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Find Two Paths of Approach from Which One Can Conclude f(x,y,z)f ( x , y , z )

Question 89

Essay

Find two paths of approach from which one can conclude that the function has no limit as (x, y) approaches (0, 0).
-We say that a function f(x,y,z)f ( x , y , z ) approaches the limit LL as (x,y,z)( x , y , z ) approaches (x0,y0,z0)( x 0 , y 0 , z 0 ) and write
(x,y,z)(x0,y0,z0)f(x,y,z)=L( x , y , z ) \rightarrow \left( x _ { 0 } , y _ { 0 } , z _ { 0 } \right) \quad f ( x , y , z ) = L
if for every number ε>0\varepsilon > 0 , there exists a corresponding number δ>0\delta > 0 such that for all (x,y,z)( x , y , z ) in the domain of ff , 0<(xx0)2+(yy0)2+(zz0)2<δf(x,y,z)L<ε0 < \sqrt { \left( x - x _ { 0 } \right) ^ { 2 } + \left( y - y _ { 0 } \right) ^ { 2 } + \left( z - z _ { 0 } \right) ^ { 2 } } < \delta \Rightarrow | f ( x , y , z ) - L | < \varepsilon . Show that the δε\delta - \varepsilon requirement in this definition is equivalent to 0<xx0<δ,0<yy0<δ0 < \left| x - x _ { 0 } \right| < \delta , 0 < \left| y - y _ { 0 } \right| < \delta , and 0<zz0<δf(x,y,z)L<ε0 < \left| \mathrm { z } - \mathrm { z } _ { 0 } \right| < \delta \Rightarrow | \mathrm { f } ( \mathrm { x } , \mathrm { y } , \mathrm { z } ) - \mathrm { L } | < \varepsilon .

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