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







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$endgroup$












  • $begingroup$
    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

248k23224463




248k23224463










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

      1,13916




      1,13916












      • $begingroup$
        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

          53.8k23990




          53.8k23990























              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.

              1,8602520




              1,8602520












              • $begingroup$
                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

                  12.2k1826




                  12.2k1826






























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                      張江高科駅

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                      張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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