Can we multiply both sides of a limit equation?












3












$begingroup$


Compute $lim_{x to -1} f(x)$ for a function $f: mathbb R to mathbb R$ such that



$$4 = lim_{x to -1} frac{f(x)+2}{x+1} - frac{x}{x^2-1} tag{1}$$



$$ = lim_{x to -1} frac{f(x)+2}{x+1} - frac{frac{x}{x-1}}{x+1}$$



$$ = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1}$$




  • Solution 1: My approach is that the numerator $f(x)+2 - frac{x}{x-1}$ must approach zero as $x to -1$ because the denominator approaches zero as $x to -1$ and so $lim_{x to -1} f(x) = -frac{3}{2}$.


Are we allowed to do the following, which seems to be the required solution, instead? This does not seem very rigorous, and I have a feeling there are obvious counterexamples. Of course, we can use $varepsilon-delta$ to check our answer, but I would like to know if and how this can be generalised for any function $f: mathbb R to mathbb R$.




  • Solution 2: Observe


$$0 = lim_{x to -1} x+1$$



Then



$$4 = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1}$$



$$implies 0 = 0 cdot 4 = (lim_{x to -1} x+1) cdot 4 = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1} lim_{x to -1} x+1$$



$$ = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1} x+1 = lim_{x to -1} (f(x)+2 - frac{x}{x-1})$$



$$ = lim_{x to -1} f(x) + lim_{x to -1} 2 - lim_{x to -1} frac{x}{x-1} = lim_{x to -1} f(x)+2 - frac12 = lim_{x to -1} f(x)+ frac32$$



$$implies - frac32 = lim_{x to -1} f(x) tag{2}$$



I suspect the preceding solution is not rigorous, and this is a case where we have only a guess and must check our answer by proving $(1)$, by computation since $varepsilon-delta$ is actually not yet allowed, assuming $(2)$, so if one does then preceding solution, then one must follow up with a computation. I'm not quite sure what the problem is, but it might be that we don't know $lim_{x to -1} f(x)$ exists in the first place.



Is the fact that the domain and range are both $mathbb R$ relevant? I think of a counterexample like $f: {7,8,10} to {1,2}$ or $f: C to mathbb Q^c$ where $C$ is the Cantor set.






real-analysis calculus limits alternative-proof







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Compute $lim_{x to -1} f(x)$ for a function $f: mathbb R to mathbb R$ such that



    $$4 = lim_{x to -1} frac{f(x)+2}{x+1} - frac{x}{x^2-1} tag{1}$$



    $$ = lim_{x to -1} frac{f(x)+2}{x+1} - frac{frac{x}{x-1}}{x+1}$$



    $$ = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1}$$




    • Solution 1: My approach is that the numerator $f(x)+2 - frac{x}{x-1}$ must approach zero as $x to -1$ because the denominator approaches zero as $x to -1$ and so $lim_{x to -1} f(x) = -frac{3}{2}$.


    Are we allowed to do the following, which seems to be the required solution, instead? This does not seem very rigorous, and I have a feeling there are obvious counterexamples. Of course, we can use $varepsilon-delta$ to check our answer, but I would like to know if and how this can be generalised for any function $f: mathbb R to mathbb R$.




    • Solution 2: Observe


    $$0 = lim_{x to -1} x+1$$



    Then



    $$4 = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1}$$



    $$implies 0 = 0 cdot 4 = (lim_{x to -1} x+1) cdot 4 = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1} lim_{x to -1} x+1$$



    $$ = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1} x+1 = lim_{x to -1} (f(x)+2 - frac{x}{x-1})$$



    $$ = lim_{x to -1} f(x) + lim_{x to -1} 2 - lim_{x to -1} frac{x}{x-1} = lim_{x to -1} f(x)+2 - frac12 = lim_{x to -1} f(x)+ frac32$$



    $$implies - frac32 = lim_{x to -1} f(x) tag{2}$$



    I suspect the preceding solution is not rigorous, and this is a case where we have only a guess and must check our answer by proving $(1)$, by computation since $varepsilon-delta$ is actually not yet allowed, assuming $(2)$, so if one does then preceding solution, then one must follow up with a computation. I'm not quite sure what the problem is, but it might be that we don't know $lim_{x to -1} f(x)$ exists in the first place.



    Is the fact that the domain and range are both $mathbb R$ relevant? I think of a counterexample like $f: {7,8,10} to {1,2}$ or $f: C to mathbb Q^c$ where $C$ is the Cantor set.






    real-analysis calculus limits alternative-proof







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Compute $lim_{x to -1} f(x)$ for a function $f: mathbb R to mathbb R$ such that



      $$4 = lim_{x to -1} frac{f(x)+2}{x+1} - frac{x}{x^2-1} tag{1}$$



      $$ = lim_{x to -1} frac{f(x)+2}{x+1} - frac{frac{x}{x-1}}{x+1}$$



      $$ = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1}$$




      • Solution 1: My approach is that the numerator $f(x)+2 - frac{x}{x-1}$ must approach zero as $x to -1$ because the denominator approaches zero as $x to -1$ and so $lim_{x to -1} f(x) = -frac{3}{2}$.


      Are we allowed to do the following, which seems to be the required solution, instead? This does not seem very rigorous, and I have a feeling there are obvious counterexamples. Of course, we can use $varepsilon-delta$ to check our answer, but I would like to know if and how this can be generalised for any function $f: mathbb R to mathbb R$.




      • Solution 2: Observe


      $$0 = lim_{x to -1} x+1$$



      Then



      $$4 = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1}$$



      $$implies 0 = 0 cdot 4 = (lim_{x to -1} x+1) cdot 4 = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1} lim_{x to -1} x+1$$



      $$ = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1} x+1 = lim_{x to -1} (f(x)+2 - frac{x}{x-1})$$



      $$ = lim_{x to -1} f(x) + lim_{x to -1} 2 - lim_{x to -1} frac{x}{x-1} = lim_{x to -1} f(x)+2 - frac12 = lim_{x to -1} f(x)+ frac32$$



      $$implies - frac32 = lim_{x to -1} f(x) tag{2}$$



      I suspect the preceding solution is not rigorous, and this is a case where we have only a guess and must check our answer by proving $(1)$, by computation since $varepsilon-delta$ is actually not yet allowed, assuming $(2)$, so if one does then preceding solution, then one must follow up with a computation. I'm not quite sure what the problem is, but it might be that we don't know $lim_{x to -1} f(x)$ exists in the first place.



      Is the fact that the domain and range are both $mathbb R$ relevant? I think of a counterexample like $f: {7,8,10} to {1,2}$ or $f: C to mathbb Q^c$ where $C$ is the Cantor set.






      real-analysis calculus limits alternative-proof







      share|cite|improve this question











      $endgroup$




      Compute $lim_{x to -1} f(x)$ for a function $f: mathbb R to mathbb R$ such that



      $$4 = lim_{x to -1} frac{f(x)+2}{x+1} - frac{x}{x^2-1} tag{1}$$



      $$ = lim_{x to -1} frac{f(x)+2}{x+1} - frac{frac{x}{x-1}}{x+1}$$



      $$ = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1}$$




      • Solution 1: My approach is that the numerator $f(x)+2 - frac{x}{x-1}$ must approach zero as $x to -1$ because the denominator approaches zero as $x to -1$ and so $lim_{x to -1} f(x) = -frac{3}{2}$.


      Are we allowed to do the following, which seems to be the required solution, instead? This does not seem very rigorous, and I have a feeling there are obvious counterexamples. Of course, we can use $varepsilon-delta$ to check our answer, but I would like to know if and how this can be generalised for any function $f: mathbb R to mathbb R$.




      • Solution 2: Observe


      $$0 = lim_{x to -1} x+1$$



      Then



      $$4 = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1}$$



      $$implies 0 = 0 cdot 4 = (lim_{x to -1} x+1) cdot 4 = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1} lim_{x to -1} x+1$$



      $$ = lim_{x to -1} frac{f(x)+2 - frac{x}{x-1}}{x+1} x+1 = lim_{x to -1} (f(x)+2 - frac{x}{x-1})$$



      $$ = lim_{x to -1} f(x) + lim_{x to -1} 2 - lim_{x to -1} frac{x}{x-1} = lim_{x to -1} f(x)+2 - frac12 = lim_{x to -1} f(x)+ frac32$$



      $$implies - frac32 = lim_{x to -1} f(x) tag{2}$$



      I suspect the preceding solution is not rigorous, and this is a case where we have only a guess and must check our answer by proving $(1)$, by computation since $varepsilon-delta$ is actually not yet allowed, assuming $(2)$, so if one does then preceding solution, then one must follow up with a computation. I'm not quite sure what the problem is, but it might be that we don't know $lim_{x to -1} f(x)$ exists in the first place.



      Is the fact that the domain and range are both $mathbb R$ relevant? I think of a counterexample like $f: {7,8,10} to {1,2}$ or $f: C to mathbb Q^c$ where $C$ is the Cantor set.






      real-analysis calculus limits alternative-proof




      real-analysis calculus limits alternative-proof






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 19 at 15:43







      Mitjackson

















      asked Jan 19 at 15:15









      MitjacksonMitjackson

      365




      365






















          1 Answer
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          $begingroup$

          If $l:=lim_{xto a}g(x),,m:=lim_{xto a}h(x)$ exist and aren't $pminfty$, you can indeed write $lm:=lim_{xto a}g(x)h(x)$. In the case $l=0$ note that, when $x$ is close enough to $a$ to constrain $h$ to some interval around $m$, this upper-bounds $|h|$ (say with $|h|ge U,,U<infty$) so $|g|lefrac{epsilon}{U}to|g|leepsilon$. Note that as long as the domain and range are such that $|g|,,|h|$ can be bounded and one is arbitrarily small in the limit, we're fine.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is solution 2 considered complete? What's an example where $|g|$ or $|h|$ isn't bounded please? What about an example of an $f: A to B$ where the above doesn't work? Thanks!
            $endgroup$
            – Mitjackson
            Jan 19 at 15:57













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          $begingroup$

          If $l:=lim_{xto a}g(x),,m:=lim_{xto a}h(x)$ exist and aren't $pminfty$, you can indeed write $lm:=lim_{xto a}g(x)h(x)$. In the case $l=0$ note that, when $x$ is close enough to $a$ to constrain $h$ to some interval around $m$, this upper-bounds $|h|$ (say with $|h|ge U,,U<infty$) so $|g|lefrac{epsilon}{U}to|g|leepsilon$. Note that as long as the domain and range are such that $|g|,,|h|$ can be bounded and one is arbitrarily small in the limit, we're fine.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is solution 2 considered complete? What's an example where $|g|$ or $|h|$ isn't bounded please? What about an example of an $f: A to B$ where the above doesn't work? Thanks!
            $endgroup$
            – Mitjackson
            Jan 19 at 15:57


















          1












          $begingroup$

          If $l:=lim_{xto a}g(x),,m:=lim_{xto a}h(x)$ exist and aren't $pminfty$, you can indeed write $lm:=lim_{xto a}g(x)h(x)$. In the case $l=0$ note that, when $x$ is close enough to $a$ to constrain $h$ to some interval around $m$, this upper-bounds $|h|$ (say with $|h|ge U,,U<infty$) so $|g|lefrac{epsilon}{U}to|g|leepsilon$. Note that as long as the domain and range are such that $|g|,,|h|$ can be bounded and one is arbitrarily small in the limit, we're fine.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Is solution 2 considered complete? What's an example where $|g|$ or $|h|$ isn't bounded please? What about an example of an $f: A to B$ where the above doesn't work? Thanks!
            $endgroup$
            – Mitjackson
            Jan 19 at 15:57
















          1












          1








          1





          $begingroup$

          If $l:=lim_{xto a}g(x),,m:=lim_{xto a}h(x)$ exist and aren't $pminfty$, you can indeed write $lm:=lim_{xto a}g(x)h(x)$. In the case $l=0$ note that, when $x$ is close enough to $a$ to constrain $h$ to some interval around $m$, this upper-bounds $|h|$ (say with $|h|ge U,,U<infty$) so $|g|lefrac{epsilon}{U}to|g|leepsilon$. Note that as long as the domain and range are such that $|g|,,|h|$ can be bounded and one is arbitrarily small in the limit, we're fine.






          share|cite|improve this answer









          $endgroup$



          If $l:=lim_{xto a}g(x),,m:=lim_{xto a}h(x)$ exist and aren't $pminfty$, you can indeed write $lm:=lim_{xto a}g(x)h(x)$. In the case $l=0$ note that, when $x$ is close enough to $a$ to constrain $h$ to some interval around $m$, this upper-bounds $|h|$ (say with $|h|ge U,,U<infty$) so $|g|lefrac{epsilon}{U}to|g|leepsilon$. Note that as long as the domain and range are such that $|g|,,|h|$ can be bounded and one is arbitrarily small in the limit, we're fine.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 19 at 15:46









          J.G.J.G.

          27k22843




          27k22843












          • $begingroup$
            Is solution 2 considered complete? What's an example where $|g|$ or $|h|$ isn't bounded please? What about an example of an $f: A to B$ where the above doesn't work? Thanks!
            $endgroup$
            – Mitjackson
            Jan 19 at 15:57




















          • $begingroup$
            Is solution 2 considered complete? What's an example where $|g|$ or $|h|$ isn't bounded please? What about an example of an $f: A to B$ where the above doesn't work? Thanks!
            $endgroup$
            – Mitjackson
            Jan 19 at 15:57


















          $begingroup$
          Is solution 2 considered complete? What's an example where $|g|$ or $|h|$ isn't bounded please? What about an example of an $f: A to B$ where the above doesn't work? Thanks!
          $endgroup$
          – Mitjackson
          Jan 19 at 15:57






          $begingroup$
          Is solution 2 considered complete? What's an example where $|g|$ or $|h|$ isn't bounded please? What about an example of an $f: A to B$ where the above doesn't work? Thanks!
          $endgroup$
          – Mitjackson
          Jan 19 at 15:57




















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