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in the polar coordinates, $\ x = r \cos(\theta) , y = r \sin(\theta) $ ,

$$f(r, \theta) = \frac{\cos(r \cos(\theta)) - 1 + \frac{(r \cos(\theta))^2}{2}}{(r \cos(\theta))^2 + (r \sin(\theta))^2}$$

$$\lim_{r \to 0+} f(r, \theta) = \lim_{r \to 0+} \frac{\cos(r \cos(\theta)) - 1 + \frac{(r \cos(\theta))^2}{2}}{r^2}=\lim_{r \to 0+} \left( \frac{\cos(r \cos(\theta)) - 1}{r^2} + \frac{\cos^2(\theta)}{2} \right).$$

$$\lim_{r \to 0+} \frac{\cos(r \cos(\theta)) - 1}{r^2} =\lim_{r \to 0+} -\frac{\cos^2(\theta)}{2}.$$ (L'Hôpital's rule)

therefore, $$\lim_{r \to 0+} f(r, \theta) =\lim_{r \to 0+} (-\frac{\cos^2(\theta)}{2}+\frac{\cos^2(\theta)}{2})=0.$$

It's my solutions... The answer to this question is DNE. I don't know how to solve this problem exactly. Please give me some feedback.

Riemann
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kradunes
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2 Answers2

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There's a mistake in this step: $$ \lim_{r \to 0+} \biggl( \frac{\cos(r \cos(\theta)) - 1}{r^2} + \frac{\cos^2(\theta)}{2} \biggr) = \lim_{r \to 0+} \frac{\cos(r \cos(\theta)) - 1}{r^2} + \lim_{r \to 0+} \frac{\cos^2(\theta)}{2}. $$ The arithmetic rules of limits say that $\lim (f(r)+g(r)) = \lim f(r) + \lim g(r)$ if both $\lim f(r)$ and $\lim g(r)$ exist. If we don't know that both new limits exist (and here they do not), the splitting step is invalid.

Nevertheless, the limit does actually exist and equals $0$ (the numerator $\cos x - 1 + \frac{x^2}2$ can be bounded by a constant times $x^4$, for instance). I don't know why the book says the limit does not exist.

(By the way, I would recommend moving away from writing "DNE" where a number should go. The written phrase "$\lim h(r)$ does not exist" is perfectly clear and better than "$\lim h(r)=DNE$", which reinforces bad usage of the equals sign as a universal math verb.)

Greg Martin
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Use the fact that: for $x\in\mathbb R$ (cosx), we have $$1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24},$$ so $$0\leq\cos(x)-1+\frac{x^2}{2}\leq\frac{x^4}{24},\quad\forall x\in\mathbb R.$$ Hence $$0\leq\frac{\cos x-1+x^2/2}{x^2+y^2}\leq\frac{x^2}{24}\to0,\quad (x,y)\to(0,0).$$

Riemann
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