Section 14.2 Limits and Continuity
Ruipeng Shen
March 19
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Limit
Definition 1. We say that the limit of f (x, y) as (x, y) approaches (a, b) is L and use the
notation
lim
f (x, y) = L
(x,y)→(a,b)
if the values of f (x, y) approach the number L as the point (x, y) approaches the point (a, b).
Example 2. Evaluate the limit
lim
(x + y 2 ).
(x,y)→(1,1)
Solution
Since (x, y) → (1, 1), we have x → 1 and y → 1. This implies y 2 → 1 and x + y 2 → 2.
Proposition 3. If f (x, y) → L1 as (x, y) → (a, b) along a path C1 and f (x, y) → L2 as
(x, y) → (a, b) along another path C2 , when L1 6= L2 , then
lim
f (x, y) does not exist.
(x,y)→(a,b)
Example 4. Show that
Solution
x2 − y 2
does not exist.
(x,y)→(0,0) x2 + y 2
lim
If we approach (0, 0) along the x-axis, then we have
x2 − 02
= 1.
x→0 x2 + 02
lim f (x, 0) = lim
x→0
On the other hand, if we approach (0, 0) along the y-axis, then we have
02 − y 2
= −1.
y→0 02 + y 2
lim f (0, y) = lim
y→0
Thus the original limit does not exist.
Example 5. Determine whether
lim
(x,y)→(0,0)
xy
exists or not.
x2 + y 2
Solution If we approach (0, 0) along x-axis or y-axis, then the limits are both 0. However,
this does not guarantee the original limit is zero. In fact, the limit along the line y = x is
lim f (x, x) = lim
x→0
x→0 x2
As a result, the original limit does not exist.
1
x2
1
= 6= 0.
2
+x
2
y
f=1/2
x
f=0
Figure 1: Different paths of convergence
Figure 2: Graph of the function f (x, y) =
2
x2
xy
+ y2
Example 6. Determine whether
lim
xy 2
exists or not.
+ y4
(x,y)→(0,0) x2
Solution If we approach (0, 0) along x-axis or y-axis, then the limits are both 0. In addition,
we can consider the limit along any straight line y = kx:
x(kx)2
k2 x
=
lim
= 0.
x→0 x2 + k 4 x4
x→0 1 + k 4 x2
lim f (x, kx) = lim
x→0
However, this does not guarantee the original limit is zero. In fact, the limit along the parabola
x = y 2 is
1
y2 · y2
= .
lim f (y 2 , y) = lim 2 2
y→0
y→0 (y ) + y 4
2
As a result, the original limit does not exist.
Figure 3: Graph of the function f (x, y) =
xy 2
+ y4
x2
Theorem 7 (The Squeeze Theorem). If f (x, y) ≤ g(x, y) ≤ h(x, y) when (x, y) is near (a, b)
(except possibly at (a, b)) and
lim
(x,y)→(a,b)
then
lim
f (x, y) =
lim
(x,y)→(a,b)
g(x, y) = L.
(x,y)→(a,b)
3
h(x, y) = L,
Example 8. Find
Solution
3x2 y
if it exists.
+ y2
lim
(x,y)→(0,0) x2
By the inequality
3x2 y x2
x2 + y 2 = x2 + y 2 · 3|y| ≤ 3|y|,
we have
−3|y| ≤
3x2 y
≤ 3|y|.
x2 + y 2
Applying the squeeze theorem and using the fact lim |y| = 0, we have
y→0
2
lim
3x2 y
= 0.
+ y2
(x,y)→(0,0) x2
Continuity
Definition 9. A function f of two variables is called continuous at (a, b) if
lim
f (x, y) = f (a, b).
(x,y)→(a,b)
We say f is continuous on D if f is continuous at every point (a, b) in D.
Proposition 10. For continuity we have
(a) If g(x) is a continuous function of x, then the two-variable function f (x, y) = g(x) is a
continuous function.
(b) If f (x, y) and g(x, y) are continuous at (x0 , y0 ), then we have the functions f (x, y)+g(x, y),
f (x, y) − g(x, y), f (x, y)g(x, y), f (x, y)/g(x, y) are also continuous at (x0 , y0 ). We need to
assume g(x0 , y0 ) 6= 0 in the last case.
(c) The composition of continuous function is always continuous.
(d) In summary, any function you can construct with elementary functions by taking sum,
difference, product, quotient and composition is always continuous in the domain. For
example, a polynomial function of two variables is always continuous everywhere; a rational
function is always continuous whenever the denominator is nonzero.
Example 11. Evaluate
Solution
x2 − y 2
.
(x,y)→(1,2) x2 + y 2
lim
By the proposition above, the function f (x, y) above is continuous at (1, 2). Therefore
x2 − y 2
12 − 22
3
=
f
(1,
2)
=
=− .
2
2
2
2
1 +2
5
(x,y)→(1,2) x + y
lim
Example 12. Determine whether the function
 2
 x − y2
, if (x, y) 6= (0, 0);
f (x, y) =
x2 + y 2

0,
if (x, y) = (0, 0);
is continuous at (0, 0).
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Solution
Since the limit
x2 − y 2
does not exist, we know the function is not con(x,y)→(0,0) x2 + y 2
lim
tinuous at (0, 0).
Example 13. Determine whether the function

 3x2 y
, if (x, y) 6= (0, 0);
f (x, y) =
x2 + y 2

0,
if (x, y) = (0, 0);
is continuous at (0, 0).
Solution
Example 8 gives
lim
3x2 y
= 0 = f (0, 0).
+ y2
(x,y)→(0,0) x2
As a result, the function is continuous at the origin by definition.
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