Can you use LHopitals rule when solving for a limit that is

Solution

yes f(x,y) = [ysin(x)] / [x^2 + y^2 ] convert to polar f(r,?) = [r sin(?)sin(r cos(?))] / [r^2 ] f(r,?) = [sin(?)sin(r cos(?))] / r To test limit approaching (x,y)?(0,0) in rectangular, we need only test r?0, for all ?, in polar. Notice that as r? 0, we get a 0/0 indeterminate even in polar. apply lollipops rule with respect to r f(r,?) = sin(?)cos(r cos(?)) cos(?) As r ? 0 f(r,?) = sin(?)cos(?) f(r,?) = ½sin(2?) Limit is divergent because it doesnt converge to the same value for all ?