Limit $limlimits_{x to 0^{+}}frac{x^{cos x}}{x}$












1












$begingroup$


I need to calculate the limit $limlimits_{x to 0^{+}}frac{x^{cos x}}{x}$.



I tried to form it as $limlimits_{x to 0^{+}}frac{e^{ln (x)cdot cos x}}{x} $ and do L'Hôpital's rule but it doesn't solve it.






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you try with $x^{cos x -1}$?
    $endgroup$
    – ecrin
    Jan 8 at 16:19






  • 2




    $begingroup$
    Hint: $(ln{x})(cos{x}-1) rightarrow 0$.
    $endgroup$
    – Mindlack
    Jan 8 at 16:20










  • $begingroup$
    But the limit is still $0^0$
    $endgroup$
    – violettagold
    Jan 8 at 16:23
















1












$begingroup$


I need to calculate the limit $limlimits_{x to 0^{+}}frac{x^{cos x}}{x}$.



I tried to form it as $limlimits_{x to 0^{+}}frac{e^{ln (x)cdot cos x}}{x} $ and do L'Hôpital's rule but it doesn't solve it.






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you try with $x^{cos x -1}$?
    $endgroup$
    – ecrin
    Jan 8 at 16:19






  • 2




    $begingroup$
    Hint: $(ln{x})(cos{x}-1) rightarrow 0$.
    $endgroup$
    – Mindlack
    Jan 8 at 16:20










  • $begingroup$
    But the limit is still $0^0$
    $endgroup$
    – violettagold
    Jan 8 at 16:23














1












1








1





$begingroup$


I need to calculate the limit $limlimits_{x to 0^{+}}frac{x^{cos x}}{x}$.



I tried to form it as $limlimits_{x to 0^{+}}frac{e^{ln (x)cdot cos x}}{x} $ and do L'Hôpital's rule but it doesn't solve it.






limits







share|cite|improve this question











$endgroup$




I need to calculate the limit $limlimits_{x to 0^{+}}frac{x^{cos x}}{x}$.



I tried to form it as $limlimits_{x to 0^{+}}frac{e^{ln (x)cdot cos x}}{x} $ and do L'Hôpital's rule but it doesn't solve it.






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 16:50









Did

247k23223459




247k23223459










asked Jan 8 at 16:15









violettagoldviolettagold

266




266












  • $begingroup$
    Did you try with $x^{cos x -1}$?
    $endgroup$
    – ecrin
    Jan 8 at 16:19






  • 2




    $begingroup$
    Hint: $(ln{x})(cos{x}-1) rightarrow 0$.
    $endgroup$
    – Mindlack
    Jan 8 at 16:20










  • $begingroup$
    But the limit is still $0^0$
    $endgroup$
    – violettagold
    Jan 8 at 16:23


















  • $begingroup$
    Did you try with $x^{cos x -1}$?
    $endgroup$
    – ecrin
    Jan 8 at 16:19






  • 2




    $begingroup$
    Hint: $(ln{x})(cos{x}-1) rightarrow 0$.
    $endgroup$
    – Mindlack
    Jan 8 at 16:20










  • $begingroup$
    But the limit is still $0^0$
    $endgroup$
    – violettagold
    Jan 8 at 16:23
















$begingroup$
Did you try with $x^{cos x -1}$?
$endgroup$
– ecrin
Jan 8 at 16:19




$begingroup$
Did you try with $x^{cos x -1}$?
$endgroup$
– ecrin
Jan 8 at 16:19




2




2




$begingroup$
Hint: $(ln{x})(cos{x}-1) rightarrow 0$.
$endgroup$
– Mindlack
Jan 8 at 16:20




$begingroup$
Hint: $(ln{x})(cos{x}-1) rightarrow 0$.
$endgroup$
– Mindlack
Jan 8 at 16:20












$begingroup$
But the limit is still $0^0$
$endgroup$
– violettagold
Jan 8 at 16:23




$begingroup$
But the limit is still $0^0$
$endgroup$
– violettagold
Jan 8 at 16:23










4 Answers
4






active

oldest

votes


















2












$begingroup$

We have :



$$frac{x^{cos x}}{x} = x^{cos x -1} = e^{ln x (cos x -1)}$$



Now using the fact that in a neighborhood of $0$ we have :



$$cos x - 1 = -frac{x^2}{2} + o(x^2)$$



Then we can easily deduce that :



$$ln x cdot (cos x -1) to 0$$



Hence the desired limit is $1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much.
    $endgroup$
    – violettagold
    Jan 8 at 16:48



















1












$begingroup$

$$begin{align}
lim_{xto0^+}x^{cos x-1}&=lim_{xto0^+}e^{ln x(cos x-1)}\
&=expleft(lim_{xto0^+}frac{cos x-1}{dfrac1{ln x}}right)\
&=expleft(lim_{xto0^+}frac{-sin x}{-dfrac1{x(ln x)^2}}right)\
&=expleft(lim_{xto0^+}(xln x)(sin xln x)right)=e^0
end{align}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $cos x= 1+o(1)$ when $xto 0$ (first order of Taylor expansion). Then
    $$frac{x^{cos x}}x =frac{x^{1+o(1)}}x=frac{xcdot x^{o(1)}}x=x^{o(1)}longrightarrow_{xto0}1$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      This is quite wrong: $y=x+o(1)$ does not imply $x^y=x+o(1)$, and $frac{o(1)}x$ does not necessarily converge to $0$.
      $endgroup$
      – Did
      Jan 8 at 16:48












    • $begingroup$
      I should have corrected it now
      $endgroup$
      – Lorenzo B.
      Jan 8 at 16:56










    • $begingroup$
      Yeah, except that now, you pushed everything in the last step $x^{o(1)}longrightarrow_{xto0}1$, which happens to be wrong in general. Sorry but this is not a guessing game...
      $endgroup$
      – Did
      Jan 8 at 16:59



















    0












    $begingroup$

    You may also use the following facts:




    • $lim_{xto 0}frac{cos x-1}{x} = cos'(0) = -sin(0) = 0$

    • $lim_{xto 0}xln x = 0$


    So, you get
    $$ln frac{x^{cos x}}{x} = (cos x - 1)cdot ln x = frac{(cos x - 1)}{x}cdot x ln x stackrel{xto 0^+}{longrightarrow}0cdot 0 = 0$$



    Hence, $limlimits_{x to 0^{+}}frac{x^{cos x}}{x} = e^0 = 1$.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      We have :



      $$frac{x^{cos x}}{x} = x^{cos x -1} = e^{ln x (cos x -1)}$$



      Now using the fact that in a neighborhood of $0$ we have :



      $$cos x - 1 = -frac{x^2}{2} + o(x^2)$$



      Then we can easily deduce that :



      $$ln x cdot (cos x -1) to 0$$



      Hence the desired limit is $1$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you very much.
        $endgroup$
        – violettagold
        Jan 8 at 16:48
















      2












      $begingroup$

      We have :



      $$frac{x^{cos x}}{x} = x^{cos x -1} = e^{ln x (cos x -1)}$$



      Now using the fact that in a neighborhood of $0$ we have :



      $$cos x - 1 = -frac{x^2}{2} + o(x^2)$$



      Then we can easily deduce that :



      $$ln x cdot (cos x -1) to 0$$



      Hence the desired limit is $1$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thank you very much.
        $endgroup$
        – violettagold
        Jan 8 at 16:48














      2












      2








      2





      $begingroup$

      We have :



      $$frac{x^{cos x}}{x} = x^{cos x -1} = e^{ln x (cos x -1)}$$



      Now using the fact that in a neighborhood of $0$ we have :



      $$cos x - 1 = -frac{x^2}{2} + o(x^2)$$



      Then we can easily deduce that :



      $$ln x cdot (cos x -1) to 0$$



      Hence the desired limit is $1$.






      share|cite|improve this answer









      $endgroup$



      We have :



      $$frac{x^{cos x}}{x} = x^{cos x -1} = e^{ln x (cos x -1)}$$



      Now using the fact that in a neighborhood of $0$ we have :



      $$cos x - 1 = -frac{x^2}{2} + o(x^2)$$



      Then we can easily deduce that :



      $$ln x cdot (cos x -1) to 0$$



      Hence the desired limit is $1$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 8 at 16:30









      ThinkingThinking

      1,03916




      1,03916












      • $begingroup$
        Thank you very much.
        $endgroup$
        – violettagold
        Jan 8 at 16:48


















      • $begingroup$
        Thank you very much.
        $endgroup$
        – violettagold
        Jan 8 at 16:48
















      $begingroup$
      Thank you very much.
      $endgroup$
      – violettagold
      Jan 8 at 16:48




      $begingroup$
      Thank you very much.
      $endgroup$
      – violettagold
      Jan 8 at 16:48











      1












      $begingroup$

      $$begin{align}
      lim_{xto0^+}x^{cos x-1}&=lim_{xto0^+}e^{ln x(cos x-1)}\
      &=expleft(lim_{xto0^+}frac{cos x-1}{dfrac1{ln x}}right)\
      &=expleft(lim_{xto0^+}frac{-sin x}{-dfrac1{x(ln x)^2}}right)\
      &=expleft(lim_{xto0^+}(xln x)(sin xln x)right)=e^0
      end{align}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $$begin{align}
        lim_{xto0^+}x^{cos x-1}&=lim_{xto0^+}e^{ln x(cos x-1)}\
        &=expleft(lim_{xto0^+}frac{cos x-1}{dfrac1{ln x}}right)\
        &=expleft(lim_{xto0^+}frac{-sin x}{-dfrac1{x(ln x)^2}}right)\
        &=expleft(lim_{xto0^+}(xln x)(sin xln x)right)=e^0
        end{align}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $$begin{align}
          lim_{xto0^+}x^{cos x-1}&=lim_{xto0^+}e^{ln x(cos x-1)}\
          &=expleft(lim_{xto0^+}frac{cos x-1}{dfrac1{ln x}}right)\
          &=expleft(lim_{xto0^+}frac{-sin x}{-dfrac1{x(ln x)^2}}right)\
          &=expleft(lim_{xto0^+}(xln x)(sin xln x)right)=e^0
          end{align}$$






          share|cite|improve this answer









          $endgroup$



          $$begin{align}
          lim_{xto0^+}x^{cos x-1}&=lim_{xto0^+}e^{ln x(cos x-1)}\
          &=expleft(lim_{xto0^+}frac{cos x-1}{dfrac1{ln x}}right)\
          &=expleft(lim_{xto0^+}frac{-sin x}{-dfrac1{x(ln x)^2}}right)\
          &=expleft(lim_{xto0^+}(xln x)(sin xln x)right)=e^0
          end{align}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 16:29









          ajotatxeajotatxe

          53.8k23890




          53.8k23890























              0












              $begingroup$

              $cos x= 1+o(1)$ when $xto 0$ (first order of Taylor expansion). Then
              $$frac{x^{cos x}}x =frac{x^{1+o(1)}}x=frac{xcdot x^{o(1)}}x=x^{o(1)}longrightarrow_{xto0}1$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                This is quite wrong: $y=x+o(1)$ does not imply $x^y=x+o(1)$, and $frac{o(1)}x$ does not necessarily converge to $0$.
                $endgroup$
                – Did
                Jan 8 at 16:48












              • $begingroup$
                I should have corrected it now
                $endgroup$
                – Lorenzo B.
                Jan 8 at 16:56










              • $begingroup$
                Yeah, except that now, you pushed everything in the last step $x^{o(1)}longrightarrow_{xto0}1$, which happens to be wrong in general. Sorry but this is not a guessing game...
                $endgroup$
                – Did
                Jan 8 at 16:59
















              0












              $begingroup$

              $cos x= 1+o(1)$ when $xto 0$ (first order of Taylor expansion). Then
              $$frac{x^{cos x}}x =frac{x^{1+o(1)}}x=frac{xcdot x^{o(1)}}x=x^{o(1)}longrightarrow_{xto0}1$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                This is quite wrong: $y=x+o(1)$ does not imply $x^y=x+o(1)$, and $frac{o(1)}x$ does not necessarily converge to $0$.
                $endgroup$
                – Did
                Jan 8 at 16:48












              • $begingroup$
                I should have corrected it now
                $endgroup$
                – Lorenzo B.
                Jan 8 at 16:56










              • $begingroup$
                Yeah, except that now, you pushed everything in the last step $x^{o(1)}longrightarrow_{xto0}1$, which happens to be wrong in general. Sorry but this is not a guessing game...
                $endgroup$
                – Did
                Jan 8 at 16:59














              0












              0








              0





              $begingroup$

              $cos x= 1+o(1)$ when $xto 0$ (first order of Taylor expansion). Then
              $$frac{x^{cos x}}x =frac{x^{1+o(1)}}x=frac{xcdot x^{o(1)}}x=x^{o(1)}longrightarrow_{xto0}1$$






              share|cite|improve this answer











              $endgroup$



              $cos x= 1+o(1)$ when $xto 0$ (first order of Taylor expansion). Then
              $$frac{x^{cos x}}x =frac{x^{1+o(1)}}x=frac{xcdot x^{o(1)}}x=x^{o(1)}longrightarrow_{xto0}1$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 8 at 16:55

























              answered Jan 8 at 16:45









              Lorenzo B.Lorenzo B.

              1,8402520




              1,8402520












              • $begingroup$
                This is quite wrong: $y=x+o(1)$ does not imply $x^y=x+o(1)$, and $frac{o(1)}x$ does not necessarily converge to $0$.
                $endgroup$
                – Did
                Jan 8 at 16:48












              • $begingroup$
                I should have corrected it now
                $endgroup$
                – Lorenzo B.
                Jan 8 at 16:56










              • $begingroup$
                Yeah, except that now, you pushed everything in the last step $x^{o(1)}longrightarrow_{xto0}1$, which happens to be wrong in general. Sorry but this is not a guessing game...
                $endgroup$
                – Did
                Jan 8 at 16:59


















              • $begingroup$
                This is quite wrong: $y=x+o(1)$ does not imply $x^y=x+o(1)$, and $frac{o(1)}x$ does not necessarily converge to $0$.
                $endgroup$
                – Did
                Jan 8 at 16:48












              • $begingroup$
                I should have corrected it now
                $endgroup$
                – Lorenzo B.
                Jan 8 at 16:56










              • $begingroup$
                Yeah, except that now, you pushed everything in the last step $x^{o(1)}longrightarrow_{xto0}1$, which happens to be wrong in general. Sorry but this is not a guessing game...
                $endgroup$
                – Did
                Jan 8 at 16:59
















              $begingroup$
              This is quite wrong: $y=x+o(1)$ does not imply $x^y=x+o(1)$, and $frac{o(1)}x$ does not necessarily converge to $0$.
              $endgroup$
              – Did
              Jan 8 at 16:48






              $begingroup$
              This is quite wrong: $y=x+o(1)$ does not imply $x^y=x+o(1)$, and $frac{o(1)}x$ does not necessarily converge to $0$.
              $endgroup$
              – Did
              Jan 8 at 16:48














              $begingroup$
              I should have corrected it now
              $endgroup$
              – Lorenzo B.
              Jan 8 at 16:56




              $begingroup$
              I should have corrected it now
              $endgroup$
              – Lorenzo B.
              Jan 8 at 16:56












              $begingroup$
              Yeah, except that now, you pushed everything in the last step $x^{o(1)}longrightarrow_{xto0}1$, which happens to be wrong in general. Sorry but this is not a guessing game...
              $endgroup$
              – Did
              Jan 8 at 16:59




              $begingroup$
              Yeah, except that now, you pushed everything in the last step $x^{o(1)}longrightarrow_{xto0}1$, which happens to be wrong in general. Sorry but this is not a guessing game...
              $endgroup$
              – Did
              Jan 8 at 16:59











              0












              $begingroup$

              You may also use the following facts:




              • $lim_{xto 0}frac{cos x-1}{x} = cos'(0) = -sin(0) = 0$

              • $lim_{xto 0}xln x = 0$


              So, you get
              $$ln frac{x^{cos x}}{x} = (cos x - 1)cdot ln x = frac{(cos x - 1)}{x}cdot x ln x stackrel{xto 0^+}{longrightarrow}0cdot 0 = 0$$



              Hence, $limlimits_{x to 0^{+}}frac{x^{cos x}}{x} = e^0 = 1$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You may also use the following facts:




                • $lim_{xto 0}frac{cos x-1}{x} = cos'(0) = -sin(0) = 0$

                • $lim_{xto 0}xln x = 0$


                So, you get
                $$ln frac{x^{cos x}}{x} = (cos x - 1)cdot ln x = frac{(cos x - 1)}{x}cdot x ln x stackrel{xto 0^+}{longrightarrow}0cdot 0 = 0$$



                Hence, $limlimits_{x to 0^{+}}frac{x^{cos x}}{x} = e^0 = 1$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You may also use the following facts:




                  • $lim_{xto 0}frac{cos x-1}{x} = cos'(0) = -sin(0) = 0$

                  • $lim_{xto 0}xln x = 0$


                  So, you get
                  $$ln frac{x^{cos x}}{x} = (cos x - 1)cdot ln x = frac{(cos x - 1)}{x}cdot x ln x stackrel{xto 0^+}{longrightarrow}0cdot 0 = 0$$



                  Hence, $limlimits_{x to 0^{+}}frac{x^{cos x}}{x} = e^0 = 1$.






                  share|cite|improve this answer









                  $endgroup$



                  You may also use the following facts:




                  • $lim_{xto 0}frac{cos x-1}{x} = cos'(0) = -sin(0) = 0$

                  • $lim_{xto 0}xln x = 0$


                  So, you get
                  $$ln frac{x^{cos x}}{x} = (cos x - 1)cdot ln x = frac{(cos x - 1)}{x}cdot x ln x stackrel{xto 0^+}{longrightarrow}0cdot 0 = 0$$



                  Hence, $limlimits_{x to 0^{+}}frac{x^{cos x}}{x} = e^0 = 1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 17:11









                  trancelocationtrancelocation

                  10.5k1722




                  10.5k1722






























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                      Has there ever been an instance of an active nuclear power plant within or near a war zone?

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                      36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
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