Limit by polar coordinates: $frac{2x(1-cos(x-y))}{x^2+y^2}$
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
asked Dec 28 '18 at 9:00
Seven
989
add a comment |
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
asked Dec 28 '18 at 9:00
Seven
989
add a comment |
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
asked Dec 28 '18 at 9:00
Seven
989
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
asked Dec 28 '18 at 9:00
Seven
989
asked Dec 28 '18 at 9:00
Seven
989
asked Dec 28 '18 at 9:00
Seven
989
asked Dec 28 '18 at 9:00
Seven
989
asked Dec 28 '18 at 9:00
Seven
989
989
add a comment |
add a comment |
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 $theta$.
answered Dec 28 '18 at 9:16
Kavi Rama Murthy
51.2k31855
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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
$$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0.$$
answered Dec 28 '18 at 16:09
user376343
2,8932823
add a comment |
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 $left[0,frac{pi}2right]$.
answered Dec 28 '18 at 13:59
mechanodroid
26.9k62447
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
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active
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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 $theta$.
answered Dec 28 '18 at 9:16
Kavi Rama Murthy
51.2k31855
add a comment |
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 $theta$.
answered Dec 28 '18 at 9:16
Kavi Rama Murthy
51.2k31855
add a comment |
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 $theta$.
answered Dec 28 '18 at 9:16
Kavi Rama Murthy
51.2k31855
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 $theta$.
answered Dec 28 '18 at 9:16
Kavi Rama Murthy
51.2k31855
answered Dec 28 '18 at 9:16
Kavi Rama Murthy
51.2k31855
answered Dec 28 '18 at 9:16
Kavi Rama Murthy
51.2k31855
answered Dec 28 '18 at 9:16
Kavi Rama Murthy
51.2k31855
51.2k31855
add a comment |
add a comment |
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
$$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0.$$
answered Dec 28 '18 at 16:09
user376343
2,8932823
add a comment |
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
$$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0.$$
answered Dec 28 '18 at 16:09
user376343
2,8932823
add a comment |
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
$$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0.$$
answered Dec 28 '18 at 16:09
user376343
2,8932823
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
$$lim_{(x,y)to (0,0)}frac{2x(1-cos(x-y))}{x^2+y^2} = 0.$$
answered Dec 28 '18 at 16:09
user376343
2,8932823
answered Dec 28 '18 at 16:09
user376343
2,8932823
answered Dec 28 '18 at 16:09
user376343
2,8932823
answered Dec 28 '18 at 16:09
user376343
2,8932823
2,8932823
add a comment |
add a comment |
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 $left[0,frac{pi}2right]$.
answered Dec 28 '18 at 13:59
mechanodroid
26.9k62447
add a comment |
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 $left[0,frac{pi}2right]$.
answered Dec 28 '18 at 13:59
mechanodroid
26.9k62447
add a comment |
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 $left[0,frac{pi}2right]$.
answered Dec 28 '18 at 13:59
mechanodroid
26.9k62447
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 $left[0,frac{pi}2right]$.
answered Dec 28 '18 at 13:59
mechanodroid
26.9k62447
answered Dec 28 '18 at 13:59
mechanodroid
26.9k62447
answered Dec 28 '18 at 13:59
mechanodroid
26.9k62447
answered Dec 28 '18 at 13:59
mechanodroid
26.9k62447
26.9k62447
add a comment |
add a comment |
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