Using the Squeeze Theorem on $lim_{xto 0}frac{sin^2x}{x^2}$












1












$begingroup$


$$lim_{xto 0}frac{sin^2x}{x^2}$$



I'm trying to evaluate this limit using Squeeze Theorem. However, looking at the graph I know it approaches $1$, but I am getting $0$ using the Squeeze Theorem.



$$-frac{1}{x^2} < frac{sin^2x}{x^2} < frac{1}{x^2}$$



when I sub in $0$ it's just $0$. What am I doing wrong?



Edit: Wait, it's not zero! The upper and lower bounds are indeterminate. So I can't use squeeze theorem, correct?






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $1/x^2to +infty$
    $endgroup$
    – A.Γ.
    Dec 16 '18 at 18:54










  • $begingroup$
    You can only use squeeze theorem when upper and lower bound limits exist and are equal.
    $endgroup$
    – D.B.
    Dec 16 '18 at 18:57










  • $begingroup$
    Use L'Hopital's rule.
    $endgroup$
    – D.B.
    Dec 16 '18 at 18:59
















1












$begingroup$


$$lim_{xto 0}frac{sin^2x}{x^2}$$



I'm trying to evaluate this limit using Squeeze Theorem. However, looking at the graph I know it approaches $1$, but I am getting $0$ using the Squeeze Theorem.



$$-frac{1}{x^2} < frac{sin^2x}{x^2} < frac{1}{x^2}$$



when I sub in $0$ it's just $0$. What am I doing wrong?



Edit: Wait, it's not zero! The upper and lower bounds are indeterminate. So I can't use squeeze theorem, correct?






limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $1/x^2to +infty$
    $endgroup$
    – A.Γ.
    Dec 16 '18 at 18:54










  • $begingroup$
    You can only use squeeze theorem when upper and lower bound limits exist and are equal.
    $endgroup$
    – D.B.
    Dec 16 '18 at 18:57










  • $begingroup$
    Use L'Hopital's rule.
    $endgroup$
    – D.B.
    Dec 16 '18 at 18:59














1












1








1


1



$begingroup$


$$lim_{xto 0}frac{sin^2x}{x^2}$$



I'm trying to evaluate this limit using Squeeze Theorem. However, looking at the graph I know it approaches $1$, but I am getting $0$ using the Squeeze Theorem.



$$-frac{1}{x^2} < frac{sin^2x}{x^2} < frac{1}{x^2}$$



when I sub in $0$ it's just $0$. What am I doing wrong?



Edit: Wait, it's not zero! The upper and lower bounds are indeterminate. So I can't use squeeze theorem, correct?






limits







share|cite|improve this question











$endgroup$




$$lim_{xto 0}frac{sin^2x}{x^2}$$



I'm trying to evaluate this limit using Squeeze Theorem. However, looking at the graph I know it approaches $1$, but I am getting $0$ using the Squeeze Theorem.



$$-frac{1}{x^2} < frac{sin^2x}{x^2} < frac{1}{x^2}$$



when I sub in $0$ it's just $0$. What am I doing wrong?



Edit: Wait, it's not zero! The upper and lower bounds are indeterminate. So I can't use squeeze theorem, correct?






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 18:58









Blue

49.1k870156




49.1k870156










asked Dec 16 '18 at 18:52









J.W.J.W.

544




544








  • 1




    $begingroup$
    $1/x^2to +infty$
    $endgroup$
    – A.Γ.
    Dec 16 '18 at 18:54










  • $begingroup$
    You can only use squeeze theorem when upper and lower bound limits exist and are equal.
    $endgroup$
    – D.B.
    Dec 16 '18 at 18:57










  • $begingroup$
    Use L'Hopital's rule.
    $endgroup$
    – D.B.
    Dec 16 '18 at 18:59














  • 1




    $begingroup$
    $1/x^2to +infty$
    $endgroup$
    – A.Γ.
    Dec 16 '18 at 18:54










  • $begingroup$
    You can only use squeeze theorem when upper and lower bound limits exist and are equal.
    $endgroup$
    – D.B.
    Dec 16 '18 at 18:57










  • $begingroup$
    Use L'Hopital's rule.
    $endgroup$
    – D.B.
    Dec 16 '18 at 18:59








1




1




$begingroup$
$1/x^2to +infty$
$endgroup$
– A.Γ.
Dec 16 '18 at 18:54




$begingroup$
$1/x^2to +infty$
$endgroup$
– A.Γ.
Dec 16 '18 at 18:54












$begingroup$
You can only use squeeze theorem when upper and lower bound limits exist and are equal.
$endgroup$
– D.B.
Dec 16 '18 at 18:57




$begingroup$
You can only use squeeze theorem when upper and lower bound limits exist and are equal.
$endgroup$
– D.B.
Dec 16 '18 at 18:57












$begingroup$
Use L'Hopital's rule.
$endgroup$
– D.B.
Dec 16 '18 at 18:59




$begingroup$
Use L'Hopital's rule.
$endgroup$
– D.B.
Dec 16 '18 at 18:59










3 Answers
3






active

oldest

votes


















4












$begingroup$

The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-infty$ and the upper bound has limit $infty$, so they can't be used to determine the given limit.



If you want to apply squeezing, you can prove geometrically that
$$
cos^2x<frac{sin^2x}{x^2}<frac{1}{cos^2x}
$$
which is basically the usual proof that
$$
lim_{xto0}frac{sin x}{x}=1
$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The function is even, we assume $0<x<1$.



    By MVT,
    $$sin(x)=xcos(c)$$ with $0<c<x$.



    thus



    $$cos^2(x)<frac{sin^2(x)}{x^2}=cos^2(c)<1$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit



      $$lim_{x to 0}frac{sin x}{x} = 1$$



      If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)






      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









        4












        $begingroup$

        The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-infty$ and the upper bound has limit $infty$, so they can't be used to determine the given limit.



        If you want to apply squeezing, you can prove geometrically that
        $$
        cos^2x<frac{sin^2x}{x^2}<frac{1}{cos^2x}
        $$
        which is basically the usual proof that
        $$
        lim_{xto0}frac{sin x}{x}=1
        $$






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-infty$ and the upper bound has limit $infty$, so they can't be used to determine the given limit.



          If you want to apply squeezing, you can prove geometrically that
          $$
          cos^2x<frac{sin^2x}{x^2}<frac{1}{cos^2x}
          $$
          which is basically the usual proof that
          $$
          lim_{xto0}frac{sin x}{x}=1
          $$






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-infty$ and the upper bound has limit $infty$, so they can't be used to determine the given limit.



            If you want to apply squeezing, you can prove geometrically that
            $$
            cos^2x<frac{sin^2x}{x^2}<frac{1}{cos^2x}
            $$
            which is basically the usual proof that
            $$
            lim_{xto0}frac{sin x}{x}=1
            $$






            share|cite|improve this answer









            $endgroup$



            The lower and upper bounds you write are right, but unfortunately the lower bound has limit $-infty$ and the upper bound has limit $infty$, so they can't be used to determine the given limit.



            If you want to apply squeezing, you can prove geometrically that
            $$
            cos^2x<frac{sin^2x}{x^2}<frac{1}{cos^2x}
            $$
            which is basically the usual proof that
            $$
            lim_{xto0}frac{sin x}{x}=1
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 16 '18 at 18:59









            egregegreg

            184k1486205




            184k1486205























                1












                $begingroup$

                The function is even, we assume $0<x<1$.



                By MVT,
                $$sin(x)=xcos(c)$$ with $0<c<x$.



                thus



                $$cos^2(x)<frac{sin^2(x)}{x^2}=cos^2(c)<1$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  The function is even, we assume $0<x<1$.



                  By MVT,
                  $$sin(x)=xcos(c)$$ with $0<c<x$.



                  thus



                  $$cos^2(x)<frac{sin^2(x)}{x^2}=cos^2(c)<1$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    The function is even, we assume $0<x<1$.



                    By MVT,
                    $$sin(x)=xcos(c)$$ with $0<c<x$.



                    thus



                    $$cos^2(x)<frac{sin^2(x)}{x^2}=cos^2(c)<1$$






                    share|cite|improve this answer









                    $endgroup$



                    The function is even, we assume $0<x<1$.



                    By MVT,
                    $$sin(x)=xcos(c)$$ with $0<c<x$.



                    thus



                    $$cos^2(x)<frac{sin^2(x)}{x^2}=cos^2(c)<1$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 16 '18 at 19:02









                    hamam_Abdallahhamam_Abdallah

                    38.2k21634




                    38.2k21634























                        0












                        $begingroup$

                        As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit



                        $$lim_{x to 0}frac{sin x}{x} = 1$$



                        If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit



                          $$lim_{x to 0}frac{sin x}{x} = 1$$



                          If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit



                            $$lim_{x to 0}frac{sin x}{x} = 1$$



                            If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)






                            share|cite|improve this answer











                            $endgroup$



                            As you found out, the upper and lower bounds are indeterminate. Therefore, you cannot determine the limit that way. Instead, you can refer to the following well-known limit



                            $$lim_{x to 0}frac{sin x}{x} = 1$$



                            If you want to use the Squeeze Theorem, you can refer to the geometric proof of the limit above. (Then, you can apply the geometric proof here and reach a conclusion by the Squeeze Theorem.)







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 16 '18 at 19:15

























                            answered Dec 16 '18 at 19:03









                            KM101KM101

                            6,0901525




                            6,0901525






























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