Question

\[\lim_{(x,y) \to (0,0)} {{x} \over {\sqrt {x^2 + y^2}}}\]

Answer 1

Write this in polar coordinates

Answer 2

\[\lim_{(x,y) \to (0,0)} {{r \cos \theta} \over {\sqrt {r \cos^2 \theta + r \sin^2 \theta}}}\]

Answer 3

\[\lim_{r \to 0+} {{r \cos \theta} \over 1}\]

Answer 4

careful, the denominator is r.

Answer 5

\[\lim_{r \to 0+} {{r \cos \theta} \over {\sqrt {r^2 \cos^2 \theta + r^2 \sin^2 \theta}}} = \lim_{r \to 0+} {{r \cos \theta} \over {r}}\]

Answer 6

the r cancels, so the the limit 1

Answer 7

No. It depends on theta.

Answer 8

oh, and since theta can be anything, the limit doesn't exist?

Answer 9

yes