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: $$|f(\rho\cos\theta,\rho\sin\theta)| = \frac{2\cos\theta(1-\cos(\rho(\cos\theta-\sin\theta))}{\rho} \leq \frac{2(1-\cos(\sqrt{2}\rho))}{\rho}$$ 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?
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$\begingroup$ If you use L'Hospital's Rule on that last expression you'll be done. $\endgroup$– Thomas AhleMar 11, 2019 at 15:25
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3 Answers
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The inequality you have written is wrong. Cosine is not an incerasing function so $x\leq y$ 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, 2018 at 9:16
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Use the half-angle formula $$1-\cos(x-y) = 2\sin^2\left(\frac{x-y}2\right) = 2\sin^2 \left(\frac{r(\cos\theta-\sin\theta)}{2}\right)$$
so
$$\left|\frac{2x(1-\cos(x-y))}{x^2+y^2}\right| = 4\left|\cos\theta\right| \frac1{r}\sin^2\left(\frac{r(\cos\theta-\sin\theta)}{2}\right) \le \frac4{r}\sin^2\left(\frac{r\sqrt2}{2}\right) \xrightarrow{r\to 0} 0$$
because $\sin^2$ is increasing on $\left[0,\frac{\pi}2\right]$.
answered Dec 28, 2018 at 13:59
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$\begingroup$ You probably also need that $\sin^2(r) \approx r^2$, since you have that factor $1/r$ to take care of. $\endgroup$ Mar 11, 2019 at 15:24
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1$\begingroup$ @ThomasAhle It follows from $\sin x \le x, \forall x \ge 0$ like this: $$\frac4r \sin^2\left(\frac{r\sqrt{2}}2\right) \le \frac4r \left(\frac{r\sqrt{2}}2\right)^2 = 2r \xrightarrow{r\to 0} 0$$ $\endgroup$ Mar 11, 2019 at 15:54
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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 $$\begin{aligned}\frac{2x(1-\cos(x-y))}{x^2+y^2}&= 2x\cdot \frac{1-\cos x}{x^2} \cdot\frac{x^2}{x^2+y^2}\\ &+ 2x\cos x \cdot \frac{1-\cos y}{y^2}\cdot\frac{y^2}{x^2+y^2}\\&- 2y\cdot \frac{\sin x}{x}\cdot\frac{\sin y}{y}\cdot\frac{x^2}{x^2+y^2}\end{aligned}$$ From where is $$\lim_{(x,y)\to (0,0)}\frac{2x(1-\cos(x-y))}{x^2+y^2} = 0.$$
answered Dec 28, 2018 at 16:09