Check my proof that right-sided limit doesn't exist here












1














Let $f:mathbb{R^*} - >mathbb{R}, f(x) =sin (frac{1+x}{sqrt x}) $. Prove that $lim_{x to 0^+} f(x) $ doesn't exist.

My solution : Since $f$ is continuous, we have
$lim_{x to 0^+} f(x)=f(lim_{xto 0^+} (frac{1+x}{sqrt x}))=lim_{xto infty} f(x) $

However, the last limit doesn't exist and hence the result.






real-analysis limits







share|cite|improve this question



























    1














    Let $f:mathbb{R^*} - >mathbb{R}, f(x) =sin (frac{1+x}{sqrt x}) $. Prove that $lim_{x to 0^+} f(x) $ doesn't exist.

    My solution : Since $f$ is continuous, we have
    $lim_{x to 0^+} f(x)=f(lim_{xto 0^+} (frac{1+x}{sqrt x}))=lim_{xto infty} f(x) $

    However, the last limit doesn't exist and hence the result.






    real-analysis limits







    share|cite|improve this question

























      1












      1








      1







      Let $f:mathbb{R^*} - >mathbb{R}, f(x) =sin (frac{1+x}{sqrt x}) $. Prove that $lim_{x to 0^+} f(x) $ doesn't exist.

      My solution : Since $f$ is continuous, we have
      $lim_{x to 0^+} f(x)=f(lim_{xto 0^+} (frac{1+x}{sqrt x}))=lim_{xto infty} f(x) $

      However, the last limit doesn't exist and hence the result.






      real-analysis limits







      share|cite|improve this question













      Let $f:mathbb{R^*} - >mathbb{R}, f(x) =sin (frac{1+x}{sqrt x}) $. Prove that $lim_{x to 0^+} f(x) $ doesn't exist.

      My solution : Since $f$ is continuous, we have
      $lim_{x to 0^+} f(x)=f(lim_{xto 0^+} (frac{1+x}{sqrt x}))=lim_{xto infty} f(x) $

      However, the last limit doesn't exist and hence the result.






      real-analysis limits




      real-analysis limits






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 at 19:58









      user69503

      626




      626






















          1 Answer
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          1














          We have that



          $$frac{1+x}{sqrt x} to infty$$



          then the limit doesn't exist, to prove that we can consider




          • $x_n=frac1{(2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 0$


          • $x_n=frac1{(pi/2+2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim frac{pi}2+2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 1$







          share|cite|improve this answer























          • Thank you! What about my proof? Is it all right or does it have any flaws?
            – user69503
            Nov 25 at 20:05






          • 1




            @user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
            – gimusi
            Nov 25 at 20:06










          • Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
            – user69503
            Nov 25 at 20:14












          • @user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
            – gimusi
            Nov 25 at 20:19






          • 1




            @user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
            – gimusi
            Nov 25 at 20:27











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          We have that



          $$frac{1+x}{sqrt x} to infty$$



          then the limit doesn't exist, to prove that we can consider




          • $x_n=frac1{(2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 0$


          • $x_n=frac1{(pi/2+2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim frac{pi}2+2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 1$







          share|cite|improve this answer























          • Thank you! What about my proof? Is it all right or does it have any flaws?
            – user69503
            Nov 25 at 20:05






          • 1




            @user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
            – gimusi
            Nov 25 at 20:06










          • Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
            – user69503
            Nov 25 at 20:14












          • @user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
            – gimusi
            Nov 25 at 20:19






          • 1




            @user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
            – gimusi
            Nov 25 at 20:27
















          1














          We have that



          $$frac{1+x}{sqrt x} to infty$$



          then the limit doesn't exist, to prove that we can consider




          • $x_n=frac1{(2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 0$


          • $x_n=frac1{(pi/2+2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim frac{pi}2+2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 1$







          share|cite|improve this answer























          • Thank you! What about my proof? Is it all right or does it have any flaws?
            – user69503
            Nov 25 at 20:05






          • 1




            @user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
            – gimusi
            Nov 25 at 20:06










          • Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
            – user69503
            Nov 25 at 20:14












          • @user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
            – gimusi
            Nov 25 at 20:19






          • 1




            @user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
            – gimusi
            Nov 25 at 20:27














          1












          1








          1






          We have that



          $$frac{1+x}{sqrt x} to infty$$



          then the limit doesn't exist, to prove that we can consider




          • $x_n=frac1{(2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 0$


          • $x_n=frac1{(pi/2+2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim frac{pi}2+2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 1$







          share|cite|improve this answer














          We have that



          $$frac{1+x}{sqrt x} to infty$$



          then the limit doesn't exist, to prove that we can consider




          • $x_n=frac1{(2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 0$


          • $x_n=frac1{(pi/2+2pi n)^2}to 0^+implies frac{1+x_n}{sqrt x_n}sim frac{pi}2+2pi nto infty implies sin (frac{1+x_n}{sqrt x_n})to 1$








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 25 at 20:05

























          answered Nov 25 at 20:04









          gimusi

          1




          1












          • Thank you! What about my proof? Is it all right or does it have any flaws?
            – user69503
            Nov 25 at 20:05






          • 1




            @user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
            – gimusi
            Nov 25 at 20:06










          • Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
            – user69503
            Nov 25 at 20:14












          • @user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
            – gimusi
            Nov 25 at 20:19






          • 1




            @user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
            – gimusi
            Nov 25 at 20:27


















          • Thank you! What about my proof? Is it all right or does it have any flaws?
            – user69503
            Nov 25 at 20:05






          • 1




            @user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
            – gimusi
            Nov 25 at 20:06










          • Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
            – user69503
            Nov 25 at 20:14












          • @user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
            – gimusi
            Nov 25 at 20:19






          • 1




            @user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
            – gimusi
            Nov 25 at 20:27
















          Thank you! What about my proof? Is it all right or does it have any flaws?
          – user69503
          Nov 25 at 20:05




          Thank you! What about my proof? Is it all right or does it have any flaws?
          – user69503
          Nov 25 at 20:05




          1




          1




          @user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
          – gimusi
          Nov 25 at 20:06




          @user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits.
          – gimusi
          Nov 25 at 20:06












          Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
          – user69503
          Nov 25 at 20:14






          Why doesn't it work?. Didn't I correctly show that the given limit is the same as $lim_{xto infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work?
          – user69503
          Nov 25 at 20:14














          @user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
          – gimusi
          Nov 25 at 20:19




          @user69503 Essentially we are proving that $lim_{yto infty} sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $frac{1+x}{sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that.
          – gimusi
          Nov 25 at 20:19




          1




          1




          @user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
          – gimusi
          Nov 25 at 20:27




          @user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint.
          – gimusi
          Nov 25 at 20:27


















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