What is the value of $lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1}$? The Next CEO...

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What is the value of $lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1}$?



The Next CEO of Stack OverflowFinding the limit of $frac{sqrt{x}}{sqrt{x}+sinsqrt{x}}$Calculate $lim_{x to infty} x - sqrt{x^2 + 2x}$ without derivationsFind $lim_{xto-infty}frac{x}{sqrt{x^2+2}}$Find $lim_{nto infty}sqrt[n]{frac{sum_{i=1}^p a_i^n}{p}}$Find the value of : $lim_{xtoinfty}frac{sqrt{x-1} - sqrt{x-2}}{sqrt{x-2} - sqrt{x-3}}$Find the limit $lim_{t to 9} frac{3-sqrt{t}}{9-t}$$lim_{n to infty} frac{sqrt{1} + sqrt{2} + … + sqrt{n}}{nsqrt{n}}$Find: $lim_{xtoinfty} frac{sqrt{x}}{sqrt{x+sqrt{x+sqrt{x}}}}.$How to solve $lim_{xto1}=frac{x^2+x-2}{1-sqrt{x}}$?finding value of $lim_{nrightarrow infty}sqrt[n]{frac{(27)^n(n!)^3}{(3n)!}}$












1












$begingroup$


$$lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} = ?$$



I have done these steps to find the answer:




  1. $x^2-1=(x+1)(x-1)$


  2. $sqrt{4x-4}=2sqrt{x-1}$


  3. $displaystylelim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} = lim_{x to 1^+} frac{sqrt{ (x+1)(x-1) }+x-1}{ 2sqrt{x-1} + (x+1)(x-1) }$



So how do I remove what causes the hole function not to become $frac{0}{0}$ and solve the limit?






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    The numerator tends to $2$, the denominator to $0$.
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 8:04










  • $begingroup$
    Did you mean $$lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} ?$$
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 8:05










  • $begingroup$
    Yes I have edited it now!
    $endgroup$
    – Aquaman
    Mar 16 at 8:09
















1












$begingroup$


$$lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} = ?$$



I have done these steps to find the answer:




  1. $x^2-1=(x+1)(x-1)$


  2. $sqrt{4x-4}=2sqrt{x-1}$


  3. $displaystylelim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} = lim_{x to 1^+} frac{sqrt{ (x+1)(x-1) }+x-1}{ 2sqrt{x-1} + (x+1)(x-1) }$



So how do I remove what causes the hole function not to become $frac{0}{0}$ and solve the limit?






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    The numerator tends to $2$, the denominator to $0$.
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 8:04










  • $begingroup$
    Did you mean $$lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} ?$$
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 8:05










  • $begingroup$
    Yes I have edited it now!
    $endgroup$
    – Aquaman
    Mar 16 at 8:09














1












1








1





$begingroup$


$$lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} = ?$$



I have done these steps to find the answer:




  1. $x^2-1=(x+1)(x-1)$


  2. $sqrt{4x-4}=2sqrt{x-1}$


  3. $displaystylelim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} = lim_{x to 1^+} frac{sqrt{ (x+1)(x-1) }+x-1}{ 2sqrt{x-1} + (x+1)(x-1) }$



So how do I remove what causes the hole function not to become $frac{0}{0}$ and solve the limit?






limits







share|cite|improve this question











$endgroup$




$$lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} = ?$$



I have done these steps to find the answer:




  1. $x^2-1=(x+1)(x-1)$


  2. $sqrt{4x-4}=2sqrt{x-1}$


  3. $displaystylelim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} = lim_{x to 1^+} frac{sqrt{ (x+1)(x-1) }+x-1}{ 2sqrt{x-1} + (x+1)(x-1) }$



So how do I remove what causes the hole function not to become $frac{0}{0}$ and solve the limit?






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 16 at 11:36









egreg

185k1486206




185k1486206










asked Mar 16 at 8:01









AquamanAquaman

133




133












  • $begingroup$
    The numerator tends to $2$, the denominator to $0$.
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 8:04










  • $begingroup$
    Did you mean $$lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} ?$$
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 8:05










  • $begingroup$
    Yes I have edited it now!
    $endgroup$
    – Aquaman
    Mar 16 at 8:09


















  • $begingroup$
    The numerator tends to $2$, the denominator to $0$.
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 8:04










  • $begingroup$
    Did you mean $$lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} ?$$
    $endgroup$
    – Lord Shark the Unknown
    Mar 16 at 8:05










  • $begingroup$
    Yes I have edited it now!
    $endgroup$
    – Aquaman
    Mar 16 at 8:09
















$begingroup$
The numerator tends to $2$, the denominator to $0$.
$endgroup$
– Lord Shark the Unknown
Mar 16 at 8:04




$begingroup$
The numerator tends to $2$, the denominator to $0$.
$endgroup$
– Lord Shark the Unknown
Mar 16 at 8:04












$begingroup$
Did you mean $$lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} ?$$
$endgroup$
– Lord Shark the Unknown
Mar 16 at 8:05




$begingroup$
Did you mean $$lim_{x to 1^+} frac{sqrt{x^2-1}+x-1}{ sqrt{4x-4}+x^2-1} ?$$
$endgroup$
– Lord Shark the Unknown
Mar 16 at 8:05












$begingroup$
Yes I have edited it now!
$endgroup$
– Aquaman
Mar 16 at 8:09




$begingroup$
Yes I have edited it now!
$endgroup$
– Aquaman
Mar 16 at 8:09










3 Answers
3






active

oldest

votes


















2












$begingroup$

The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.



$$lim_{xto 1^+}dfrac{sqrt{x^2-1}+x+1}{sqrt{4x-4}+x^2-1}to infty$$



This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf




After the OP's edit: $$dfrac{sqrt{x^2-1}+x-1}{2sqrt{x-1}+x^2-1}=dfrac{sqrt{x-1}}{sqrt{x-1}}cdotdfrac{sqrt{x+1}+sqrt{x-1}}{2+(x-1)^{3/2}(x+1)} \ implies lim_{xto 1^+}dfrac{sqrt{x+1}+sqrt{x-1}}{2+(x-1)^{3/2}(x+1)}to dfrac{sqrt{2}}{2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Are you sure? I am not!
    $endgroup$
    – Aquaman
    Mar 16 at 8:04










  • $begingroup$
    I think that Zeros are limit kinds of zero! I mean they are $0^+$ & $0^-$
    $endgroup$
    – Aquaman
    Mar 16 at 8:06










  • $begingroup$
    I think that we have to simplify the function then solve it!
    $endgroup$
    – Aquaman
    Mar 16 at 8:07










  • $begingroup$
    AFTER REMOVING + and replacing - it will change . Sorry ive edited the question!
    $endgroup$
    – Aquaman
    Mar 16 at 8:12










  • $begingroup$
    Ok @Aquaman I've edited my answer too. Cheers:- ))
    $endgroup$
    – Paras Khosla
    Mar 16 at 8:14



















2












$begingroup$

I assume:
$$lim_{x to 1^+} frac{sqrt{x^2-1}+xcolor{red}-1}{ sqrt{4x-4}+x^2-1} = lim_{x to 1^+} frac{sqrt{x-1}cdot sqrt{x+1}+(sqrt{x-1})^2}{ sqrt{4x-4}+(sqrt{x-1})^2(x+1)} =\
lim_{x to 1^+} frac{sqrt{x-1}cdot (sqrt{x+1}+sqrt{x-1})}{sqrt{x-1}cdot (2+(sqrt{x-1})(x+1))} =frac{sqrt{2}}{2}.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Set $x-1=t^2$, which you can because the limit is for $xto1^+$. Then the limit becomes
    $$
    lim_{tto0^+}frac{tsqrt{t^2+2}+t^2}{2t+t^2(t^2+2)}=
    lim_{tto0^+}frac{sqrt{t^2+2}+t}{2+t(t^2+2)}
    $$






    share|cite|improve this answer









    $endgroup$














      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.



      $$lim_{xto 1^+}dfrac{sqrt{x^2-1}+x+1}{sqrt{4x-4}+x^2-1}to infty$$



      This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf




      After the OP's edit: $$dfrac{sqrt{x^2-1}+x-1}{2sqrt{x-1}+x^2-1}=dfrac{sqrt{x-1}}{sqrt{x-1}}cdotdfrac{sqrt{x+1}+sqrt{x-1}}{2+(x-1)^{3/2}(x+1)} \ implies lim_{xto 1^+}dfrac{sqrt{x+1}+sqrt{x-1}}{2+(x-1)^{3/2}(x+1)}to dfrac{sqrt{2}}{2}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Are you sure? I am not!
        $endgroup$
        – Aquaman
        Mar 16 at 8:04










      • $begingroup$
        I think that Zeros are limit kinds of zero! I mean they are $0^+$ & $0^-$
        $endgroup$
        – Aquaman
        Mar 16 at 8:06










      • $begingroup$
        I think that we have to simplify the function then solve it!
        $endgroup$
        – Aquaman
        Mar 16 at 8:07










      • $begingroup$
        AFTER REMOVING + and replacing - it will change . Sorry ive edited the question!
        $endgroup$
        – Aquaman
        Mar 16 at 8:12










      • $begingroup$
        Ok @Aquaman I've edited my answer too. Cheers:- ))
        $endgroup$
        – Paras Khosla
        Mar 16 at 8:14
















      2












      $begingroup$

      The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.



      $$lim_{xto 1^+}dfrac{sqrt{x^2-1}+x+1}{sqrt{4x-4}+x^2-1}to infty$$



      This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf




      After the OP's edit: $$dfrac{sqrt{x^2-1}+x-1}{2sqrt{x-1}+x^2-1}=dfrac{sqrt{x-1}}{sqrt{x-1}}cdotdfrac{sqrt{x+1}+sqrt{x-1}}{2+(x-1)^{3/2}(x+1)} \ implies lim_{xto 1^+}dfrac{sqrt{x+1}+sqrt{x-1}}{2+(x-1)^{3/2}(x+1)}to dfrac{sqrt{2}}{2}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Are you sure? I am not!
        $endgroup$
        – Aquaman
        Mar 16 at 8:04










      • $begingroup$
        I think that Zeros are limit kinds of zero! I mean they are $0^+$ & $0^-$
        $endgroup$
        – Aquaman
        Mar 16 at 8:06










      • $begingroup$
        I think that we have to simplify the function then solve it!
        $endgroup$
        – Aquaman
        Mar 16 at 8:07










      • $begingroup$
        AFTER REMOVING + and replacing - it will change . Sorry ive edited the question!
        $endgroup$
        – Aquaman
        Mar 16 at 8:12










      • $begingroup$
        Ok @Aquaman I've edited my answer too. Cheers:- ))
        $endgroup$
        – Paras Khosla
        Mar 16 at 8:14














      2












      2








      2





      $begingroup$

      The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.



      $$lim_{xto 1^+}dfrac{sqrt{x^2-1}+x+1}{sqrt{4x-4}+x^2-1}to infty$$



      This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf




      After the OP's edit: $$dfrac{sqrt{x^2-1}+x-1}{2sqrt{x-1}+x^2-1}=dfrac{sqrt{x-1}}{sqrt{x-1}}cdotdfrac{sqrt{x+1}+sqrt{x-1}}{2+(x-1)^{3/2}(x+1)} \ implies lim_{xto 1^+}dfrac{sqrt{x+1}+sqrt{x-1}}{2+(x-1)^{3/2}(x+1)}to dfrac{sqrt{2}}{2}$$






      share|cite|improve this answer











      $endgroup$



      The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.



      $$lim_{xto 1^+}dfrac{sqrt{x^2-1}+x+1}{sqrt{4x-4}+x^2-1}to infty$$



      This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf




      After the OP's edit: $$dfrac{sqrt{x^2-1}+x-1}{2sqrt{x-1}+x^2-1}=dfrac{sqrt{x-1}}{sqrt{x-1}}cdotdfrac{sqrt{x+1}+sqrt{x-1}}{2+(x-1)^{3/2}(x+1)} \ implies lim_{xto 1^+}dfrac{sqrt{x+1}+sqrt{x-1}}{2+(x-1)^{3/2}(x+1)}to dfrac{sqrt{2}}{2}$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 17 at 6:13

























      answered Mar 16 at 8:04









      Paras KhoslaParas Khosla

      2,691423




      2,691423












      • $begingroup$
        Are you sure? I am not!
        $endgroup$
        – Aquaman
        Mar 16 at 8:04










      • $begingroup$
        I think that Zeros are limit kinds of zero! I mean they are $0^+$ & $0^-$
        $endgroup$
        – Aquaman
        Mar 16 at 8:06










      • $begingroup$
        I think that we have to simplify the function then solve it!
        $endgroup$
        – Aquaman
        Mar 16 at 8:07










      • $begingroup$
        AFTER REMOVING + and replacing - it will change . Sorry ive edited the question!
        $endgroup$
        – Aquaman
        Mar 16 at 8:12










      • $begingroup$
        Ok @Aquaman I've edited my answer too. Cheers:- ))
        $endgroup$
        – Paras Khosla
        Mar 16 at 8:14


















      • $begingroup$
        Are you sure? I am not!
        $endgroup$
        – Aquaman
        Mar 16 at 8:04










      • $begingroup$
        I think that Zeros are limit kinds of zero! I mean they are $0^+$ & $0^-$
        $endgroup$
        – Aquaman
        Mar 16 at 8:06










      • $begingroup$
        I think that we have to simplify the function then solve it!
        $endgroup$
        – Aquaman
        Mar 16 at 8:07










      • $begingroup$
        AFTER REMOVING + and replacing - it will change . Sorry ive edited the question!
        $endgroup$
        – Aquaman
        Mar 16 at 8:12










      • $begingroup$
        Ok @Aquaman I've edited my answer too. Cheers:- ))
        $endgroup$
        – Paras Khosla
        Mar 16 at 8:14
















      $begingroup$
      Are you sure? I am not!
      $endgroup$
      – Aquaman
      Mar 16 at 8:04




      $begingroup$
      Are you sure? I am not!
      $endgroup$
      – Aquaman
      Mar 16 at 8:04












      $begingroup$
      I think that Zeros are limit kinds of zero! I mean they are $0^+$ & $0^-$
      $endgroup$
      – Aquaman
      Mar 16 at 8:06




      $begingroup$
      I think that Zeros are limit kinds of zero! I mean they are $0^+$ & $0^-$
      $endgroup$
      – Aquaman
      Mar 16 at 8:06












      $begingroup$
      I think that we have to simplify the function then solve it!
      $endgroup$
      – Aquaman
      Mar 16 at 8:07




      $begingroup$
      I think that we have to simplify the function then solve it!
      $endgroup$
      – Aquaman
      Mar 16 at 8:07












      $begingroup$
      AFTER REMOVING + and replacing - it will change . Sorry ive edited the question!
      $endgroup$
      – Aquaman
      Mar 16 at 8:12




      $begingroup$
      AFTER REMOVING + and replacing - it will change . Sorry ive edited the question!
      $endgroup$
      – Aquaman
      Mar 16 at 8:12












      $begingroup$
      Ok @Aquaman I've edited my answer too. Cheers:- ))
      $endgroup$
      – Paras Khosla
      Mar 16 at 8:14




      $begingroup$
      Ok @Aquaman I've edited my answer too. Cheers:- ))
      $endgroup$
      – Paras Khosla
      Mar 16 at 8:14











      2












      $begingroup$

      I assume:
      $$lim_{x to 1^+} frac{sqrt{x^2-1}+xcolor{red}-1}{ sqrt{4x-4}+x^2-1} = lim_{x to 1^+} frac{sqrt{x-1}cdot sqrt{x+1}+(sqrt{x-1})^2}{ sqrt{4x-4}+(sqrt{x-1})^2(x+1)} =\
      lim_{x to 1^+} frac{sqrt{x-1}cdot (sqrt{x+1}+sqrt{x-1})}{sqrt{x-1}cdot (2+(sqrt{x-1})(x+1))} =frac{sqrt{2}}{2}.$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        I assume:
        $$lim_{x to 1^+} frac{sqrt{x^2-1}+xcolor{red}-1}{ sqrt{4x-4}+x^2-1} = lim_{x to 1^+} frac{sqrt{x-1}cdot sqrt{x+1}+(sqrt{x-1})^2}{ sqrt{4x-4}+(sqrt{x-1})^2(x+1)} =\
        lim_{x to 1^+} frac{sqrt{x-1}cdot (sqrt{x+1}+sqrt{x-1})}{sqrt{x-1}cdot (2+(sqrt{x-1})(x+1))} =frac{sqrt{2}}{2}.$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          I assume:
          $$lim_{x to 1^+} frac{sqrt{x^2-1}+xcolor{red}-1}{ sqrt{4x-4}+x^2-1} = lim_{x to 1^+} frac{sqrt{x-1}cdot sqrt{x+1}+(sqrt{x-1})^2}{ sqrt{4x-4}+(sqrt{x-1})^2(x+1)} =\
          lim_{x to 1^+} frac{sqrt{x-1}cdot (sqrt{x+1}+sqrt{x-1})}{sqrt{x-1}cdot (2+(sqrt{x-1})(x+1))} =frac{sqrt{2}}{2}.$$






          share|cite|improve this answer









          $endgroup$



          I assume:
          $$lim_{x to 1^+} frac{sqrt{x^2-1}+xcolor{red}-1}{ sqrt{4x-4}+x^2-1} = lim_{x to 1^+} frac{sqrt{x-1}cdot sqrt{x+1}+(sqrt{x-1})^2}{ sqrt{4x-4}+(sqrt{x-1})^2(x+1)} =\
          lim_{x to 1^+} frac{sqrt{x-1}cdot (sqrt{x+1}+sqrt{x-1})}{sqrt{x-1}cdot (2+(sqrt{x-1})(x+1))} =frac{sqrt{2}}{2}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 16 at 8:08









          farruhotafarruhota

          21.6k2842




          21.6k2842























              0












              $begingroup$

              Set $x-1=t^2$, which you can because the limit is for $xto1^+$. Then the limit becomes
              $$
              lim_{tto0^+}frac{tsqrt{t^2+2}+t^2}{2t+t^2(t^2+2)}=
              lim_{tto0^+}frac{sqrt{t^2+2}+t}{2+t(t^2+2)}
              $$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Set $x-1=t^2$, which you can because the limit is for $xto1^+$. Then the limit becomes
                $$
                lim_{tto0^+}frac{tsqrt{t^2+2}+t^2}{2t+t^2(t^2+2)}=
                lim_{tto0^+}frac{sqrt{t^2+2}+t}{2+t(t^2+2)}
                $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Set $x-1=t^2$, which you can because the limit is for $xto1^+$. Then the limit becomes
                  $$
                  lim_{tto0^+}frac{tsqrt{t^2+2}+t^2}{2t+t^2(t^2+2)}=
                  lim_{tto0^+}frac{sqrt{t^2+2}+t}{2+t(t^2+2)}
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  Set $x-1=t^2$, which you can because the limit is for $xto1^+$. Then the limit becomes
                  $$
                  lim_{tto0^+}frac{tsqrt{t^2+2}+t^2}{2t+t^2(t^2+2)}=
                  lim_{tto0^+}frac{sqrt{t^2+2}+t}{2+t(t^2+2)}
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 16 at 11:39









                  egregegreg

                  185k1486206




                  185k1486206






























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