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:



enter image description here



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?







share|cite|improve this question
















  • 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.
    – Mauro ALLEGRANZA
    Aug 28 at 13:00






  • 1




    While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
    – 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?
    – 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.
    – 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".
    – Mauro ALLEGRANZA
    Aug 28 at 13:11














up vote
1
down vote

favorite
1












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:



enter image description here



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?







share|cite|improve this question
















  • 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.
    – Mauro ALLEGRANZA
    Aug 28 at 13:00






  • 1




    While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
    – 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?
    – 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.
    – 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".
    – Mauro ALLEGRANZA
    Aug 28 at 13:11












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





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:



enter image description here



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?







share|cite|improve this question












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:



enter image description here



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?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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.
    – Mauro ALLEGRANZA
    Aug 28 at 13:00






  • 1




    While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
    – 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?
    – 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.
    – 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".
    – Mauro ALLEGRANZA
    Aug 28 at 13:11












  • 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.
    – Mauro ALLEGRANZA
    Aug 28 at 13:00






  • 1




    While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
    – 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?
    – 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.
    – 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".
    – 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.
– 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.
– Mauro ALLEGRANZA
Aug 28 at 13:00




1




1




While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
– Mauro ALLEGRANZA
Aug 28 at 13:01





While in the example $(1+ldots)$ the first term : $1$, does not depends on $x$.
– 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?
– 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?
– 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.
– 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.
– 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".
– 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".
– Mauro ALLEGRANZA
Aug 28 at 13:11










3 Answers
3






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 $0$.






share|cite|improve this answer




















  • Ah, I see. But why can't you do (x+2) * 1/(x-1)? Why would that yield a wrong answer?
    – user577730
    Aug 28 at 13:02






  • 2




    @Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
    – 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?
    – 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.
    – Vasya
    Aug 28 at 14:04

















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 “anything times zero is zero.” However, $lim_xtoinfty f(x)$ is not a “thing”—it 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 “is” 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.






share|cite|improve this answer



























    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



    $$lim_xtoinftyfrac2x+33x-5=lim_xtoinftyfrac2x3x=frac25.$$






    share|cite|improve this answer




















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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 $0$.






      share|cite|improve this answer




















      • Ah, I see. But why can't you do (x+2) * 1/(x-1)? Why would that yield a wrong answer?
        – user577730
        Aug 28 at 13:02






      • 2




        @Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
        – 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?
        – 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.
        – Vasya
        Aug 28 at 14:04














      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 $0$.






      share|cite|improve this answer




















      • Ah, I see. But why can't you do (x+2) * 1/(x-1)? Why would that yield a wrong answer?
        – user577730
        Aug 28 at 13:02






      • 2




        @Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
        – 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?
        – 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.
        – Vasya
        Aug 28 at 14:04












      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 $0$.






      share|cite|improve this answer












      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 $0$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      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?
        – user577730
        Aug 28 at 13:02






      • 2




        @Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
        – 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?
        – 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.
        – 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?
        – user577730
        Aug 28 at 13:02






      • 2




        @Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
        – 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?
        – 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.
        – 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?
      – 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?
      – 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.
      – Vasya
      Aug 28 at 13:10





      @Mathguy: because $x+2$ goes to infinity and you get infinity multiplied by zero, which is uncertainty.
      – 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?
      – 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?
      – 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.
      – 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.
      – Vasya
      Aug 28 at 14:04










      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 “anything times zero is zero.” However, $lim_xtoinfty f(x)$ is not a “thing”—it 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 “is” 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.






      share|cite|improve this answer
























        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 “anything times zero is zero.” However, $lim_xtoinfty f(x)$ is not a “thing”—it 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 “is” 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.






        share|cite|improve this answer






















          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 “anything times zero is zero.” However, $lim_xtoinfty f(x)$ is not a “thing”—it 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 “is” 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.






          share|cite|improve this answer













          [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 “anything times zero is zero.” However, $lim_xtoinfty f(x)$ is not a “thing”—it 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 “is” 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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 28 at 13:10









          Matthew Leingang

          15.6k12144




          15.6k12144




















              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



              $$lim_xtoinftyfrac2x+33x-5=lim_xtoinftyfrac2x3x=frac25.$$






              share|cite|improve this answer
























                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



                $$lim_xtoinftyfrac2x+33x-5=lim_xtoinftyfrac2x3x=frac25.$$






                share|cite|improve this answer






















                  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



                  $$lim_xtoinftyfrac2x+33x-5=lim_xtoinftyfrac2x3x=frac25.$$






                  share|cite|improve this answer












                  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



                  $$lim_xtoinftyfrac2x+33x-5=lim_xtoinftyfrac2x3x=frac25.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 28 at 13:18









                  Yves Daoust

                  113k665208




                  113k665208



























                       

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