How to find a function $f(x)$ such that $lim_{xtoinfty}isqrt{1-x^2}f(x)=-1$ by maintaining the $x$...












0














As the question itself asks, how can I find a function $f(x)$ such that $lim_{xtoinfty}isqrt{1-x^2}f(x)=-1$ by maintaining the $x$ dipendence. Is there a method or is it different case by case?



I've no idea where to start and have not found anything online which I could comprehend. I'm new to this stuff so I excuse myself if the question is silly.



Thank you for your time.



EDIT:



Thanks to @qbert for his answer, he said that there are many functions that I could choose, which is the method to find them?










share|cite|improve this question





























    0














    As the question itself asks, how can I find a function $f(x)$ such that $lim_{xtoinfty}isqrt{1-x^2}f(x)=-1$ by maintaining the $x$ dipendence. Is there a method or is it different case by case?



    I've no idea where to start and have not found anything online which I could comprehend. I'm new to this stuff so I excuse myself if the question is silly.



    Thank you for your time.



    EDIT:



    Thanks to @qbert for his answer, he said that there are many functions that I could choose, which is the method to find them?










    share|cite|improve this question



























      0












      0








      0







      As the question itself asks, how can I find a function $f(x)$ such that $lim_{xtoinfty}isqrt{1-x^2}f(x)=-1$ by maintaining the $x$ dipendence. Is there a method or is it different case by case?



      I've no idea where to start and have not found anything online which I could comprehend. I'm new to this stuff so I excuse myself if the question is silly.



      Thank you for your time.



      EDIT:



      Thanks to @qbert for his answer, he said that there are many functions that I could choose, which is the method to find them?










      share|cite|improve this question















      As the question itself asks, how can I find a function $f(x)$ such that $lim_{xtoinfty}isqrt{1-x^2}f(x)=-1$ by maintaining the $x$ dipendence. Is there a method or is it different case by case?



      I've no idea where to start and have not found anything online which I could comprehend. I'm new to this stuff so I excuse myself if the question is silly.



      Thank you for your time.



      EDIT:



      Thanks to @qbert for his answer, he said that there are many functions that I could choose, which is the method to find them?







      limits






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      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 20 at 18:08

























      asked Oct 9 at 21:25









      Edoardo Serra

      32




      32






















          1 Answer
          1






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          votes


















          0














          By inspection,
          $$
          f(x)=frac{i}{sqrt{1-x^2}}
          $$

          does the job.



          The way I found it was by looking for a function which makes the limit trivial by cancelling any $x$ dependence (while giving a product of $-1$), making my life easier.






          share|cite|improve this answer





















          • Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
            – Edoardo Serra
            Oct 9 at 21:34










          • ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
            – qbert
            Oct 9 at 21:36












          • are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
            – Edoardo Serra
            Oct 9 at 21:43






          • 1




            sorry, it works, thanks a lot. I've misread the function
            – Edoardo Serra
            Oct 9 at 21:44













          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0














          By inspection,
          $$
          f(x)=frac{i}{sqrt{1-x^2}}
          $$

          does the job.



          The way I found it was by looking for a function which makes the limit trivial by cancelling any $x$ dependence (while giving a product of $-1$), making my life easier.






          share|cite|improve this answer





















          • Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
            – Edoardo Serra
            Oct 9 at 21:34










          • ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
            – qbert
            Oct 9 at 21:36












          • are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
            – Edoardo Serra
            Oct 9 at 21:43






          • 1




            sorry, it works, thanks a lot. I've misread the function
            – Edoardo Serra
            Oct 9 at 21:44


















          0














          By inspection,
          $$
          f(x)=frac{i}{sqrt{1-x^2}}
          $$

          does the job.



          The way I found it was by looking for a function which makes the limit trivial by cancelling any $x$ dependence (while giving a product of $-1$), making my life easier.






          share|cite|improve this answer





















          • Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
            – Edoardo Serra
            Oct 9 at 21:34










          • ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
            – qbert
            Oct 9 at 21:36












          • are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
            – Edoardo Serra
            Oct 9 at 21:43






          • 1




            sorry, it works, thanks a lot. I've misread the function
            – Edoardo Serra
            Oct 9 at 21:44
















          0












          0








          0






          By inspection,
          $$
          f(x)=frac{i}{sqrt{1-x^2}}
          $$

          does the job.



          The way I found it was by looking for a function which makes the limit trivial by cancelling any $x$ dependence (while giving a product of $-1$), making my life easier.






          share|cite|improve this answer












          By inspection,
          $$
          f(x)=frac{i}{sqrt{1-x^2}}
          $$

          does the job.



          The way I found it was by looking for a function which makes the limit trivial by cancelling any $x$ dependence (while giving a product of $-1$), making my life easier.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 9 at 21:30









          qbert

          22.1k32460




          22.1k32460












          • Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
            – Edoardo Serra
            Oct 9 at 21:34










          • ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
            – qbert
            Oct 9 at 21:36












          • are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
            – Edoardo Serra
            Oct 9 at 21:43






          • 1




            sorry, it works, thanks a lot. I've misread the function
            – Edoardo Serra
            Oct 9 at 21:44




















          • Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
            – Edoardo Serra
            Oct 9 at 21:34










          • ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
            – qbert
            Oct 9 at 21:36












          • are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
            – Edoardo Serra
            Oct 9 at 21:43






          • 1




            sorry, it works, thanks a lot. I've misread the function
            – Edoardo Serra
            Oct 9 at 21:44


















          Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
          – Edoardo Serra
          Oct 9 at 21:34




          Thanks, I meant by maintaining the $x$ dependence, I'll edit the question right now.
          – Edoardo Serra
          Oct 9 at 21:34












          ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
          – qbert
          Oct 9 at 21:36






          ok, $f(x)=frac{i}{sqrt{1-x-x^2}}$ there are many functions you can choose.
          – qbert
          Oct 9 at 21:36














          are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
          – Edoardo Serra
          Oct 9 at 21:43




          are you sure? I've tried graphing it in wolfram alpha and the real part goes to zero
          – Edoardo Serra
          Oct 9 at 21:43




          1




          1




          sorry, it works, thanks a lot. I've misread the function
          – Edoardo Serra
          Oct 9 at 21:44






          sorry, it works, thanks a lot. I've misread the function
          – Edoardo Serra
          Oct 9 at 21:44




















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