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enter image description here Can anyone calculate the value of K.

It seems 0 is the value as I have seen it in many questions but not sure.

If anyone can arrive at this value, then please.

Jean Marie
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shadow kh
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  • is this$$\frac{xy}{(x^2+y^2)\frac{5}{2}}(1-\cos(x^2+y^2))$$ ? – Dr. Sonnhard Graubner Jan 21 '17 at 19:53
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    Yes Dr. Sonnhard Graubner 5/2 is exponent of the denominator – shadow kh Jan 21 '17 at 19:56
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    I was thinking if it was indeed exponent why would they put × sign when there is no term before it in exponent . but if we consider 5/2 not as an exponent then the term in the bracket would be the term before it. Hence it makes sense. And accordingly when both x and y tends to zero the cosine (0)will take 1 and hence the limit is 0 so K will be 0 . am I right ? – shadow kh Jan 21 '17 at 20:02
  • the limit is indeed zero – Dr. Sonnhard Graubner Jan 21 '17 at 20:11

2 Answers2

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For $x=r\cos \theta,\;y=r\sin \theta$: $$\frac{xy}{\left(x^2+y^2\right)^{5/2}}\left(1-\cos (x^2+y^2)\right)=\frac{r^2\cos \theta\sin\theta}{\left(r^2\right)^{5/2}}\left(1-\cos r^2\right)$$ $$\sim \frac{r^2\cos \theta\sin\theta}{\left(r^2\right)^{5/2}}\cdot \frac{(r^2)^2}{2}=\frac{1}{2}r\cos \theta\sin\theta,$$ and $$\left|\frac{1}{2}r\cos \theta\sin\theta\right|\le\frac{1}{2}r\to 0\text{ as }r\to 0.$$ So, $f$ is continuous at $(0,0)$ iff $$K=\lim_{(x,y)\to (0,0)}f(x,y)=0$$

Fernando Revilla
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Assuming that $5/2$ is an exponent, we can rewrite the expression as $$ \frac{xy}{\sqrt{x^2+y^2}}\cdot \frac{1-\cos(x^2+y^2)}{(x^2+y^2)^2} $$ Using L'Hopital's Rule or Taylor series, we get $$ \lim_{u\rightarrow 0} \frac{1-\cos(u)}{u^2} = \frac{1}{2} $$ For the other factor, we have $$ \frac{xy}{\sqrt{x^2+y^2}} = \frac{\pm 1}{\sqrt{\frac{1}{x^2}+\frac{1}{y^2}}} $$ which obviously approaches $0$ when either $x$ or $y$ approaches $0$. Therefore, $K=0$.

Reinhard Meier
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