Limit with arctan: $lim_{xrightarrow 0} frac{xsin 3x}{arctan x^2}$












1












$begingroup$


$$lim_{xrightarrow 0} frac{xsin 3x}{arctan x^2}$$



NB! I haven't learnt about L'Hôpital's rule yet, so I'm still solving limits using common limits.



What I've done so far



$$lim_{xrightarrow0}left[frac{x}{arctan x^2}cdot 3xcdot frac{sin 3x}{3x}right] = 0cdot1cdotlim_{xrightarrow0}left[frac{x}{arctan x^2}right] = 0??$$
Obviously I'm wrong, but I thought I'd show what I tried.






calculus limits







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$endgroup$












  • $begingroup$
    Do you have this inequality: $sin x leq x leq tan x$ for $x geq 0$? And the limits that follow from it: $sin x / x rightarrow 0$, etc.?
    $endgroup$
    – Simon S
    Nov 3 '14 at 23:46


















1












$begingroup$


$$lim_{xrightarrow 0} frac{xsin 3x}{arctan x^2}$$



NB! I haven't learnt about L'Hôpital's rule yet, so I'm still solving limits using common limits.



What I've done so far



$$lim_{xrightarrow0}left[frac{x}{arctan x^2}cdot 3xcdot frac{sin 3x}{3x}right] = 0cdot1cdotlim_{xrightarrow0}left[frac{x}{arctan x^2}right] = 0??$$
Obviously I'm wrong, but I thought I'd show what I tried.






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you have this inequality: $sin x leq x leq tan x$ for $x geq 0$? And the limits that follow from it: $sin x / x rightarrow 0$, etc.?
    $endgroup$
    – Simon S
    Nov 3 '14 at 23:46
















1












1








1





$begingroup$


$$lim_{xrightarrow 0} frac{xsin 3x}{arctan x^2}$$



NB! I haven't learnt about L'Hôpital's rule yet, so I'm still solving limits using common limits.



What I've done so far



$$lim_{xrightarrow0}left[frac{x}{arctan x^2}cdot 3xcdot frac{sin 3x}{3x}right] = 0cdot1cdotlim_{xrightarrow0}left[frac{x}{arctan x^2}right] = 0??$$
Obviously I'm wrong, but I thought I'd show what I tried.






calculus limits







share|cite|improve this question











$endgroup$




$$lim_{xrightarrow 0} frac{xsin 3x}{arctan x^2}$$



NB! I haven't learnt about L'Hôpital's rule yet, so I'm still solving limits using common limits.



What I've done so far



$$lim_{xrightarrow0}left[frac{x}{arctan x^2}cdot 3xcdot frac{sin 3x}{3x}right] = 0cdot1cdotlim_{xrightarrow0}left[frac{x}{arctan x^2}right] = 0??$$
Obviously I'm wrong, but I thought I'd show what I tried.






calculus limits




calculus limits






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share|cite|improve this question













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edited Dec 21 '18 at 17:10









Larry

2,39131129




2,39131129










asked Nov 3 '14 at 23:36









B. LeeB. Lee

895915




895915












  • $begingroup$
    Do you have this inequality: $sin x leq x leq tan x$ for $x geq 0$? And the limits that follow from it: $sin x / x rightarrow 0$, etc.?
    $endgroup$
    – Simon S
    Nov 3 '14 at 23:46




















  • $begingroup$
    Do you have this inequality: $sin x leq x leq tan x$ for $x geq 0$? And the limits that follow from it: $sin x / x rightarrow 0$, etc.?
    $endgroup$
    – Simon S
    Nov 3 '14 at 23:46


















$begingroup$
Do you have this inequality: $sin x leq x leq tan x$ for $x geq 0$? And the limits that follow from it: $sin x / x rightarrow 0$, etc.?
$endgroup$
– Simon S
Nov 3 '14 at 23:46






$begingroup$
Do you have this inequality: $sin x leq x leq tan x$ for $x geq 0$? And the limits that follow from it: $sin x / x rightarrow 0$, etc.?
$endgroup$
– Simon S
Nov 3 '14 at 23:46












4 Answers
4






active

oldest

votes


















2












$begingroup$

You should put in
$$
frac{x^2}{arctan x^2}
$$
that has limit $1$:
$$
lim_{xto0}frac{xsin3x}{arctan x^2}=
lim_{xto0}3frac{sin3x}{3x}frac{x^2}{arctan x^2}=dots
$$
If you don't know the limit above, just substitute $t=arctan x^2$, so $x^2=tan t$ and the limit is
$$
lim_{xto0}frac{x^2}{arctan x^2}=lim_{tto0}frac{tan t}{t}
$$
that you should be able to manage.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Starting as you did, we want to find the limit of
    $$frac{3x^2}{arctan(x^2)}cdotfrac{sin(3x)}{3x}.$$
    Only the first term gives any trouble. Let $x$ be not too large, and let $x^2=tan w$, You want to find
    $$lim_{wto 0} frac{3tan w}{w},$$
    which is not difficult. Replace $tan w$ by $frac{sin w}{cos w}$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Since $u sim tan u$, then $y sim arctan y$, so $x^2 sim arctan (x^2)$.



      This means that
      $$lim_{x to 0} frac{x sin 3x}{arctan x^2} = lim_{x to 0} frac{x3x}{x^2} = 3$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        This feels unsatisfactory, as you are still using the derivatives of the functions in the numerator and denominator.
        $endgroup$
        – Simon S
        Nov 3 '14 at 23:42



















      0












      $begingroup$

      Using the Asymptotic expansion:

      $sin(x)approx_0 x$

      $arctan(x^2) approx_0 x^2$

      So
      $$lim _{xrightarrow :0}:frac{xsin :3x}{arctan :x^2}=lim :_{xrightarrow 0}:frac{3x^2}{x^2}=color{red}{3}$$






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        You should put in
        $$
        frac{x^2}{arctan x^2}
        $$
        that has limit $1$:
        $$
        lim_{xto0}frac{xsin3x}{arctan x^2}=
        lim_{xto0}3frac{sin3x}{3x}frac{x^2}{arctan x^2}=dots
        $$
        If you don't know the limit above, just substitute $t=arctan x^2$, so $x^2=tan t$ and the limit is
        $$
        lim_{xto0}frac{x^2}{arctan x^2}=lim_{tto0}frac{tan t}{t}
        $$
        that you should be able to manage.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          You should put in
          $$
          frac{x^2}{arctan x^2}
          $$
          that has limit $1$:
          $$
          lim_{xto0}frac{xsin3x}{arctan x^2}=
          lim_{xto0}3frac{sin3x}{3x}frac{x^2}{arctan x^2}=dots
          $$
          If you don't know the limit above, just substitute $t=arctan x^2$, so $x^2=tan t$ and the limit is
          $$
          lim_{xto0}frac{x^2}{arctan x^2}=lim_{tto0}frac{tan t}{t}
          $$
          that you should be able to manage.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            You should put in
            $$
            frac{x^2}{arctan x^2}
            $$
            that has limit $1$:
            $$
            lim_{xto0}frac{xsin3x}{arctan x^2}=
            lim_{xto0}3frac{sin3x}{3x}frac{x^2}{arctan x^2}=dots
            $$
            If you don't know the limit above, just substitute $t=arctan x^2$, so $x^2=tan t$ and the limit is
            $$
            lim_{xto0}frac{x^2}{arctan x^2}=lim_{tto0}frac{tan t}{t}
            $$
            that you should be able to manage.






            share|cite|improve this answer









            $endgroup$



            You should put in
            $$
            frac{x^2}{arctan x^2}
            $$
            that has limit $1$:
            $$
            lim_{xto0}frac{xsin3x}{arctan x^2}=
            lim_{xto0}3frac{sin3x}{3x}frac{x^2}{arctan x^2}=dots
            $$
            If you don't know the limit above, just substitute $t=arctan x^2$, so $x^2=tan t$ and the limit is
            $$
            lim_{xto0}frac{x^2}{arctan x^2}=lim_{tto0}frac{tan t}{t}
            $$
            that you should be able to manage.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 3 '14 at 23:47









            egregegreg

            181k1485203




            181k1485203























                4












                $begingroup$

                Starting as you did, we want to find the limit of
                $$frac{3x^2}{arctan(x^2)}cdotfrac{sin(3x)}{3x}.$$
                Only the first term gives any trouble. Let $x$ be not too large, and let $x^2=tan w$, You want to find
                $$lim_{wto 0} frac{3tan w}{w},$$
                which is not difficult. Replace $tan w$ by $frac{sin w}{cos w}$.






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  Starting as you did, we want to find the limit of
                  $$frac{3x^2}{arctan(x^2)}cdotfrac{sin(3x)}{3x}.$$
                  Only the first term gives any trouble. Let $x$ be not too large, and let $x^2=tan w$, You want to find
                  $$lim_{wto 0} frac{3tan w}{w},$$
                  which is not difficult. Replace $tan w$ by $frac{sin w}{cos w}$.






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    Starting as you did, we want to find the limit of
                    $$frac{3x^2}{arctan(x^2)}cdotfrac{sin(3x)}{3x}.$$
                    Only the first term gives any trouble. Let $x$ be not too large, and let $x^2=tan w$, You want to find
                    $$lim_{wto 0} frac{3tan w}{w},$$
                    which is not difficult. Replace $tan w$ by $frac{sin w}{cos w}$.






                    share|cite|improve this answer









                    $endgroup$



                    Starting as you did, we want to find the limit of
                    $$frac{3x^2}{arctan(x^2)}cdotfrac{sin(3x)}{3x}.$$
                    Only the first term gives any trouble. Let $x$ be not too large, and let $x^2=tan w$, You want to find
                    $$lim_{wto 0} frac{3tan w}{w},$$
                    which is not difficult. Replace $tan w$ by $frac{sin w}{cos w}$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 3 '14 at 23:50









                    André NicolasAndré Nicolas

                    452k36425810




                    452k36425810























                        1












                        $begingroup$

                        Since $u sim tan u$, then $y sim arctan y$, so $x^2 sim arctan (x^2)$.



                        This means that
                        $$lim_{x to 0} frac{x sin 3x}{arctan x^2} = lim_{x to 0} frac{x3x}{x^2} = 3$$






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          This feels unsatisfactory, as you are still using the derivatives of the functions in the numerator and denominator.
                          $endgroup$
                          – Simon S
                          Nov 3 '14 at 23:42
















                        1












                        $begingroup$

                        Since $u sim tan u$, then $y sim arctan y$, so $x^2 sim arctan (x^2)$.



                        This means that
                        $$lim_{x to 0} frac{x sin 3x}{arctan x^2} = lim_{x to 0} frac{x3x}{x^2} = 3$$






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          This feels unsatisfactory, as you are still using the derivatives of the functions in the numerator and denominator.
                          $endgroup$
                          – Simon S
                          Nov 3 '14 at 23:42














                        1












                        1








                        1





                        $begingroup$

                        Since $u sim tan u$, then $y sim arctan y$, so $x^2 sim arctan (x^2)$.



                        This means that
                        $$lim_{x to 0} frac{x sin 3x}{arctan x^2} = lim_{x to 0} frac{x3x}{x^2} = 3$$






                        share|cite|improve this answer









                        $endgroup$



                        Since $u sim tan u$, then $y sim arctan y$, so $x^2 sim arctan (x^2)$.



                        This means that
                        $$lim_{x to 0} frac{x sin 3x}{arctan x^2} = lim_{x to 0} frac{x3x}{x^2} = 3$$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 3 '14 at 23:41









                        CrostulCrostul

                        27.9k22352




                        27.9k22352












                        • $begingroup$
                          This feels unsatisfactory, as you are still using the derivatives of the functions in the numerator and denominator.
                          $endgroup$
                          – Simon S
                          Nov 3 '14 at 23:42


















                        • $begingroup$
                          This feels unsatisfactory, as you are still using the derivatives of the functions in the numerator and denominator.
                          $endgroup$
                          – Simon S
                          Nov 3 '14 at 23:42
















                        $begingroup$
                        This feels unsatisfactory, as you are still using the derivatives of the functions in the numerator and denominator.
                        $endgroup$
                        – Simon S
                        Nov 3 '14 at 23:42




                        $begingroup$
                        This feels unsatisfactory, as you are still using the derivatives of the functions in the numerator and denominator.
                        $endgroup$
                        – Simon S
                        Nov 3 '14 at 23:42











                        0












                        $begingroup$

                        Using the Asymptotic expansion:

                        $sin(x)approx_0 x$

                        $arctan(x^2) approx_0 x^2$

                        So
                        $$lim _{xrightarrow :0}:frac{xsin :3x}{arctan :x^2}=lim :_{xrightarrow 0}:frac{3x^2}{x^2}=color{red}{3}$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Using the Asymptotic expansion:

                          $sin(x)approx_0 x$

                          $arctan(x^2) approx_0 x^2$

                          So
                          $$lim _{xrightarrow :0}:frac{xsin :3x}{arctan :x^2}=lim :_{xrightarrow 0}:frac{3x^2}{x^2}=color{red}{3}$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Using the Asymptotic expansion:

                            $sin(x)approx_0 x$

                            $arctan(x^2) approx_0 x^2$

                            So
                            $$lim _{xrightarrow :0}:frac{xsin :3x}{arctan :x^2}=lim :_{xrightarrow 0}:frac{3x^2}{x^2}=color{red}{3}$$






                            share|cite|improve this answer









                            $endgroup$



                            Using the Asymptotic expansion:

                            $sin(x)approx_0 x$

                            $arctan(x^2) approx_0 x^2$

                            So
                            $$lim _{xrightarrow :0}:frac{xsin :3x}{arctan :x^2}=lim :_{xrightarrow 0}:frac{3x^2}{x^2}=color{red}{3}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Oct 27 '16 at 20:49









                            AmarildoAmarildo

                            1,769816




                            1,769816






























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