limit is of the "form" $fracinftyinfty,$ that does not mean that the limit does not exst.
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
Alternatively you can first apply multiplication and addition laws to see that $$ lim_ntoinftyfrac3n+12n−1=frac32lim_ntoinftyfrac6n+26n−3 =frac32+frac32lim_ntoinftyfrac56n−3 $$
$endgroup$
– LutzL
8 hours ago
add a comment |
$begingroup$
The fact that, when solving limits of sequences ($n in mathbbN$), it strikes me as very bizarre that, even though I follow all the (elementary) rules for solving limits (of sequences), I can get different results for the same example! Consider the following:
$$
lim_n to infty left( frac3n + 12n - 1 right) = fraclim_n to infty(3n + 1)lim_n to infty (2n - 1) = fraclim_n to infty(3n) + lim_n to infty(1)lim_n to infty (2n) - lim_n to infty(1) = fracinfty + 1infty - 1 Longrightarrow textThis sequence does not have a limit.
$$
I followed all of the rules for solving limits, yet this conclusion is wrong. The correct approach is the following.
$$
lim_n to infty left( frac3n + 12n - 1 right) = lim_n to infty left( fracn(3 + frac1n)n(2 - frac1n) right) = lim_n to infty left( frac3 + frac1n2 - frac1n right) =fraclim_n to infty(3 + frac1n)lim_n to infty (2 - frac1n) = fraclim_n to infty(3) + lim_n to infty(frac1n)lim_n to infty (2) - lim_n to infty(frac1n) = frac3 + 02 - 0 = frac32.
$$
My question is: why do I get the correct answer only if I arrange the sequence a certain way? I don't understand that at all. Thank you for your answers in advance and God, Jesus Christ, bless and keep you always.
sequences-and-series limits
asked 8 hours ago
Gregor PerčičGregor Perčič
6691 gold badge5 silver badges12 bronze badges
$endgroup$
The fact that, when solving limits of sequences ($n in mathbbN$), it strikes me as very bizarre that, even though I follow all the (elementary) rules for solving limits (of sequences), I can get different results for the same example! Consider the following:
$$
lim_n to infty left( frac3n + 12n - 1 right) = fraclim_n to infty(3n + 1)lim_n to infty (2n - 1) = fraclim_n to infty(3n) + lim_n to infty(1)lim_n to infty (2n) - lim_n to infty(1) = fracinfty + 1infty - 1 Longrightarrow textThis sequence does not have a limit.
$$
I followed all of the rules for solving limits, yet this conclusion is wrong. The correct approach is the following.
$$
lim_n to infty left( frac3n + 12n - 1 right) = lim_n to infty left( fracn(3 + frac1n)n(2 - frac1n) right) = lim_n to infty left( frac3 + frac1n2 - frac1n right) =fraclim_n to infty(3 + frac1n)lim_n to infty (2 - frac1n) = fraclim_n to infty(3) + lim_n to infty(frac1n)lim_n to infty (2) - lim_n to infty(frac1n) = frac3 + 02 - 0 = frac32.
$$
My question is: why do I get the correct answer only if I arrange the sequence a certain way? I don't understand that at all. Thank you for your answers in advance and God, Jesus Christ, bless and keep you always.
sequences-and-series limits
sequences-and-series limits
asked 8 hours ago
Gregor PerčičGregor Perčič
6691 gold badge5 silver badges12 bronze badges
asked 8 hours ago
Gregor PerčičGregor Perčič
6691 gold badge5 silver badges12 bronze badges
asked 8 hours ago
Gregor PerčičGregor Perčič
6691 gold badge5 silver badges12 bronze badges
asked 8 hours ago
Gregor PerčičGregor Perčič
6691 gold badge5 silver badges12 bronze badges
asked 8 hours ago
Gregor PerčičGregor Perčič
6691 gold badge5 silver badges12 bronze badges
6691 gold badge5 silver badges12 bronze badges
6
$begingroup$
The error is in the first part: you are assuming that if you get $fracinfty+1infty-1$, then the limit doesn't exist. This is wrong. You cannot conclude anything about the existence of the limit if you get a so-called "indeterminate form," such as $fracinftyinfty$ or $frac00$.
$endgroup$
– symplectomorphic
8 hours ago
$begingroup$
@symplectomorphic Could you elaborate on that, please?
$endgroup$
– Gregor Perčič
8 hours ago
1
$begingroup$
Take the limit of x/x as x approaches zero. Obviously the answer is 1, but taking the limit of the top and bottom separately gives 0/0. This is an indeterminate form, not a limit that doesn't exist.
$endgroup$
– Gabe
8 hours ago
$begingroup$
If a limit is of the "form" $fracinftyinfty,$ that does not mean that the limit does not exst.
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
Alternatively you can first apply multiplication and addition laws to see that $$ lim_ntoinftyfrac3n+12n−1=frac32lim_ntoinftyfrac6n+26n−3 =frac32+frac32lim_ntoinftyfrac56n−3 $$
$endgroup$
– LutzL
8 hours ago
add a comment |
6
$begingroup$
The error is in the first part: you are assuming that if you get $fracinfty+1infty-1$, then the limit doesn't exist. This is wrong. You cannot conclude anything about the existence of the limit if you get a so-called "indeterminate form," such as $fracinftyinfty$ or $frac00$.
$endgroup$
– symplectomorphic
8 hours ago
$begingroup$
@symplectomorphic Could you elaborate on that, please?
$endgroup$
– Gregor Perčič
8 hours ago
1
$begingroup$
Take the limit of x/x as x approaches zero. Obviously the answer is 1, but taking the limit of the top and bottom separately gives 0/0. This is an indeterminate form, not a limit that doesn't exist.
$endgroup$
– Gabe
8 hours ago
$begingroup$
If a limit is of the "form" $fracinftyinfty,$ that does not mean that the limit does not exst.
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
Alternatively you can first apply multiplication and addition laws to see that $$ lim_ntoinftyfrac3n+12n−1=frac32lim_ntoinftyfrac6n+26n−3 =frac32+frac32lim_ntoinftyfrac56n−3 $$
$endgroup$
– LutzL
8 hours ago
6
6
$begingroup$
The error is in the first part: you are assuming that if you get $fracinfty+1infty-1$, then the limit doesn't exist. This is wrong. You cannot conclude anything about the existence of the limit if you get a so-called "indeterminate form," such as $fracinftyinfty$ or $frac00$.
$endgroup$
– symplectomorphic
8 hours ago
$begingroup$
The error is in the first part: you are assuming that if you get $fracinfty+1infty-1$, then the limit doesn't exist. This is wrong. You cannot conclude anything about the existence of the limit if you get a so-called "indeterminate form," such as $fracinftyinfty$ or $frac00$.
$endgroup$
– symplectomorphic
8 hours ago
$begingroup$
@symplectomorphic Could you elaborate on that, please?
$endgroup$
– Gregor Perčič
8 hours ago
$begingroup$
@symplectomorphic Could you elaborate on that, please?
$endgroup$
– Gregor Perčič
8 hours ago
1
1
$begingroup$
Take the limit of x/x as x approaches zero. Obviously the answer is 1, but taking the limit of the top and bottom separately gives 0/0. This is an indeterminate form, not a limit that doesn't exist.
$endgroup$
– Gabe
8 hours ago
$begingroup$
Take the limit of x/x as x approaches zero. Obviously the answer is 1, but taking the limit of the top and bottom separately gives 0/0. This is an indeterminate form, not a limit that doesn't exist.
$endgroup$
– Gabe
8 hours ago
$begingroup$
If a limit is of the "form" $fracinftyinfty,$ that does not mean that the limit does not exst.
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
If a limit is of the "form" $fracinftyinfty,$ that does not mean that the limit does not exst.
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
Alternatively you can first apply multiplication and addition laws to see that $$ lim_ntoinftyfrac3n+12n−1=frac32lim_ntoinftyfrac6n+26n−3 =frac32+frac32lim_ntoinftyfrac56n−3 $$
$endgroup$
– LutzL
8 hours ago
$begingroup$
Alternatively you can first apply multiplication and addition laws to see that $$ lim_ntoinftyfrac3n+12n−1=frac32lim_ntoinftyfrac6n+26n−3 =frac32+frac32lim_ntoinftyfrac56n−3 $$
$endgroup$
– LutzL
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You did not in fact follow all the rules (and this is a good example of why understanding why the rules are true is important). The relevant rule is:
If both $lim_nrightarrow af(n)$ and $lim_nrightarrow ag(n)$ exist and are finite (and $lim_nrightarrow ag(n)not=0$), then $$lim_nrightarrow af(n)over g(n)=lim_nrightarrow af(n)over lim_nrightarrow ag(n).$$
There are various other rules of similar flavor. But the point is that the "rule" you've tried to apply is not one of them. (And it's a good exercise at this point to go through your textbook and look at what the rules do in fact say, and note that none of them actually get you what you want.)
Indeed, the example you give is a good example of the limitations of these rules: while we can often manipulate limits "algebraically," we can't do this in all cases, and some hypotheses are needed. It's also a good example of why proofs are important, since there are plenty of plausible-sounding statements which are in fact false.
answered 8 hours ago
Noah SchweberNoah Schweber
140k10 gold badges168 silver badges318 bronze badges
$endgroup$
$begingroup$
Hmmm, thank you very much, I didn't know about this caveat! We only scratched the surface of what we can do with limits, as I only studied them in the context of sequences and series. Jesus bless and keep you always. Amen.
$endgroup$
– Gregor Perčič
8 hours ago
add a comment |
$begingroup$
Your reference page here starts with this introduction (emphasis mine):
Listed here are a couple of basic limits and the standard limit laws which, when used in conjunction, can find most limits. They are listed for standard, two-sided limits, but they work for all forms of limits. However, note that if a limit is infinite, then the limit does not exist.
So everywhere below that warning, if there is a statement that a limit must exist, then it's implicit that the limit must not be $infty$.
answered 8 hours ago
Chris CulterChris Culter
22.7k4 gold badges40 silver badges92 bronze badges
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You did not in fact follow all the rules (and this is a good example of why understanding why the rules are true is important). The relevant rule is:
If both $lim_nrightarrow af(n)$ and $lim_nrightarrow ag(n)$ exist and are finite (and $lim_nrightarrow ag(n)not=0$), then $$lim_nrightarrow af(n)over g(n)=lim_nrightarrow af(n)over lim_nrightarrow ag(n).$$
There are various other rules of similar flavor. But the point is that the "rule" you've tried to apply is not one of them. (And it's a good exercise at this point to go through your textbook and look at what the rules do in fact say, and note that none of them actually get you what you want.)
Indeed, the example you give is a good example of the limitations of these rules: while we can often manipulate limits "algebraically," we can't do this in all cases, and some hypotheses are needed. It's also a good example of why proofs are important, since there are plenty of plausible-sounding statements which are in fact false.
answered 8 hours ago
Noah SchweberNoah Schweber
140k10 gold badges168 silver badges318 bronze badges
$endgroup$
$begingroup$
Hmmm, thank you very much, I didn't know about this caveat! We only scratched the surface of what we can do with limits, as I only studied them in the context of sequences and series. Jesus bless and keep you always. Amen.
$endgroup$
– Gregor Perčič
8 hours ago
add a comment |
$begingroup$
You did not in fact follow all the rules (and this is a good example of why understanding why the rules are true is important). The relevant rule is:
If both $lim_nrightarrow af(n)$ and $lim_nrightarrow ag(n)$ exist and are finite (and $lim_nrightarrow ag(n)not=0$), then $$lim_nrightarrow af(n)over g(n)=lim_nrightarrow af(n)over lim_nrightarrow ag(n).$$
There are various other rules of similar flavor. But the point is that the "rule" you've tried to apply is not one of them. (And it's a good exercise at this point to go through your textbook and look at what the rules do in fact say, and note that none of them actually get you what you want.)
Indeed, the example you give is a good example of the limitations of these rules: while we can often manipulate limits "algebraically," we can't do this in all cases, and some hypotheses are needed. It's also a good example of why proofs are important, since there are plenty of plausible-sounding statements which are in fact false.
answered 8 hours ago
Noah SchweberNoah Schweber
140k10 gold badges168 silver badges318 bronze badges
$endgroup$
$begingroup$
Hmmm, thank you very much, I didn't know about this caveat! We only scratched the surface of what we can do with limits, as I only studied them in the context of sequences and series. Jesus bless and keep you always. Amen.
$endgroup$
– Gregor Perčič
8 hours ago
add a comment |
$begingroup$
You did not in fact follow all the rules (and this is a good example of why understanding why the rules are true is important). The relevant rule is:
If both $lim_nrightarrow af(n)$ and $lim_nrightarrow ag(n)$ exist and are finite (and $lim_nrightarrow ag(n)not=0$), then $$lim_nrightarrow af(n)over g(n)=lim_nrightarrow af(n)over lim_nrightarrow ag(n).$$
There are various other rules of similar flavor. But the point is that the "rule" you've tried to apply is not one of them. (And it's a good exercise at this point to go through your textbook and look at what the rules do in fact say, and note that none of them actually get you what you want.)
Indeed, the example you give is a good example of the limitations of these rules: while we can often manipulate limits "algebraically," we can't do this in all cases, and some hypotheses are needed. It's also a good example of why proofs are important, since there are plenty of plausible-sounding statements which are in fact false.
answered 8 hours ago
Noah SchweberNoah Schweber
140k10 gold badges168 silver badges318 bronze badges
$endgroup$
You did not in fact follow all the rules (and this is a good example of why understanding why the rules are true is important). The relevant rule is:
If both $lim_nrightarrow af(n)$ and $lim_nrightarrow ag(n)$ exist and are finite (and $lim_nrightarrow ag(n)not=0$), then $$lim_nrightarrow af(n)over g(n)=lim_nrightarrow af(n)over lim_nrightarrow ag(n).$$
There are various other rules of similar flavor. But the point is that the "rule" you've tried to apply is not one of them. (And it's a good exercise at this point to go through your textbook and look at what the rules do in fact say, and note that none of them actually get you what you want.)
Indeed, the example you give is a good example of the limitations of these rules: while we can often manipulate limits "algebraically," we can't do this in all cases, and some hypotheses are needed. It's also a good example of why proofs are important, since there are plenty of plausible-sounding statements which are in fact false.
answered 8 hours ago
Noah SchweberNoah Schweber
140k10 gold badges168 silver badges318 bronze badges
answered 8 hours ago
Noah SchweberNoah Schweber
140k10 gold badges168 silver badges318 bronze badges
answered 8 hours ago
Noah SchweberNoah Schweber
140k10 gold badges168 silver badges318 bronze badges
answered 8 hours ago
Noah SchweberNoah Schweber
140k10 gold badges168 silver badges318 bronze badges
140k10 gold badges168 silver badges318 bronze badges
$begingroup$
Hmmm, thank you very much, I didn't know about this caveat! We only scratched the surface of what we can do with limits, as I only studied them in the context of sequences and series. Jesus bless and keep you always. Amen.
$endgroup$
– Gregor Perčič
8 hours ago
add a comment |
$begingroup$
Hmmm, thank you very much, I didn't know about this caveat! We only scratched the surface of what we can do with limits, as I only studied them in the context of sequences and series. Jesus bless and keep you always. Amen.
$endgroup$
– Gregor Perčič
8 hours ago
$begingroup$
Hmmm, thank you very much, I didn't know about this caveat! We only scratched the surface of what we can do with limits, as I only studied them in the context of sequences and series. Jesus bless and keep you always. Amen.
$endgroup$
– Gregor Perčič
8 hours ago
$begingroup$
Hmmm, thank you very much, I didn't know about this caveat! We only scratched the surface of what we can do with limits, as I only studied them in the context of sequences and series. Jesus bless and keep you always. Amen.
$endgroup$
– Gregor Perčič
8 hours ago
add a comment |
$begingroup$
Your reference page here starts with this introduction (emphasis mine):
Listed here are a couple of basic limits and the standard limit laws which, when used in conjunction, can find most limits. They are listed for standard, two-sided limits, but they work for all forms of limits. However, note that if a limit is infinite, then the limit does not exist.
So everywhere below that warning, if there is a statement that a limit must exist, then it's implicit that the limit must not be $infty$.
answered 8 hours ago
Chris CulterChris Culter
22.7k4 gold badges40 silver badges92 bronze badges
$endgroup$
add a comment |
$begingroup$
Your reference page here starts with this introduction (emphasis mine):
Listed here are a couple of basic limits and the standard limit laws which, when used in conjunction, can find most limits. They are listed for standard, two-sided limits, but they work for all forms of limits. However, note that if a limit is infinite, then the limit does not exist.
So everywhere below that warning, if there is a statement that a limit must exist, then it's implicit that the limit must not be $infty$.
answered 8 hours ago
Chris CulterChris Culter
22.7k4 gold badges40 silver badges92 bronze badges
$endgroup$
add a comment |
$begingroup$
Your reference page here starts with this introduction (emphasis mine):
Listed here are a couple of basic limits and the standard limit laws which, when used in conjunction, can find most limits. They are listed for standard, two-sided limits, but they work for all forms of limits. However, note that if a limit is infinite, then the limit does not exist.
So everywhere below that warning, if there is a statement that a limit must exist, then it's implicit that the limit must not be $infty$.
answered 8 hours ago
Chris CulterChris Culter
22.7k4 gold badges40 silver badges92 bronze badges
$endgroup$
Your reference page here starts with this introduction (emphasis mine):
Listed here are a couple of basic limits and the standard limit laws which, when used in conjunction, can find most limits. They are listed for standard, two-sided limits, but they work for all forms of limits. However, note that if a limit is infinite, then the limit does not exist.
So everywhere below that warning, if there is a statement that a limit must exist, then it's implicit that the limit must not be $infty$.
answered 8 hours ago
Chris CulterChris Culter
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answered 8 hours ago
Chris CulterChris Culter
22.7k4 gold badges40 silver badges92 bronze badges
answered 8 hours ago
Chris CulterChris Culter
22.7k4 gold badges40 silver badges92 bronze badges
answered 8 hours ago
Chris CulterChris Culter
22.7k4 gold badges40 silver badges92 bronze badges
22.7k4 gold badges40 silver badges92 bronze badges
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6
$begingroup$
The error is in the first part: you are assuming that if you get $fracinfty+1infty-1$, then the limit doesn't exist. This is wrong. You cannot conclude anything about the existence of the limit if you get a so-called "indeterminate form," such as $fracinftyinfty$ or $frac00$.
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– symplectomorphic
8 hours ago
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@symplectomorphic Could you elaborate on that, please?
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– Gregor Perčič
8 hours ago
1
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Take the limit of x/x as x approaches zero. Obviously the answer is 1, but taking the limit of the top and bottom separately gives 0/0. This is an indeterminate form, not a limit that doesn't exist.
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– Gabe
8 hours ago
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If a limit is of the "form" $fracinftyinfty,$ that does not mean that the limit does not exst.
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– Thomas Andrews
8 hours ago
$begingroup$
Alternatively you can first apply multiplication and addition laws to see that $$ lim_ntoinftyfrac3n+12n−1=frac32lim_ntoinftyfrac6n+26n−3 =frac32+frac32lim_ntoinftyfrac56n−3 $$
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– LutzL
8 hours ago