Limit by polar coordinates: $frac{2x(1-cos(x-y))}{x^2+y^2}$












2














I need to prove that:
$$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0$$
I have tried using polar coordinates, but I'm not comfortable with my approach. What I did was saying:
$$|f(rhocostheta,rhosintheta)| = frac{2costheta(1-cos(rho(costheta-sintheta))}{rho} leq frac{2(1-cos(sqrt{2}rho))}{rho}$$
and this last expression converges to $0$ when $rho$ goes to $0$ uniformly in $theta$. Is this correct? How can I do it better? I'm also looking for a way to solve this limit without using polar coordinates ( the easier, the better) . Any ideas?






calculus limits







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    2














    I need to prove that:
    $$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0$$
    I have tried using polar coordinates, but I'm not comfortable with my approach. What I did was saying:
    $$|f(rhocostheta,rhosintheta)| = frac{2costheta(1-cos(rho(costheta-sintheta))}{rho} leq frac{2(1-cos(sqrt{2}rho))}{rho}$$
    and this last expression converges to $0$ when $rho$ goes to $0$ uniformly in $theta$. Is this correct? How can I do it better? I'm also looking for a way to solve this limit without using polar coordinates ( the easier, the better) . Any ideas?






    calculus limits







    share|cite|improve this question

























      2












      2








      2







      I need to prove that:
      $$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0$$
      I have tried using polar coordinates, but I'm not comfortable with my approach. What I did was saying:
      $$|f(rhocostheta,rhosintheta)| = frac{2costheta(1-cos(rho(costheta-sintheta))}{rho} leq frac{2(1-cos(sqrt{2}rho))}{rho}$$
      and this last expression converges to $0$ when $rho$ goes to $0$ uniformly in $theta$. Is this correct? How can I do it better? I'm also looking for a way to solve this limit without using polar coordinates ( the easier, the better) . Any ideas?






      calculus limits







      share|cite|improve this question













      I need to prove that:
      $$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0$$
      I have tried using polar coordinates, but I'm not comfortable with my approach. What I did was saying:
      $$|f(rhocostheta,rhosintheta)| = frac{2costheta(1-cos(rho(costheta-sintheta))}{rho} leq frac{2(1-cos(sqrt{2}rho))}{rho}$$
      and this last expression converges to $0$ when $rho$ goes to $0$ uniformly in $theta$. Is this correct? How can I do it better? I'm also looking for a way to solve this limit without using polar coordinates ( the easier, the better) . Any ideas?






      calculus limits




      calculus limits






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      asked Dec 28 '18 at 9:00









      Seven

      989




      989






















          3 Answers
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          The inequality you have written is wrong. Cosine is not an incerasing function so $xleq y$ does not imply $cos, x leq cos , y$. Instead, use the inequality $1-cos theta leq frac {theta^{2}} 2$ valid for all real $theta$. Proof of this inequality: $1-cos theta -frac {theta^{2}} 2$ vanishes at $0$ and its derivative is $sin, theta - theta$ which is negative for all $theta >0$. Hence the inequality holds for all positive $theta$. Since both sides are even functions the inequality holds for all $theta$.






          share|cite|improve this answer





























            2














            Solution without the use of polar coordinates, as asked by OP at the end of their post.



            Rewrite $$begin{aligned}1-cos(x-y)&=1-cos x cos y - sin x sin y \ &=1-cos x + cos x (1-cos y)- sin x sin y, end{aligned}$$
            then $$begin{aligned}frac{2x(1-cos(x-y))}{x^2+y^2}&= 2xcdot frac{1-cos x}{x^2} cdotfrac{x^2}{x^2+y^2}\ &+ 2xcos x cdot frac{1-cos y}{y^2}cdotfrac{y^2}{x^2+y^2}\&- 2ycdot frac{sin x}{x}cdotfrac{sin y}{y}cdotfrac{x^2}{x^2+y^2}end{aligned}$$
            From where is
            $$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0.$$






            share|cite|improve this answer





























              1














              Use the half-angle formula $$1-cos(x-y) = 2sin^2left(frac{x-y}2right) = 2sin^2 left(frac{r(costheta-sintheta)}{2}right)$$



              so



              $$left|frac{2x(1-cos(x-y))}{x^2+y^2}right| = 4left|costhetaright| frac1{r}sin^2left(frac{r(costheta-sintheta)}{2}right) le frac4{r}sin^2left(frac{rsqrt2}{2}right) xrightarrow{rto 0} 0$$



              because $sin^2$ is increasing on $left[0,frac{pi}2right]$.






              share|cite|improve this answer





















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                3 Answers
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                3 Answers
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                active

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                active

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                active

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                2














                The inequality you have written is wrong. Cosine is not an incerasing function so $xleq y$ does not imply $cos, x leq cos , y$. Instead, use the inequality $1-cos theta leq frac {theta^{2}} 2$ valid for all real $theta$. Proof of this inequality: $1-cos theta -frac {theta^{2}} 2$ vanishes at $0$ and its derivative is $sin, theta - theta$ which is negative for all $theta >0$. Hence the inequality holds for all positive $theta$. Since both sides are even functions the inequality holds for all $theta$.






                share|cite|improve this answer


























                  2














                  The inequality you have written is wrong. Cosine is not an incerasing function so $xleq y$ does not imply $cos, x leq cos , y$. Instead, use the inequality $1-cos theta leq frac {theta^{2}} 2$ valid for all real $theta$. Proof of this inequality: $1-cos theta -frac {theta^{2}} 2$ vanishes at $0$ and its derivative is $sin, theta - theta$ which is negative for all $theta >0$. Hence the inequality holds for all positive $theta$. Since both sides are even functions the inequality holds for all $theta$.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    The inequality you have written is wrong. Cosine is not an incerasing function so $xleq y$ does not imply $cos, x leq cos , y$. Instead, use the inequality $1-cos theta leq frac {theta^{2}} 2$ valid for all real $theta$. Proof of this inequality: $1-cos theta -frac {theta^{2}} 2$ vanishes at $0$ and its derivative is $sin, theta - theta$ which is negative for all $theta >0$. Hence the inequality holds for all positive $theta$. Since both sides are even functions the inequality holds for all $theta$.






                    share|cite|improve this answer












                    The inequality you have written is wrong. Cosine is not an incerasing function so $xleq y$ does not imply $cos, x leq cos , y$. Instead, use the inequality $1-cos theta leq frac {theta^{2}} 2$ valid for all real $theta$. Proof of this inequality: $1-cos theta -frac {theta^{2}} 2$ vanishes at $0$ and its derivative is $sin, theta - theta$ which is negative for all $theta >0$. Hence the inequality holds for all positive $theta$. Since both sides are even functions the inequality holds for all $theta$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 28 '18 at 9:16









                    Kavi Rama Murthy

                    51.2k31855




                    51.2k31855























                        2














                        Solution without the use of polar coordinates, as asked by OP at the end of their post.



                        Rewrite $$begin{aligned}1-cos(x-y)&=1-cos x cos y - sin x sin y \ &=1-cos x + cos x (1-cos y)- sin x sin y, end{aligned}$$
                        then $$begin{aligned}frac{2x(1-cos(x-y))}{x^2+y^2}&= 2xcdot frac{1-cos x}{x^2} cdotfrac{x^2}{x^2+y^2}\ &+ 2xcos x cdot frac{1-cos y}{y^2}cdotfrac{y^2}{x^2+y^2}\&- 2ycdot frac{sin x}{x}cdotfrac{sin y}{y}cdotfrac{x^2}{x^2+y^2}end{aligned}$$
                        From where is
                        $$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0.$$






                        share|cite|improve this answer


























                          2














                          Solution without the use of polar coordinates, as asked by OP at the end of their post.



                          Rewrite $$begin{aligned}1-cos(x-y)&=1-cos x cos y - sin x sin y \ &=1-cos x + cos x (1-cos y)- sin x sin y, end{aligned}$$
                          then $$begin{aligned}frac{2x(1-cos(x-y))}{x^2+y^2}&= 2xcdot frac{1-cos x}{x^2} cdotfrac{x^2}{x^2+y^2}\ &+ 2xcos x cdot frac{1-cos y}{y^2}cdotfrac{y^2}{x^2+y^2}\&- 2ycdot frac{sin x}{x}cdotfrac{sin y}{y}cdotfrac{x^2}{x^2+y^2}end{aligned}$$
                          From where is
                          $$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0.$$






                          share|cite|improve this answer
























                            2












                            2








                            2






                            Solution without the use of polar coordinates, as asked by OP at the end of their post.



                            Rewrite $$begin{aligned}1-cos(x-y)&=1-cos x cos y - sin x sin y \ &=1-cos x + cos x (1-cos y)- sin x sin y, end{aligned}$$
                            then $$begin{aligned}frac{2x(1-cos(x-y))}{x^2+y^2}&= 2xcdot frac{1-cos x}{x^2} cdotfrac{x^2}{x^2+y^2}\ &+ 2xcos x cdot frac{1-cos y}{y^2}cdotfrac{y^2}{x^2+y^2}\&- 2ycdot frac{sin x}{x}cdotfrac{sin y}{y}cdotfrac{x^2}{x^2+y^2}end{aligned}$$
                            From where is
                            $$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0.$$






                            share|cite|improve this answer












                            Solution without the use of polar coordinates, as asked by OP at the end of their post.



                            Rewrite $$begin{aligned}1-cos(x-y)&=1-cos x cos y - sin x sin y \ &=1-cos x + cos x (1-cos y)- sin x sin y, end{aligned}$$
                            then $$begin{aligned}frac{2x(1-cos(x-y))}{x^2+y^2}&= 2xcdot frac{1-cos x}{x^2} cdotfrac{x^2}{x^2+y^2}\ &+ 2xcos x cdot frac{1-cos y}{y^2}cdotfrac{y^2}{x^2+y^2}\&- 2ycdot frac{sin x}{x}cdotfrac{sin y}{y}cdotfrac{x^2}{x^2+y^2}end{aligned}$$
                            From where is
                            $$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 28 '18 at 16:09









                            user376343

                            2,8932823




                            2,8932823























                                1














                                Use the half-angle formula $$1-cos(x-y) = 2sin^2left(frac{x-y}2right) = 2sin^2 left(frac{r(costheta-sintheta)}{2}right)$$



                                so



                                $$left|frac{2x(1-cos(x-y))}{x^2+y^2}right| = 4left|costhetaright| frac1{r}sin^2left(frac{r(costheta-sintheta)}{2}right) le frac4{r}sin^2left(frac{rsqrt2}{2}right) xrightarrow{rto 0} 0$$



                                because $sin^2$ is increasing on $left[0,frac{pi}2right]$.






                                share|cite|improve this answer


























                                  1














                                  Use the half-angle formula $$1-cos(x-y) = 2sin^2left(frac{x-y}2right) = 2sin^2 left(frac{r(costheta-sintheta)}{2}right)$$



                                  so



                                  $$left|frac{2x(1-cos(x-y))}{x^2+y^2}right| = 4left|costhetaright| frac1{r}sin^2left(frac{r(costheta-sintheta)}{2}right) le frac4{r}sin^2left(frac{rsqrt2}{2}right) xrightarrow{rto 0} 0$$



                                  because $sin^2$ is increasing on $left[0,frac{pi}2right]$.






                                  share|cite|improve this answer
























                                    1












                                    1








                                    1






                                    Use the half-angle formula $$1-cos(x-y) = 2sin^2left(frac{x-y}2right) = 2sin^2 left(frac{r(costheta-sintheta)}{2}right)$$



                                    so



                                    $$left|frac{2x(1-cos(x-y))}{x^2+y^2}right| = 4left|costhetaright| frac1{r}sin^2left(frac{r(costheta-sintheta)}{2}right) le frac4{r}sin^2left(frac{rsqrt2}{2}right) xrightarrow{rto 0} 0$$



                                    because $sin^2$ is increasing on $left[0,frac{pi}2right]$.






                                    share|cite|improve this answer












                                    Use the half-angle formula $$1-cos(x-y) = 2sin^2left(frac{x-y}2right) = 2sin^2 left(frac{r(costheta-sintheta)}{2}right)$$



                                    so



                                    $$left|frac{2x(1-cos(x-y))}{x^2+y^2}right| = 4left|costhetaright| frac1{r}sin^2left(frac{r(costheta-sintheta)}{2}right) le frac4{r}sin^2left(frac{rsqrt2}{2}right) xrightarrow{rto 0} 0$$



                                    because $sin^2$ is increasing on $left[0,frac{pi}2right]$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 28 '18 at 13:59









                                    mechanodroid

                                    26.9k62447




                                    26.9k62447






























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