explain why ${left(frac{{1}}{{2}}right)}^{infty}=0$












1












$begingroup$


Mathematica shows
${left(frac{{1}}{{2}}right)}^{infty}=0$, anyone can explain why ?



I know we can get $limlimits_{{{x}toinfty}}{left(frac{{1}}{{2}}right)}^{{x}}={0}$ by taking limit , does ${left(frac{{1}}{{2}}right)}^{infty}$ is just an abbreviated expression of $limlimits_{{{x}toinfty}}{left(frac{{1}}{{2}}right)}^{{x}}$



P.S. $infty = +infty$ here.






calculus limits nonstandard-analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Yes. (words to fill the comment)
    $endgroup$
    – Vinícius Novelli
    May 3 '15 at 7:48










  • $begingroup$
    As long as you agree that $infty = +infty$. I personally don't, and in calculus I usually teach that $infty$ means nothing. But Mathematica agrees with this shorthand.
    $endgroup$
    – Siminore
    May 3 '15 at 7:50










  • $begingroup$
    Well exactly infinity would not be defined, because it's a concept not a number. However, the notation that $lim_{xto infty}$ implies that $x$ is approaching infinity. If you directly say that it's an abbreviation, it won't be a good idea. It's the limit as $x$ approaches it.
    $endgroup$
    – Mann
    May 3 '15 at 7:51










  • $begingroup$
    @Siminore yes, $infty = +infty$ here. I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
    $endgroup$
    – iMath
    May 3 '15 at 7:55












  • $begingroup$
    @Mann I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
    $endgroup$
    – iMath
    May 3 '15 at 7:58
















1












$begingroup$


Mathematica shows
${left(frac{{1}}{{2}}right)}^{infty}=0$, anyone can explain why ?



I know we can get $limlimits_{{{x}toinfty}}{left(frac{{1}}{{2}}right)}^{{x}}={0}$ by taking limit , does ${left(frac{{1}}{{2}}right)}^{infty}$ is just an abbreviated expression of $limlimits_{{{x}toinfty}}{left(frac{{1}}{{2}}right)}^{{x}}$



P.S. $infty = +infty$ here.






calculus limits nonstandard-analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Yes. (words to fill the comment)
    $endgroup$
    – Vinícius Novelli
    May 3 '15 at 7:48










  • $begingroup$
    As long as you agree that $infty = +infty$. I personally don't, and in calculus I usually teach that $infty$ means nothing. But Mathematica agrees with this shorthand.
    $endgroup$
    – Siminore
    May 3 '15 at 7:50










  • $begingroup$
    Well exactly infinity would not be defined, because it's a concept not a number. However, the notation that $lim_{xto infty}$ implies that $x$ is approaching infinity. If you directly say that it's an abbreviation, it won't be a good idea. It's the limit as $x$ approaches it.
    $endgroup$
    – Mann
    May 3 '15 at 7:51










  • $begingroup$
    @Siminore yes, $infty = +infty$ here. I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
    $endgroup$
    – iMath
    May 3 '15 at 7:55












  • $begingroup$
    @Mann I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
    $endgroup$
    – iMath
    May 3 '15 at 7:58














1












1








1





$begingroup$


Mathematica shows
${left(frac{{1}}{{2}}right)}^{infty}=0$, anyone can explain why ?



I know we can get $limlimits_{{{x}toinfty}}{left(frac{{1}}{{2}}right)}^{{x}}={0}$ by taking limit , does ${left(frac{{1}}{{2}}right)}^{infty}$ is just an abbreviated expression of $limlimits_{{{x}toinfty}}{left(frac{{1}}{{2}}right)}^{{x}}$



P.S. $infty = +infty$ here.






calculus limits nonstandard-analysis







share|cite|improve this question











$endgroup$




Mathematica shows
${left(frac{{1}}{{2}}right)}^{infty}=0$, anyone can explain why ?



I know we can get $limlimits_{{{x}toinfty}}{left(frac{{1}}{{2}}right)}^{{x}}={0}$ by taking limit , does ${left(frac{{1}}{{2}}right)}^{infty}$ is just an abbreviated expression of $limlimits_{{{x}toinfty}}{left(frac{{1}}{{2}}right)}^{{x}}$



P.S. $infty = +infty$ here.






calculus limits nonstandard-analysis




calculus limits nonstandard-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 3 '15 at 8:40









Francesco Gramano

1476




1476










asked May 3 '15 at 7:46









iMathiMath

1,006821




1,006821








  • 2




    $begingroup$
    Yes. (words to fill the comment)
    $endgroup$
    – Vinícius Novelli
    May 3 '15 at 7:48










  • $begingroup$
    As long as you agree that $infty = +infty$. I personally don't, and in calculus I usually teach that $infty$ means nothing. But Mathematica agrees with this shorthand.
    $endgroup$
    – Siminore
    May 3 '15 at 7:50










  • $begingroup$
    Well exactly infinity would not be defined, because it's a concept not a number. However, the notation that $lim_{xto infty}$ implies that $x$ is approaching infinity. If you directly say that it's an abbreviation, it won't be a good idea. It's the limit as $x$ approaches it.
    $endgroup$
    – Mann
    May 3 '15 at 7:51










  • $begingroup$
    @Siminore yes, $infty = +infty$ here. I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
    $endgroup$
    – iMath
    May 3 '15 at 7:55












  • $begingroup$
    @Mann I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
    $endgroup$
    – iMath
    May 3 '15 at 7:58














  • 2




    $begingroup$
    Yes. (words to fill the comment)
    $endgroup$
    – Vinícius Novelli
    May 3 '15 at 7:48










  • $begingroup$
    As long as you agree that $infty = +infty$. I personally don't, and in calculus I usually teach that $infty$ means nothing. But Mathematica agrees with this shorthand.
    $endgroup$
    – Siminore
    May 3 '15 at 7:50










  • $begingroup$
    Well exactly infinity would not be defined, because it's a concept not a number. However, the notation that $lim_{xto infty}$ implies that $x$ is approaching infinity. If you directly say that it's an abbreviation, it won't be a good idea. It's the limit as $x$ approaches it.
    $endgroup$
    – Mann
    May 3 '15 at 7:51










  • $begingroup$
    @Siminore yes, $infty = +infty$ here. I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
    $endgroup$
    – iMath
    May 3 '15 at 7:55












  • $begingroup$
    @Mann I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
    $endgroup$
    – iMath
    May 3 '15 at 7:58








2




2




$begingroup$
Yes. (words to fill the comment)
$endgroup$
– Vinícius Novelli
May 3 '15 at 7:48




$begingroup$
Yes. (words to fill the comment)
$endgroup$
– Vinícius Novelli
May 3 '15 at 7:48












$begingroup$
As long as you agree that $infty = +infty$. I personally don't, and in calculus I usually teach that $infty$ means nothing. But Mathematica agrees with this shorthand.
$endgroup$
– Siminore
May 3 '15 at 7:50




$begingroup$
As long as you agree that $infty = +infty$. I personally don't, and in calculus I usually teach that $infty$ means nothing. But Mathematica agrees with this shorthand.
$endgroup$
– Siminore
May 3 '15 at 7:50












$begingroup$
Well exactly infinity would not be defined, because it's a concept not a number. However, the notation that $lim_{xto infty}$ implies that $x$ is approaching infinity. If you directly say that it's an abbreviation, it won't be a good idea. It's the limit as $x$ approaches it.
$endgroup$
– Mann
May 3 '15 at 7:51




$begingroup$
Well exactly infinity would not be defined, because it's a concept not a number. However, the notation that $lim_{xto infty}$ implies that $x$ is approaching infinity. If you directly say that it's an abbreviation, it won't be a good idea. It's the limit as $x$ approaches it.
$endgroup$
– Mann
May 3 '15 at 7:51












$begingroup$
@Siminore yes, $infty = +infty$ here. I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
$endgroup$
– iMath
May 3 '15 at 7:55






$begingroup$
@Siminore yes, $infty = +infty$ here. I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
$endgroup$
– iMath
May 3 '15 at 7:55














$begingroup$
@Mann I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
$endgroup$
– iMath
May 3 '15 at 7:58




$begingroup$
@Mann I think infinity could be used as a number to measure quantities that finite number , like real number, could not measure, infinity is greater than any assignable quantity or countable number, while “x tends to infinity”(x→∞) is actually an unending procedure.
$endgroup$
– iMath
May 3 '15 at 7:58










3 Answers
3






active

oldest

votes


















2












$begingroup$

$(1/2)^2=1/4$ and $(1/2)^{10}=1/1024$ and then $(1/2)^{20}=1/1048576$. As the power gets larger, the denominator approaches infinity. This makes the value of the fraction tend to zero.



This might help. enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have already known what you said here, but you seems doesn't answered my question
    $endgroup$
    – iMath
    May 3 '15 at 8:06



















2












$begingroup$

I think this is just a definition from Wolfram Alpha. As you see here!
If you type in $left(frac{1}{2}right)^infty == limlimits_{x to infty} left(frac{1}{2}right)^x $, then Wolfram Alpha give you the result: True.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    What you mean is probably $$lim_{xrightarrow infty} {left(frac{1}{2}right)}^{x}$$



    A limit can be defined with epsilon and delta. I.e. can we find any number $delta$ for every $epsilon > 0$ which the expression is close enough to.



    A variant of this: We can take the binary number $2^{-i}$ for an integer $i$ which is bounded below by 0, because positive numbers are closed under multiplication. We see that we can make the expression smaller than any such binary number, by just making $x$ large enough, say $x = i + 1$. But the binary number system is good enough to express any positive number if enough digits are allowed. Therefore the limit must be 0.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      $(1/2)^2=1/4$ and $(1/2)^{10}=1/1024$ and then $(1/2)^{20}=1/1048576$. As the power gets larger, the denominator approaches infinity. This makes the value of the fraction tend to zero.



      This might help. enter image description here






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I have already known what you said here, but you seems doesn't answered my question
        $endgroup$
        – iMath
        May 3 '15 at 8:06
















      2












      $begingroup$

      $(1/2)^2=1/4$ and $(1/2)^{10}=1/1024$ and then $(1/2)^{20}=1/1048576$. As the power gets larger, the denominator approaches infinity. This makes the value of the fraction tend to zero.



      This might help. enter image description here






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I have already known what you said here, but you seems doesn't answered my question
        $endgroup$
        – iMath
        May 3 '15 at 8:06














      2












      2








      2





      $begingroup$

      $(1/2)^2=1/4$ and $(1/2)^{10}=1/1024$ and then $(1/2)^{20}=1/1048576$. As the power gets larger, the denominator approaches infinity. This makes the value of the fraction tend to zero.



      This might help. enter image description here






      share|cite|improve this answer









      $endgroup$



      $(1/2)^2=1/4$ and $(1/2)^{10}=1/1024$ and then $(1/2)^{20}=1/1048576$. As the power gets larger, the denominator approaches infinity. This makes the value of the fraction tend to zero.



      This might help. enter image description here







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered May 3 '15 at 7:54









      Hritik NarayanHritik Narayan

      372312




      372312












      • $begingroup$
        I have already known what you said here, but you seems doesn't answered my question
        $endgroup$
        – iMath
        May 3 '15 at 8:06


















      • $begingroup$
        I have already known what you said here, but you seems doesn't answered my question
        $endgroup$
        – iMath
        May 3 '15 at 8:06
















      $begingroup$
      I have already known what you said here, but you seems doesn't answered my question
      $endgroup$
      – iMath
      May 3 '15 at 8:06




      $begingroup$
      I have already known what you said here, but you seems doesn't answered my question
      $endgroup$
      – iMath
      May 3 '15 at 8:06











      2












      $begingroup$

      I think this is just a definition from Wolfram Alpha. As you see here!
      If you type in $left(frac{1}{2}right)^infty == limlimits_{x to infty} left(frac{1}{2}right)^x $, then Wolfram Alpha give you the result: True.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        I think this is just a definition from Wolfram Alpha. As you see here!
        If you type in $left(frac{1}{2}right)^infty == limlimits_{x to infty} left(frac{1}{2}right)^x $, then Wolfram Alpha give you the result: True.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          I think this is just a definition from Wolfram Alpha. As you see here!
          If you type in $left(frac{1}{2}right)^infty == limlimits_{x to infty} left(frac{1}{2}right)^x $, then Wolfram Alpha give you the result: True.






          share|cite|improve this answer









          $endgroup$



          I think this is just a definition from Wolfram Alpha. As you see here!
          If you type in $left(frac{1}{2}right)^infty == limlimits_{x to infty} left(frac{1}{2}right)^x $, then Wolfram Alpha give you the result: True.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 3 '15 at 8:38









          aGeraGer

          811823




          811823























              0












              $begingroup$

              What you mean is probably $$lim_{xrightarrow infty} {left(frac{1}{2}right)}^{x}$$



              A limit can be defined with epsilon and delta. I.e. can we find any number $delta$ for every $epsilon > 0$ which the expression is close enough to.



              A variant of this: We can take the binary number $2^{-i}$ for an integer $i$ which is bounded below by 0, because positive numbers are closed under multiplication. We see that we can make the expression smaller than any such binary number, by just making $x$ large enough, say $x = i + 1$. But the binary number system is good enough to express any positive number if enough digits are allowed. Therefore the limit must be 0.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                What you mean is probably $$lim_{xrightarrow infty} {left(frac{1}{2}right)}^{x}$$



                A limit can be defined with epsilon and delta. I.e. can we find any number $delta$ for every $epsilon > 0$ which the expression is close enough to.



                A variant of this: We can take the binary number $2^{-i}$ for an integer $i$ which is bounded below by 0, because positive numbers are closed under multiplication. We see that we can make the expression smaller than any such binary number, by just making $x$ large enough, say $x = i + 1$. But the binary number system is good enough to express any positive number if enough digits are allowed. Therefore the limit must be 0.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  What you mean is probably $$lim_{xrightarrow infty} {left(frac{1}{2}right)}^{x}$$



                  A limit can be defined with epsilon and delta. I.e. can we find any number $delta$ for every $epsilon > 0$ which the expression is close enough to.



                  A variant of this: We can take the binary number $2^{-i}$ for an integer $i$ which is bounded below by 0, because positive numbers are closed under multiplication. We see that we can make the expression smaller than any such binary number, by just making $x$ large enough, say $x = i + 1$. But the binary number system is good enough to express any positive number if enough digits are allowed. Therefore the limit must be 0.






                  share|cite|improve this answer









                  $endgroup$



                  What you mean is probably $$lim_{xrightarrow infty} {left(frac{1}{2}right)}^{x}$$



                  A limit can be defined with epsilon and delta. I.e. can we find any number $delta$ for every $epsilon > 0$ which the expression is close enough to.



                  A variant of this: We can take the binary number $2^{-i}$ for an integer $i$ which is bounded below by 0, because positive numbers are closed under multiplication. We see that we can make the expression smaller than any such binary number, by just making $x$ large enough, say $x = i + 1$. But the binary number system is good enough to express any positive number if enough digits are allowed. Therefore the limit must be 0.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered May 3 '15 at 9:05









                  mathreadlermathreadler

                  14.8k72160




                  14.8k72160






























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