Does $lim_xto infty f(x)= infty$

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Let f be a twice differentiable function on R such that both $f'$ and $f''$
are strictly positive on R. Then $lim_xto infty f(x)= infty$.



I know the result is false and i can think of some example in my mind...but I can't get a concrete counterexample.



If this result is true please exaplain why? I hope it won't be




real-analysis derivatives




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    up vote
    0
    down vote

    favorite












    Let f be a twice differentiable function on R such that both $f'$ and $f''$
    are strictly positive on R. Then $lim_xto infty f(x)= infty$.



    I know the result is false and i can think of some example in my mind...but I can't get a concrete counterexample.



    If this result is true please exaplain why? I hope it won't be




    real-analysis derivatives




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let f be a twice differentiable function on R such that both $f'$ and $f''$
      are strictly positive on R. Then $lim_xto infty f(x)= infty$.



      I know the result is false and i can think of some example in my mind...but I can't get a concrete counterexample.



      If this result is true please exaplain why? I hope it won't be




      real-analysis derivatives




      share|cite|improve this question













      Let f be a twice differentiable function on R such that both $f'$ and $f''$
      are strictly positive on R. Then $lim_xto infty f(x)= infty$.



      I know the result is false and i can think of some example in my mind...but I can't get a concrete counterexample.



      If this result is true please exaplain why? I hope it won't be





      real-analysis derivatives





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      G Tony Jacobs

      25.5k43483




      25.5k43483









      asked 2 days ago









      Cloud JR

      456311




      456311




















          1 Answer
          1






          active

          oldest

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          up vote
          2
          down vote



          accepted










          Since $f'$ is strictly positive, and increasing, then we have $f'(x)>k=f'(0)$ for all positive $x$. That means that, for positive $x$, $f(x)ge g(x)=f(0)+kx$. We know that $gtoinfty$ as $x$ grows, so that's a proof.






          share|cite|improve this answer





















          • But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
            – Cloud JR
            2 days ago










          • $f, f'$ are supposed to be striclty positive, not stricly increasing.
            – Jonas
            2 days ago










          • @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
            – G Tony Jacobs
            2 days ago






          • 1




            Ups. Misread the problem. Sorry.
            – Jonas
            2 days ago






          • 1




            It's the obvious counterexample once you read the question correctly.
            – wonko
            2 days ago










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          1 Answer
          1






          active

          oldest

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          oldest

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          active

          oldest

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          up vote
          2
          down vote



          accepted










          Since $f'$ is strictly positive, and increasing, then we have $f'(x)>k=f'(0)$ for all positive $x$. That means that, for positive $x$, $f(x)ge g(x)=f(0)+kx$. We know that $gtoinfty$ as $x$ grows, so that's a proof.






          share|cite|improve this answer





















          • But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
            – Cloud JR
            2 days ago










          • $f, f'$ are supposed to be striclty positive, not stricly increasing.
            – Jonas
            2 days ago










          • @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
            – G Tony Jacobs
            2 days ago






          • 1




            Ups. Misread the problem. Sorry.
            – Jonas
            2 days ago






          • 1




            It's the obvious counterexample once you read the question correctly.
            – wonko
            2 days ago














          up vote
          2
          down vote



          accepted










          Since $f'$ is strictly positive, and increasing, then we have $f'(x)>k=f'(0)$ for all positive $x$. That means that, for positive $x$, $f(x)ge g(x)=f(0)+kx$. We know that $gtoinfty$ as $x$ grows, so that's a proof.






          share|cite|improve this answer





















          • But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
            – Cloud JR
            2 days ago










          • $f, f'$ are supposed to be striclty positive, not stricly increasing.
            – Jonas
            2 days ago










          • @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
            – G Tony Jacobs
            2 days ago






          • 1




            Ups. Misread the problem. Sorry.
            – Jonas
            2 days ago






          • 1




            It's the obvious counterexample once you read the question correctly.
            – wonko
            2 days ago












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Since $f'$ is strictly positive, and increasing, then we have $f'(x)>k=f'(0)$ for all positive $x$. That means that, for positive $x$, $f(x)ge g(x)=f(0)+kx$. We know that $gtoinfty$ as $x$ grows, so that's a proof.






          share|cite|improve this answer













          Since $f'$ is strictly positive, and increasing, then we have $f'(x)>k=f'(0)$ for all positive $x$. That means that, for positive $x$, $f(x)ge g(x)=f(0)+kx$. We know that $gtoinfty$ as $x$ grows, so that's a proof.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered 2 days ago









          G Tony Jacobs

          25.5k43483




          25.5k43483











          • But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
            – Cloud JR
            2 days ago










          • $f, f'$ are supposed to be striclty positive, not stricly increasing.
            – Jonas
            2 days ago










          • @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
            – G Tony Jacobs
            2 days ago






          • 1




            Ups. Misread the problem. Sorry.
            – Jonas
            2 days ago






          • 1




            It's the obvious counterexample once you read the question correctly.
            – wonko
            2 days ago
















          • But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
            – Cloud JR
            2 days ago










          • $f, f'$ are supposed to be striclty positive, not stricly increasing.
            – Jonas
            2 days ago










          • @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
            – G Tony Jacobs
            2 days ago






          • 1




            Ups. Misread the problem. Sorry.
            – Jonas
            2 days ago






          • 1




            It's the obvious counterexample once you read the question correctly.
            – wonko
            2 days ago















          But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
          – Cloud JR
          2 days ago




          But the answer is given as false univ.tifr.res.in/gs2018/Files/GS2018_QP_MTH.pdf. See question number 8
          – Cloud JR
          2 days ago












          $f, f'$ are supposed to be striclty positive, not stricly increasing.
          – Jonas
          2 days ago




          $f, f'$ are supposed to be striclty positive, not stricly increasing.
          – Jonas
          2 days ago












          @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
          – G Tony Jacobs
          2 days ago




          @Jonas, it's $f'$ and $f''$ that are strictly positive. That makes $f$ and $f'$ strictly increasing. Otherwise $arctan x + 2$ would be a counterexample.
          – G Tony Jacobs
          2 days ago




          1




          1




          Ups. Misread the problem. Sorry.
          – Jonas
          2 days ago




          Ups. Misread the problem. Sorry.
          – Jonas
          2 days ago




          1




          1




          It's the obvious counterexample once you read the question correctly.
          – wonko
          2 days ago




          It's the obvious counterexample once you read the question correctly.
          – wonko
          2 days ago












           

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