How do you find the limit as (x,y) approaches (0,0) of #(x+y^2) / (2x+y)#?

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Answer 1 · 2015-06-11T23:10:09

#lim_((x,y)->(0,0)) (x+ y^2)/(2x + y)#

#ln [lim_((x,y)->(0,0)) (x+ y^2)/(2x + y)]#

#lim_((x,y)->(0,0)) ln[(x+ y^2)/(2x + y)]#

#lim_((x,y)->(0,0)) ln(x+ y^2) - ln(2x + y)#

#lim_((x,y)->(0,0)) ln(x+ y^2) - lim_((x,y)->(0,0)) ln(2x + y)#

#= e^(lim_((x,y)->(0,0)) ln(x+ y^2) - lim_((x,y)->(0,0)) ln(2x + y))#

#= e^(oo - oo)#

You cannot do #oo - oo#.

There is no way you can rewrite this without it be undefined. This is the graph: