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How can I prove by definition the continuity of $f(x,y)=\frac{1}{\sqrt{x^2+y^2}}$ where $(x,y)\ne (0,0)$. How can I set the value at $(0,0)$ to make it continuous?

I think that the continuity in $(0,0)$ is not possible, using trajectories, but I'm not able to see which is the $\delta$ in the definition given any $\varepsilon>0$.

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You are right that there is no value at the origin that will make $f$ continuous since $f(x,y) \to +\infty$ as $(x,y)\to (0,0)$. Let's write $\vec x = (x,y) \ne (0,0) = \vec 0$ and let $\vec h = (h,k)$. I'll use $\|\vec x\|$ as shorthand for $\sqrt{x^2 + y^2}$. To prove continuity of $f$ at the point $\vec x$, we want to bound the difference $$ \bigg|\frac{1}{\|\vec x + \vec h\|} - \frac{1}{\|\vec x\|}\bigg| = \frac{\big| \|\vec x + \vec h\| - \|\vec x\|\big|}{\|\vec x + \vec h\|\cdot\|\vec x\|}. $$ Use the reverse triangle inequality to bound the numerator. If $\|\vec h\|$ is sufficiently small, say $\|\vec h\| < \|\vec x\|\big/2$, then $\|\vec x + \vec h\| > \|\vec x\|\big/2$. Try taking it from there.

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