On "Proving' Infinite Limits, And A Counterexample

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The textbook I am using has come up with this method to prove why in infinite limits, the term with the highest degree will dominate:

It then goes on to say that at infinity, 1000/3x and all the other terms will be virtually 0. Therefore, the function will approach 1 (since it it lim x->infinity (1-0+0-0).
However, my problem with this, is for instance, take a function x+2/x-1. On the logic of the above, when we break it up, it is (x+2)*(1/x-1). At infinity, 1/x-1=0. So therefore, the limit is 0. But the limit of x+2/x-2 is actually 1.
We see this again we we take a function sin(x)*1/x. Again, at infinity, the limit should be 0, but it's not.
So what can explain this contradiction? Why is the above proof still right in these situations despite the examples below? How can we know that the proof above is universally true? And why is my logic wrong? Why is is x+2/x-1 not equals to 0?
calculus limits self-learning limits-without-lhopital
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The textbook I am using has come up with this method to prove why in infinite limits, the term with the highest degree will dominate:

It then goes on to say that at infinity, 1000/3x and all the other terms will be virtually 0. Therefore, the function will approach 1 (since it it lim x->infinity (1-0+0-0).
However, my problem with this, is for instance, take a function x+2/x-1. On the logic of the above, when we break it up, it is (x+2)*(1/x-1). At infinity, 1/x-1=0. So therefore, the limit is 0. But the limit of x+2/x-2 is actually 1.
We see this again we we take a function sin(x)*1/x. Again, at infinity, the limit should be 0, but it's not.
So what can explain this contradiction? Why is the above proof still right in these situations despite the examples below? How can we know that the proof above is universally true? And why is my logic wrong? Why is is x+2/x-1 not equals to 0?
calculus limits self-learning limits-without-lhopital
1
The examples are different: in $dfrac (x+2)(x-1)=(x+2) dfrac 1(x-1)$ you have $x$ also in the "part" $(x+2)$ and thus you cannot consider the limit for $x$ only of the other part.
â Mauro ALLEGRANZA
Aug 28 at 13:00
1
While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
â Mauro ALLEGRANZA
Aug 28 at 13:01
@MauroALLEGRANZA Can you elaborate what you mean "you have x also in the "part" (x+2) and thus you cannot consider the limit for x only of the other part."? Why can't you consider the limit for x only of the other part?
â user577730
Aug 28 at 13:04
Consider $x/x$. We know that $lim_x to infty 1/x = 0$ and that $lim_x to infty x = infty$, but $lim_x to infty x/x = 1$. You have to look at the term as a whole.
â Bill Wallis
Aug 28 at 13:11
1
Because you cannot "compute" the limit for $dfrac1(x-1)$ for $x$ to $0$ and then multiply the result for $(x+2)$, because also $(x+2)$ is "affected" by the operation of "computing the limit".
â Mauro ALLEGRANZA
Aug 28 at 13:11
 |Â
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1
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up vote
1
down vote
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The textbook I am using has come up with this method to prove why in infinite limits, the term with the highest degree will dominate:

It then goes on to say that at infinity, 1000/3x and all the other terms will be virtually 0. Therefore, the function will approach 1 (since it it lim x->infinity (1-0+0-0).
However, my problem with this, is for instance, take a function x+2/x-1. On the logic of the above, when we break it up, it is (x+2)*(1/x-1). At infinity, 1/x-1=0. So therefore, the limit is 0. But the limit of x+2/x-2 is actually 1.
We see this again we we take a function sin(x)*1/x. Again, at infinity, the limit should be 0, but it's not.
So what can explain this contradiction? Why is the above proof still right in these situations despite the examples below? How can we know that the proof above is universally true? And why is my logic wrong? Why is is x+2/x-1 not equals to 0?
calculus limits self-learning limits-without-lhopital
The textbook I am using has come up with this method to prove why in infinite limits, the term with the highest degree will dominate:

It then goes on to say that at infinity, 1000/3x and all the other terms will be virtually 0. Therefore, the function will approach 1 (since it it lim x->infinity (1-0+0-0).
However, my problem with this, is for instance, take a function x+2/x-1. On the logic of the above, when we break it up, it is (x+2)*(1/x-1). At infinity, 1/x-1=0. So therefore, the limit is 0. But the limit of x+2/x-2 is actually 1.
We see this again we we take a function sin(x)*1/x. Again, at infinity, the limit should be 0, but it's not.
So what can explain this contradiction? Why is the above proof still right in these situations despite the examples below? How can we know that the proof above is universally true? And why is my logic wrong? Why is is x+2/x-1 not equals to 0?
calculus limits self-learning limits-without-lhopital
asked Aug 28 at 12:56
user577730
1
The examples are different: in $dfrac (x+2)(x-1)=(x+2) dfrac 1(x-1)$ you have $x$ also in the "part" $(x+2)$ and thus you cannot consider the limit for $x$ only of the other part.
â Mauro ALLEGRANZA
Aug 28 at 13:00
1
While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
â Mauro ALLEGRANZA
Aug 28 at 13:01
@MauroALLEGRANZA Can you elaborate what you mean "you have x also in the "part" (x+2) and thus you cannot consider the limit for x only of the other part."? Why can't you consider the limit for x only of the other part?
â user577730
Aug 28 at 13:04
Consider $x/x$. We know that $lim_x to infty 1/x = 0$ and that $lim_x to infty x = infty$, but $lim_x to infty x/x = 1$. You have to look at the term as a whole.
â Bill Wallis
Aug 28 at 13:11
1
Because you cannot "compute" the limit for $dfrac1(x-1)$ for $x$ to $0$ and then multiply the result for $(x+2)$, because also $(x+2)$ is "affected" by the operation of "computing the limit".
â Mauro ALLEGRANZA
Aug 28 at 13:11
 |Â
show 2 more comments
1
The examples are different: in $dfrac (x+2)(x-1)=(x+2) dfrac 1(x-1)$ you have $x$ also in the "part" $(x+2)$ and thus you cannot consider the limit for $x$ only of the other part.
â Mauro ALLEGRANZA
Aug 28 at 13:00
1
While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
â Mauro ALLEGRANZA
Aug 28 at 13:01
@MauroALLEGRANZA Can you elaborate what you mean "you have x also in the "part" (x+2) and thus you cannot consider the limit for x only of the other part."? Why can't you consider the limit for x only of the other part?
â user577730
Aug 28 at 13:04
Consider $x/x$. We know that $lim_x to infty 1/x = 0$ and that $lim_x to infty x = infty$, but $lim_x to infty x/x = 1$. You have to look at the term as a whole.
â Bill Wallis
Aug 28 at 13:11
1
Because you cannot "compute" the limit for $dfrac1(x-1)$ for $x$ to $0$ and then multiply the result for $(x+2)$, because also $(x+2)$ is "affected" by the operation of "computing the limit".
â Mauro ALLEGRANZA
Aug 28 at 13:11
1
1
The examples are different: in $dfrac (x+2)(x-1)=(x+2) dfrac 1(x-1)$ you have $x$ also in the "part" $(x+2)$ and thus you cannot consider the limit for $x$ only of the other part.
â Mauro ALLEGRANZA
Aug 28 at 13:00
The examples are different: in $dfrac (x+2)(x-1)=(x+2) dfrac 1(x-1)$ you have $x$ also in the "part" $(x+2)$ and thus you cannot consider the limit for $x$ only of the other part.
â Mauro ALLEGRANZA
Aug 28 at 13:00
1
1
While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
â Mauro ALLEGRANZA
Aug 28 at 13:01
While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
â Mauro ALLEGRANZA
Aug 28 at 13:01
@MauroALLEGRANZA Can you elaborate what you mean "you have x also in the "part" (x+2) and thus you cannot consider the limit for x only of the other part."? Why can't you consider the limit for x only of the other part?
â user577730
Aug 28 at 13:04
@MauroALLEGRANZA Can you elaborate what you mean "you have x also in the "part" (x+2) and thus you cannot consider the limit for x only of the other part."? Why can't you consider the limit for x only of the other part?
â user577730
Aug 28 at 13:04
Consider $x/x$. We know that $lim_x to infty 1/x = 0$ and that $lim_x to infty x = infty$, but $lim_x to infty x/x = 1$. You have to look at the term as a whole.
â Bill Wallis
Aug 28 at 13:11
Consider $x/x$. We know that $lim_x to infty 1/x = 0$ and that $lim_x to infty x = infty$, but $lim_x to infty x/x = 1$. You have to look at the term as a whole.
â Bill Wallis
Aug 28 at 13:11
1
1
Because you cannot "compute" the limit for $dfrac1(x-1)$ for $x$ to $0$ and then multiply the result for $(x+2)$, because also $(x+2)$ is "affected" by the operation of "computing the limit".
â Mauro ALLEGRANZA
Aug 28 at 13:11
Because you cannot "compute" the limit for $dfrac1(x-1)$ for $x$ to $0$ and then multiply the result for $(x+2)$, because also $(x+2)$ is "affected" by the operation of "computing the limit".
â Mauro ALLEGRANZA
Aug 28 at 13:11
 |Â
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3 Answers
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Applying the same method to $fracx+2x-1$, what we get isbeginalignlim_xtoinftyfracx+2x-1&=lim_xtoinftyfrac(x-1)+3x-1\&=lim_xtoinfty1+frac3x-1\&=1.endalign
And the limit $lim_xtoinftyfracsin xx$ is $0$.
Ah, I see. But why can't you do (x+2) * 1/(x-1)? Why would that yield a wrong answer?
â user577730
Aug 28 at 13:02
2
@Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
â Vasya
Aug 28 at 13:10
@Vasya Okay. But why is infinity multiplied by zero an uncertainty? And will infinity multiplied by any other value besides zero be acceptable? So will say infinity times 3 be acceptable?
â user577730
Aug 28 at 13:34
1
@Mathguy: I follow these rules when dealing with limits: 1) infinity multiplied by a non-zero value is infinity (could be negative); 2) a non-zero value divided by infinity is zero; 3) zero multiplied by non-zero value is zero; 4) a non-zero value divided by zero is infinity (could be negative). If you have any other combination of infinity and zero, you need to modify the expression to fit one of the 4 cases.
â Vasya
Aug 28 at 14:04
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[T]ake a function $fracx+2x-1$. On the logic of the above, when we break it up, it is $(x+2)frac1x-1$. At infinity, $frac1x-1$ [tends to] $0$. So therefore, the limit is 0. But the limit of $fracx+2x-1$ is actually $1$.
It's very good that you're noticing what seems to be a contradiction here. The flaw in your reasoning is that the limit laws can only be used when the constituent limits exist. For instance, the product law for limits is
If $lim_xtoinftyf(x) = L$ and $lim_xtoinfty g(x) = M$, then $lim_xtoinfty f(x)g(x) = LM$.
You're trying to apply this law to the situation $f(x) = x+2$ and $g(x) = frac1x+1$. Yes, $lim_xtoinfty g(x) = 0$ and âÂÂanything times zero is zero.â However, $lim_xtoinfty f(x)$ is not a âÂÂthingâÂÂâÂÂit doesn't exist. So we cannot apply the product law.
Students get confused because we say $lim_xtoinftyf(x) =infty$, and because there's an equals sign and a symbol on the other side, it sure looks like the limit âÂÂisâ something. But $lim_xtoinftyf(x) =infty$ is shorthand for an infinite limit. There is some arithmetic than can be done with infinity, but $inftycdot0$ is undefined.
We see this again we we take a function $sin(x)frac1x$. Again, at infinity, the limit should be 0, but it's not.
Again, the first factor has no limit as $xtoinfty$, so the product law for limits doesn't apply.
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There is no contradiction at all.
The rule says that in a sum of terms you can neglect the terms that tend to zero (unless all the terms tend to zero).
The rule never said that this holds for a product.
For instance, consider $x+dfrac1x$ vs. $xcdotdfrac1x$.
For increasing powers of $10$, we have
$$x+frac1x=2,10.1,100.01,1000.001,10000.0001cdots$$
which clearly tends to $x$, vs.
$$xcdotfrac 1x=1,1,1,1,1,cdots$$
which is intermediate between $x$ and $dfrac1x$ and doesn't vanish.
For these reasons, you can write statements such as
$$lim_xtoinftyfrac2x+33x-5=lim_xtoinftyfrac2x3x=frac25.$$
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Applying the same method to $fracx+2x-1$, what we get isbeginalignlim_xtoinftyfracx+2x-1&=lim_xtoinftyfrac(x-1)+3x-1\&=lim_xtoinfty1+frac3x-1\&=1.endalign
And the limit $lim_xtoinftyfracsin xx$ is $0$.
Ah, I see. But why can't you do (x+2) * 1/(x-1)? Why would that yield a wrong answer?
â user577730
Aug 28 at 13:02
2
@Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
â Vasya
Aug 28 at 13:10
@Vasya Okay. But why is infinity multiplied by zero an uncertainty? And will infinity multiplied by any other value besides zero be acceptable? So will say infinity times 3 be acceptable?
â user577730
Aug 28 at 13:34
1
@Mathguy: I follow these rules when dealing with limits: 1) infinity multiplied by a non-zero value is infinity (could be negative); 2) a non-zero value divided by infinity is zero; 3) zero multiplied by non-zero value is zero; 4) a non-zero value divided by zero is infinity (could be negative). If you have any other combination of infinity and zero, you need to modify the expression to fit one of the 4 cases.
â Vasya
Aug 28 at 14:04
add a comment |Â
up vote
1
down vote
accepted
Applying the same method to $fracx+2x-1$, what we get isbeginalignlim_xtoinftyfracx+2x-1&=lim_xtoinftyfrac(x-1)+3x-1\&=lim_xtoinfty1+frac3x-1\&=1.endalign
And the limit $lim_xtoinftyfracsin xx$ is $0$.
Ah, I see. But why can't you do (x+2) * 1/(x-1)? Why would that yield a wrong answer?
â user577730
Aug 28 at 13:02
2
@Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
â Vasya
Aug 28 at 13:10
@Vasya Okay. But why is infinity multiplied by zero an uncertainty? And will infinity multiplied by any other value besides zero be acceptable? So will say infinity times 3 be acceptable?
â user577730
Aug 28 at 13:34
1
@Mathguy: I follow these rules when dealing with limits: 1) infinity multiplied by a non-zero value is infinity (could be negative); 2) a non-zero value divided by infinity is zero; 3) zero multiplied by non-zero value is zero; 4) a non-zero value divided by zero is infinity (could be negative). If you have any other combination of infinity and zero, you need to modify the expression to fit one of the 4 cases.
â Vasya
Aug 28 at 14:04
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Applying the same method to $fracx+2x-1$, what we get isbeginalignlim_xtoinftyfracx+2x-1&=lim_xtoinftyfrac(x-1)+3x-1\&=lim_xtoinfty1+frac3x-1\&=1.endalign
And the limit $lim_xtoinftyfracsin xx$ is $0$.
Applying the same method to $fracx+2x-1$, what we get isbeginalignlim_xtoinftyfracx+2x-1&=lim_xtoinftyfrac(x-1)+3x-1\&=lim_xtoinfty1+frac3x-1\&=1.endalign
And the limit $lim_xtoinftyfracsin xx$ is $0$.
answered Aug 28 at 13:01
José Carlos Santos
120k16101182
120k16101182
Ah, I see. But why can't you do (x+2) * 1/(x-1)? Why would that yield a wrong answer?
â user577730
Aug 28 at 13:02
2
@Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
â Vasya
Aug 28 at 13:10
@Vasya Okay. But why is infinity multiplied by zero an uncertainty? And will infinity multiplied by any other value besides zero be acceptable? So will say infinity times 3 be acceptable?
â user577730
Aug 28 at 13:34
1
@Mathguy: I follow these rules when dealing with limits: 1) infinity multiplied by a non-zero value is infinity (could be negative); 2) a non-zero value divided by infinity is zero; 3) zero multiplied by non-zero value is zero; 4) a non-zero value divided by zero is infinity (could be negative). If you have any other combination of infinity and zero, you need to modify the expression to fit one of the 4 cases.
â Vasya
Aug 28 at 14:04
add a comment |Â
Ah, I see. But why can't you do (x+2) * 1/(x-1)? Why would that yield a wrong answer?
â user577730
Aug 28 at 13:02
2
@Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
â Vasya
Aug 28 at 13:10
@Vasya Okay. But why is infinity multiplied by zero an uncertainty? And will infinity multiplied by any other value besides zero be acceptable? So will say infinity times 3 be acceptable?
â user577730
Aug 28 at 13:34
1
@Mathguy: I follow these rules when dealing with limits: 1) infinity multiplied by a non-zero value is infinity (could be negative); 2) a non-zero value divided by infinity is zero; 3) zero multiplied by non-zero value is zero; 4) a non-zero value divided by zero is infinity (could be negative). If you have any other combination of infinity and zero, you need to modify the expression to fit one of the 4 cases.
â Vasya
Aug 28 at 14:04
Ah, I see. But why can't you do (x+2) * 1/(x-1)? Why would that yield a wrong answer?
â user577730
Aug 28 at 13:02
Ah, I see. But why can't you do (x+2) * 1/(x-1)? Why would that yield a wrong answer?
â user577730
Aug 28 at 13:02
2
2
@Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
â Vasya
Aug 28 at 13:10
@Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
â Vasya
Aug 28 at 13:10
@Vasya Okay. But why is infinity multiplied by zero an uncertainty? And will infinity multiplied by any other value besides zero be acceptable? So will say infinity times 3 be acceptable?
â user577730
Aug 28 at 13:34
@Vasya Okay. But why is infinity multiplied by zero an uncertainty? And will infinity multiplied by any other value besides zero be acceptable? So will say infinity times 3 be acceptable?
â user577730
Aug 28 at 13:34
1
1
@Mathguy: I follow these rules when dealing with limits: 1) infinity multiplied by a non-zero value is infinity (could be negative); 2) a non-zero value divided by infinity is zero; 3) zero multiplied by non-zero value is zero; 4) a non-zero value divided by zero is infinity (could be negative). If you have any other combination of infinity and zero, you need to modify the expression to fit one of the 4 cases.
â Vasya
Aug 28 at 14:04
@Mathguy: I follow these rules when dealing with limits: 1) infinity multiplied by a non-zero value is infinity (could be negative); 2) a non-zero value divided by infinity is zero; 3) zero multiplied by non-zero value is zero; 4) a non-zero value divided by zero is infinity (could be negative). If you have any other combination of infinity and zero, you need to modify the expression to fit one of the 4 cases.
â Vasya
Aug 28 at 14:04
add a comment |Â
up vote
2
down vote
[T]ake a function $fracx+2x-1$. On the logic of the above, when we break it up, it is $(x+2)frac1x-1$. At infinity, $frac1x-1$ [tends to] $0$. So therefore, the limit is 0. But the limit of $fracx+2x-1$ is actually $1$.
It's very good that you're noticing what seems to be a contradiction here. The flaw in your reasoning is that the limit laws can only be used when the constituent limits exist. For instance, the product law for limits is
If $lim_xtoinftyf(x) = L$ and $lim_xtoinfty g(x) = M$, then $lim_xtoinfty f(x)g(x) = LM$.
You're trying to apply this law to the situation $f(x) = x+2$ and $g(x) = frac1x+1$. Yes, $lim_xtoinfty g(x) = 0$ and âÂÂanything times zero is zero.â However, $lim_xtoinfty f(x)$ is not a âÂÂthingâÂÂâÂÂit doesn't exist. So we cannot apply the product law.
Students get confused because we say $lim_xtoinftyf(x) =infty$, and because there's an equals sign and a symbol on the other side, it sure looks like the limit âÂÂisâ something. But $lim_xtoinftyf(x) =infty$ is shorthand for an infinite limit. There is some arithmetic than can be done with infinity, but $inftycdot0$ is undefined.
We see this again we we take a function $sin(x)frac1x$. Again, at infinity, the limit should be 0, but it's not.
Again, the first factor has no limit as $xtoinfty$, so the product law for limits doesn't apply.
add a comment |Â
up vote
2
down vote
[T]ake a function $fracx+2x-1$. On the logic of the above, when we break it up, it is $(x+2)frac1x-1$. At infinity, $frac1x-1$ [tends to] $0$. So therefore, the limit is 0. But the limit of $fracx+2x-1$ is actually $1$.
It's very good that you're noticing what seems to be a contradiction here. The flaw in your reasoning is that the limit laws can only be used when the constituent limits exist. For instance, the product law for limits is
If $lim_xtoinftyf(x) = L$ and $lim_xtoinfty g(x) = M$, then $lim_xtoinfty f(x)g(x) = LM$.
You're trying to apply this law to the situation $f(x) = x+2$ and $g(x) = frac1x+1$. Yes, $lim_xtoinfty g(x) = 0$ and âÂÂanything times zero is zero.â However, $lim_xtoinfty f(x)$ is not a âÂÂthingâÂÂâÂÂit doesn't exist. So we cannot apply the product law.
Students get confused because we say $lim_xtoinftyf(x) =infty$, and because there's an equals sign and a symbol on the other side, it sure looks like the limit âÂÂisâ something. But $lim_xtoinftyf(x) =infty$ is shorthand for an infinite limit. There is some arithmetic than can be done with infinity, but $inftycdot0$ is undefined.
We see this again we we take a function $sin(x)frac1x$. Again, at infinity, the limit should be 0, but it's not.
Again, the first factor has no limit as $xtoinfty$, so the product law for limits doesn't apply.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
[T]ake a function $fracx+2x-1$. On the logic of the above, when we break it up, it is $(x+2)frac1x-1$. At infinity, $frac1x-1$ [tends to] $0$. So therefore, the limit is 0. But the limit of $fracx+2x-1$ is actually $1$.
It's very good that you're noticing what seems to be a contradiction here. The flaw in your reasoning is that the limit laws can only be used when the constituent limits exist. For instance, the product law for limits is
If $lim_xtoinftyf(x) = L$ and $lim_xtoinfty g(x) = M$, then $lim_xtoinfty f(x)g(x) = LM$.
You're trying to apply this law to the situation $f(x) = x+2$ and $g(x) = frac1x+1$. Yes, $lim_xtoinfty g(x) = 0$ and âÂÂanything times zero is zero.â However, $lim_xtoinfty f(x)$ is not a âÂÂthingâÂÂâÂÂit doesn't exist. So we cannot apply the product law.
Students get confused because we say $lim_xtoinftyf(x) =infty$, and because there's an equals sign and a symbol on the other side, it sure looks like the limit âÂÂisâ something. But $lim_xtoinftyf(x) =infty$ is shorthand for an infinite limit. There is some arithmetic than can be done with infinity, but $inftycdot0$ is undefined.
We see this again we we take a function $sin(x)frac1x$. Again, at infinity, the limit should be 0, but it's not.
Again, the first factor has no limit as $xtoinfty$, so the product law for limits doesn't apply.
[T]ake a function $fracx+2x-1$. On the logic of the above, when we break it up, it is $(x+2)frac1x-1$. At infinity, $frac1x-1$ [tends to] $0$. So therefore, the limit is 0. But the limit of $fracx+2x-1$ is actually $1$.
It's very good that you're noticing what seems to be a contradiction here. The flaw in your reasoning is that the limit laws can only be used when the constituent limits exist. For instance, the product law for limits is
If $lim_xtoinftyf(x) = L$ and $lim_xtoinfty g(x) = M$, then $lim_xtoinfty f(x)g(x) = LM$.
You're trying to apply this law to the situation $f(x) = x+2$ and $g(x) = frac1x+1$. Yes, $lim_xtoinfty g(x) = 0$ and âÂÂanything times zero is zero.â However, $lim_xtoinfty f(x)$ is not a âÂÂthingâÂÂâÂÂit doesn't exist. So we cannot apply the product law.
Students get confused because we say $lim_xtoinftyf(x) =infty$, and because there's an equals sign and a symbol on the other side, it sure looks like the limit âÂÂisâ something. But $lim_xtoinftyf(x) =infty$ is shorthand for an infinite limit. There is some arithmetic than can be done with infinity, but $inftycdot0$ is undefined.
We see this again we we take a function $sin(x)frac1x$. Again, at infinity, the limit should be 0, but it's not.
Again, the first factor has no limit as $xtoinfty$, so the product law for limits doesn't apply.
answered Aug 28 at 13:10
Matthew Leingang
15.6k12144
15.6k12144
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0
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There is no contradiction at all.
The rule says that in a sum of terms you can neglect the terms that tend to zero (unless all the terms tend to zero).
The rule never said that this holds for a product.
For instance, consider $x+dfrac1x$ vs. $xcdotdfrac1x$.
For increasing powers of $10$, we have
$$x+frac1x=2,10.1,100.01,1000.001,10000.0001cdots$$
which clearly tends to $x$, vs.
$$xcdotfrac 1x=1,1,1,1,1,cdots$$
which is intermediate between $x$ and $dfrac1x$ and doesn't vanish.
For these reasons, you can write statements such as
$$lim_xtoinftyfrac2x+33x-5=lim_xtoinftyfrac2x3x=frac25.$$
add a comment |Â
up vote
0
down vote
There is no contradiction at all.
The rule says that in a sum of terms you can neglect the terms that tend to zero (unless all the terms tend to zero).
The rule never said that this holds for a product.
For instance, consider $x+dfrac1x$ vs. $xcdotdfrac1x$.
For increasing powers of $10$, we have
$$x+frac1x=2,10.1,100.01,1000.001,10000.0001cdots$$
which clearly tends to $x$, vs.
$$xcdotfrac 1x=1,1,1,1,1,cdots$$
which is intermediate between $x$ and $dfrac1x$ and doesn't vanish.
For these reasons, you can write statements such as
$$lim_xtoinftyfrac2x+33x-5=lim_xtoinftyfrac2x3x=frac25.$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There is no contradiction at all.
The rule says that in a sum of terms you can neglect the terms that tend to zero (unless all the terms tend to zero).
The rule never said that this holds for a product.
For instance, consider $x+dfrac1x$ vs. $xcdotdfrac1x$.
For increasing powers of $10$, we have
$$x+frac1x=2,10.1,100.01,1000.001,10000.0001cdots$$
which clearly tends to $x$, vs.
$$xcdotfrac 1x=1,1,1,1,1,cdots$$
which is intermediate between $x$ and $dfrac1x$ and doesn't vanish.
For these reasons, you can write statements such as
$$lim_xtoinftyfrac2x+33x-5=lim_xtoinftyfrac2x3x=frac25.$$
There is no contradiction at all.
The rule says that in a sum of terms you can neglect the terms that tend to zero (unless all the terms tend to zero).
The rule never said that this holds for a product.
For instance, consider $x+dfrac1x$ vs. $xcdotdfrac1x$.
For increasing powers of $10$, we have
$$x+frac1x=2,10.1,100.01,1000.001,10000.0001cdots$$
which clearly tends to $x$, vs.
$$xcdotfrac 1x=1,1,1,1,1,cdots$$
which is intermediate between $x$ and $dfrac1x$ and doesn't vanish.
For these reasons, you can write statements such as
$$lim_xtoinftyfrac2x+33x-5=lim_xtoinftyfrac2x3x=frac25.$$
answered Aug 28 at 13:18
Yves Daoust
113k665208
113k665208
add a comment |Â
add a comment |Â
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1
The examples are different: in $dfrac (x+2)(x-1)=(x+2) dfrac 1(x-1)$ you have $x$ also in the "part" $(x+2)$ and thus you cannot consider the limit for $x$ only of the other part.
â Mauro ALLEGRANZA
Aug 28 at 13:00
1
While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
â Mauro ALLEGRANZA
Aug 28 at 13:01
@MauroALLEGRANZA Can you elaborate what you mean "you have x also in the "part" (x+2) and thus you cannot consider the limit for x only of the other part."? Why can't you consider the limit for x only of the other part?
â user577730
Aug 28 at 13:04
Consider $x/x$. We know that $lim_x to infty 1/x = 0$ and that $lim_x to infty x = infty$, but $lim_x to infty x/x = 1$. You have to look at the term as a whole.
â Bill Wallis
Aug 28 at 13:11
1
Because you cannot "compute" the limit for $dfrac1(x-1)$ for $x$ to $0$ and then multiply the result for $(x+2)$, because also $(x+2)$ is "affected" by the operation of "computing the limit".
â Mauro ALLEGRANZA
Aug 28 at 13:11