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Given the function $f(x)$, $$ f(x,y,z,w) = \frac{x+iy}{\sqrt{|w+z|}} \text{.} $$ How do I calculate the limit $$ \lim\limits_{w\rightarrow -z} f $$ under the constraint that the points $(x,y,z) \in \mathbb{R}^3$ lie on a sphere with radius $w$: $$ x^2+y^2+z^2=w^2 \text{.} $$ Note that the constraint in the limit $w \rightarrow - z$ implies $x \rightarrow 0$ and $y \rightarrow 0$. Numerically, I obtained the following conjecture: $$ \sqrt{z} (1+i) $$ How do I prove this analytically? |
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I post this as an answer because it would be a bit long for a comment. But I am still not sure to really understand the question. I don't think you can expect a limit in general. If we choose $y=0$ and $x=\sqrt{w^2-z^2}$ then the conditions you set are satisfied and (if it makes sense) $$\frac{x+iy}{\sqrt{|w+z|}}=\frac{\sqrt{w^2-z^2}\sqrt{|w-z|}}{\sqrt{|w+z|}\sqrt{|w-z|}}=\sqrt{|w-z|}\to\sqrt{2|z|}\,.$$ If we choose $x=0$ and $y=\sqrt{w^2-z^2}$ then the conditions you set are satisfied and (if it makes sense) $$\frac{x+iy}{\sqrt{|w+z|}}=\frac{i\sqrt{w^2-z^2}\sqrt{|w-z|}}{\sqrt{|w+z|}\sqrt{|w-z|}}=i\sqrt{|w-z|}\to i\sqrt{2|z|}\,.$$ |
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