Chapter 3: Functions of Several Variables
Section 3.2: Limits and Continuity
Example 3.2.12
Show that for f=x yx3+y2 the bivariate limit at the origin does not exist, but the iterated limits limy→0limx→0f and limx→0limy→0f are both zero.
Solution
Mathematical Solution
Evaluate the limit along the parabolas y=m x2, that is, evaluate
limx→0fx,m x2
=limx→0m x3x3+m2x4
=limx→0m1+m2x
=m
Since the limit depends on the direction of approach to the origin, the bivariate limit at the origin does not exist.
The "inner" limits for both iterated limits are zero, so the "outer" limits are taken of zero, with the result that both the iterated limits are zero.
Maple Solution - Interactive
Define the function fx,y
Context Panel: Assign Function
fx,y=x yx3+y2→assign as functionf
Evaluate limx→0fx,m x
Calculus palette: Limit operator
Context Panel: Evaluate and Display Inline
limx→0fx,m x2 = m
Since the limit depends on the direction of approach to the origin, the bivariate limit at the origin does not exist. Alternatively, access Maple's bivariate limit through the Context Panel.
Context Panel: Limit (Bivariate) (Fill in the Limit Point dialog as per Figure 3.2.12(a).)
Figure 3.2.12(a) Limit Point dialog
fx,y = x⁢yx3+y2→bivariate limitundefined
Maple's declaration that the limit is undefined is equivalent to the more prevalent statement that the limit does not exist.
Examine the iterated limits
Calculus palette: Limit template
limy→0limx→0fx,y = 0
limx→0limy→0fx,y = 0
Examine the "inner" limits
limx→0fx,y = 0
limy→0fx,y = 0
Maple Solution - Coded
Define the function fx,y.
f≔x,y→x yx3+y2:
Apply the limit command to fx,m x. Context Panel: Evaluate and Display Inline
limitfx,m x2,x=0 = m
Obtain the iterated limits
Apply the limit command.
limitlimitfx,y,x=0,y=0 = 0
limitlimitfx,y,y=0,x=0 = 0
It is possible to gain some insight into the near-origin behavior of this function by changing to polar coordinates.
fr cosθ,r sinθ=cos⁡θ⁢sin⁡θcos⁡θ3⁢r+sin⁡θ2
which tends to cotθ as r→0. The unbounded behavior of the cotangent in the interval 0,2 π should help explain why the bivariate limit does not exist at the origin.
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