Section 14.2 Limits and Continuity
17. Show that the limit does not exist. lim
(x,y)→(0,0) y2sin2x
x4+y4 . Solution:
SECTION 14.2 LIMITS AND CONTINUITY ¤ 1361 11. ( ) = 23− 32
2− 2 = 22( − )
( − )( + ) = −22
+ for − 6= 0; in particular, ( ) 6= (1 1). Thus,
()lim→(11) ( ) = lim
()→(11)
−22
+
= −(1)2(1)2 1 + 1 = −1
2.
12. ( ) =cos − sin 2
cos cos = cos − 2 sin cos
cos cos = 1 − 2 sin
cos for cos 6= 0; in particular, ( ) 6= ( 2). Thus,
lim
()→(2) ( ) = lim
()→(2)
1 − 2 sin
cos = 1 − 2 sin(2) cos = −1
−1 = 1.
13. ( ) = 2
2+ 2. First approach (0 0) along the axis. Then ( 0) = 02= 0for 6= 0, so ( ) → 0. Now approach (0 0) along the axis. Then (0 ) = 22= 1for 6= 0, so ( ) → 1. Since has two different limits along two different lines, the limit does not exist.
14. ( ) = 2
2+ 32. First approach (0 0) along the axis. Then ( 0) = 02= 0for 6= 0, so ( ) → 0. Now approach (0 0) along the line = . Then ( ) = 2242= 12for 6= 0. Since has two different limits along two different lines, the limit does not exist.
15. ( ) = ( + )2
2+ 2 . First approach (0 0) along the axis. Then ( 0) = 22= 1for 6= 0, so ( ) → 1. Now approach (0 0) along the line = . Then ( ) = 42(22) = 2for 6= 0, so ( ) → 2. Since has two different limits along two different lines, the limit does not exist.
16. ( ) =2+ 2
4+ 2 . First approach (0 0) along the axis. Then (0 ) = 02= 0for 6= 0, so ( ) → 0. Now approach (0 0) along the line = . Then ( ) = 2+ 3
4+ 2 = 2(1 + )
2(2+ 1) = 1 +
1 + 2 for 6= 0 so ( ) → 1. Since has two different limits along two different lines, the limit does not exist.
17. ( ) = 2sin2
4+ 4 . First approach (0 0) along the axis. Then (0 ) = 04 = 0for 6= 0, so ( ) → 0. Now
approach (0 0) along the line = . Then ( ) = 2sin2 24 = 1
2
sin
2
for 6= 0, so by Equation 3.3.5,
( ) → 12(1)2= 12. Since has two different limits along two different lines, the limit does not exist.
18. ( ) = −
1 − + ln . First approach (1 1) along the line = 1. Then (1 ) = − 1
1 − + 0 = −1 for 6= 1, so
( ) → −1. Now approach (1 1) along the line = . Then ( ) = −
1 − + ln = 0for 1 − + ln 6= 0. So
( ) → 0. Since has two different limits along two different lines, the limit does not exist.
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23. Find the limit, if it exists, or show that the limit does not exist. lim
(x,y)→(0,0) xy2cos y
x2+y4 . Solution:
398 ¤ CHAPTER 14 PARTIAL DERIVATIVES 14. ( ) = 3− 3
2+ + 2 = ( − )(2+ + 2)
2+ + 2 = − for ( ) 6= (0 0). [Note that 2+ + 2= 0only when ( ) = (0 0).] Thus lim
()→(00) ( ) = lim
()→(00)( − ) = 0 − 0 = 0.
15. Let ( ) = 2cos
2+ 4 . Then ( 0) = 0 for 6= 0, so ( ) → 0 as ( ) → (0 0) along the -axis. Approaching
(0 0)along the -axis or the line = also gives a limit of 0. But
2
= 22 cos
(2)2+ 4 = 4cos
24 = cos
2 for 6= 0, so ( ) →12cos 0 =12 as ( ) → (0 0) along the parabola = 2. Thus the limit doesn’t exist.
16. We can use the Squeeze Theorem to show that lim
()→(00)
4
4+ 4 = 0:
0 ≤ || 4
4+ 4 ≤ || since 0 ≤ 4
4+ 4 ≤ 1, and || → 0 as ( ) → (0 0), so || 4
4+ 4 → 0 ⇒ 4
4+ 4 → 0 as ( ) → (0 0).
17. Let ( ) = 2
4+ 42. Then ( 0) = 0 for 6= 0, so ( ) → 0 as ( ) → (0 0) along the -axis. Approaching
(0 0)along the -axis or the line = also gives a limit of 0. But
2
= 222
4+ 4(2)2 = 42 54 = 2
5 for 6= 0, so
( ) → 05 = 15as ( ) → (0 0) along the parabola = 2. Thus the limit doesn’t exist.
18. We can use the Squeeze Theorem to show that lim
()→(00)
2sin2
2+ 22 = 0:
0 ≤ 2sin2
2+ 22 ≤ sin2since 2
2+ 22 ≤ 1, and sin2 → 0 as ( ) → (0 0), so lim
()→(00)
2sin2
2+ 22 = 0.
19. 2is a composition of continuous functions and hence continuous. is a continuous function and tan is continuous for
6= 2 + ( an integer), so the composition tan() is continuous for 6= 2 + . Thus the product
( ) = 2tan()is a continuous function for 6= 2 + . If = and = 13 then 6= 2 + , so
()→(013)lim ( ) = ( 0 13) = 02tan( · 13) = 1 · tan(3) =√ 3.
20. ( ) =
2+ 42+ 92. Then ( 0 0) = 0 for 6= 0, so as ( ) → (0 0 0) along the -axis, ( ) → 0.
But (0 ) = 2(132) = 131 for 6= 0, so as ( ) → (0 0 0) along the line = , = 0, ( ) → 131. Thus the limit doesn’t exist.
21. ( ) = + 2+ 2
2+ 2+ 4 . Then ( 0 0) = 02= 0for 6= 0, so as ( ) → (0 0 0) along the -axis,
( ) → 0. But ( 0) = 2(22) = 12 for 6= 0, so as ( ) → (0 0 0) along the line = , = 0,
( ) → 12. Thus the limit doesn’t exist.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
33. Use the Squeeze Theorem to find the limit. lim
(x,y)→(0,0) xy4 x4+y4. Solution:
SECTION 14.2 LIMITS AND CONTINUITY ¤ 1363 28. ( ) = +
2+ 2+ 2. Then ( 0 0) = 02 = 0for 6= 0, so as ( ) → (0 0 0) along the axis,
( ) → 0. But ( 0) = 2(22) = 12 for 6= 0, so as ( ) → (0 0 0) along the line = , = 0,
( ) → 12. Thus the limit doesn’t exist.
29. ( ) = + 2+ 2
2+ 2+ 4 . Then ( 0 0) = 02= 0for 6= 0, so as ( ) → (0 0 0) along the axis,
( ) → 0. But ( 0) = 2(22) = 12 for 6= 0, so as ( ) → (0 0 0) along the line = , = 0,
( ) → 12. Thus the limit doesn’t exist.
30. ( ) = 4+ 2+ 3
4+ 22+ . Then ( 0 0) = 44= 1for 6= 0, so ( ) → 1 as ( ) → (0 0 0) along the
axis. But (0 0) = 2(22) = 12for 6= 0, so ( ) → 12 as ( ) → (0 0 0) along the axis. Since has two different limits along two different lines, the limit does not exist.
31. −1 ≤ sin
1
2+ 2
≤ 1 ⇒ − ≤ sin
1
2+ 2
≤ for 0. If 0, we have
− ≥ sin
1
2+ 2
≥ . In either case, lim
()→(00) = 0and lim
()→(00)(−) = 0. Thus,
()lim→(00) sin
1
2+ 2
= 0by the Squeeze Theorem.
32. ( ) =
2+ 2. We can see that the limit along any line through (0 0) is 0, as well as along other paths through
(0 0)such as = 2and = 2. So we suspect that the limit exists and equals 0; we use the Squeeze Theorem to prove our
assertion. Since || ≤
2+ 2, we have ||
2+ 2 ≤ 1 and so 0 ≤
2+ 2
≤ ||. Now || → 0 as ( ) → (0 0),
so
2+ 2
→ 0 and hence lim
()→(00) ( ) = 0.
33. We use the Squeeze Theorem to show that lim
()→(00)
4
4+ 4 = 0:
0 ≤ || 4
4+ 4 ≤ || since 0 ≤ 4
4+ 4 ≤ 1, and || → 0 as ( ) → (0 0), so || 4
4+ 4 → 0 ⇒ 4
4+ 4 → 0 as ( ) → (0 0).
34. We use the Squeeze Theorem to show that lim
()→(000)
222
2+ 2+ 2 = 0:
0 ≤ 222
2+ 2+ 2 ≤ 22 since 0 ≤ 2
2+ 2+ 2 ≤ 1, and 22→ 0 as ( ) → (0 0 0), so 222
2+ 2+ 2 → 0 as ( ) → (0 0 0).
° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
49. Determine the set of points at which the function is continuous. f (x, y) =
x2y3
2x2+y2 if (x, y) 6= (0, 0) 1 if (x, y) = (0, 0) Solution:
SECTION 14.2 LIMITS AND CONTINUITY ¤ 401
37. ( ) =
23
22+ 2 if ( ) 6= (0 0) 1 if ( ) = (0 0)
The first piece of is a rational function defined everywhere except at the
origin, so is continuous on R2except possibly at the origin. Since 2≤ 22+ 2, we have23(22+ 2)
≤3.
We know that3
→ 0 as ( ) → (0 0). So, by the Squeeze Theorem, lim
()→(00) ( ) = lim
()→(00)
23 22+ 2 = 0.
But (0 0) = 1, so is discontinuous at (0 0). Therefore, is continuous on the set {( ) | ( ) 6= (0 0)}.
38. ( ) =
2+ + 2 if ( ) 6= (0 0)
0 if ( ) = (0 0)
The first piece of is a rational function defined everywhere except
at the origin, so is continuous on R2except possibly at the origin. ( 0) = 02= 0for 6= 0, so ( ) → 0 as ( ) → (0 0) along the -axis. But ( ) = 2(32) = 13 for 6= 0, so ( ) → 13 as ( ) → (0 0) along the line = . Thus lim
()→(00) ( )doesn’t exist, so is not continuous at (0 0) and the largest set on which is continuous is {( ) | ( ) 6= (0 0)}.
39. lim
()→(00)
3+ 3
2+ 2 = lim
→0+
( cos )3+ ( sin )3
2 = lim
→0+( cos3 + sin3) = 0
40. lim
()→(00)(2+ 2) ln(2+ 2) = lim
→0+2ln 2= lim
→0+
ln 2 12 = lim
→0+
(12)(2)
−23 [using l’Hospital’s Rule]
= lim
→0+(−2) = 0
41. lim
()→(00)
−2−2− 1
2+ 2 = lim
→0+
−2− 1
2 = lim
→0+
−2(−2)
2 [using l’Hospital’s Rule]
= lim
→0+−−2= −0= −1 42. lim
()→(00)
sin(2+ 2)
2+ 2 = lim
→0+
sin(2)
2 , which is an indeterminate form of type 00. Using l’Hospital’s Rule, we get
lim
→0+
sin(2)
2
= limH
→0+
2 cos(2)
2 = lim
→0+cos(2) = 1.
Or: Use the fact that lim
→0
sin
= 1.
43. ( ) =
sin()
if ( ) 6= (0 0) 1 if ( ) = (0 0)
From the graph, it appears that is continuous everywhere. We know
is continuous on R2and sin is continuous everywhere, so sin()is continuous on R2and sin()
is continuous on R2
[continued]
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52. Use polar coordinates to find the limit. [If (r, θ) are polar coordinates of the point (x, y) with r ≥ 0, note that r → 0+ as (x, y) → (0, 0).]
lim
(x,y)→(0,0)(x2+ y2) ln(x2+ y2) Solution:
SECTION 14.2 LIMITS AND CONTINUITY ¤ 401
37. ( ) =
23
22+ 2 if ( ) 6= (0 0) 1 if ( ) = (0 0)
The first piece of is a rational function defined everywhere except at the
origin, so is continuous on R2except possibly at the origin. Since 2≤ 22+ 2, we have23(22+ 2)
≤3.
We know that3
→ 0 as ( ) → (0 0). So, by the Squeeze Theorem, lim
()→(00) ( ) = lim
()→(00)
23 22+ 2 = 0.
But (0 0) = 1, so is discontinuous at (0 0). Therefore, is continuous on the set {( ) | ( ) 6= (0 0)}.
38. ( ) =
2+ + 2 if ( ) 6= (0 0)
0 if ( ) = (0 0)
The first piece of is a rational function defined everywhere except
at the origin, so is continuous on R2except possibly at the origin. ( 0) = 02= 0for 6= 0, so ( ) → 0 as ( ) → (0 0) along the -axis. But ( ) = 2(32) = 13 for 6= 0, so ( ) → 13 as ( ) → (0 0) along the line = . Thus lim
()→(00) ( )doesn’t exist, so is not continuous at (0 0) and the largest set on which is continuous is {( ) | ( ) 6= (0 0)}.
39. lim
()→(00)
3+ 3
2+ 2 = lim
→0+
( cos )3+ ( sin )3
2 = lim
→0+( cos3 + sin3) = 0
40. lim
()→(00)(2+ 2) ln(2+ 2) = lim
→0+
2ln 2= lim
→0+
ln 2 12 = lim
→0+
(12)(2)
−23 [using l’Hospital’s Rule]
= lim
→0+(−2) = 0
41. lim
()→(00)
−2−2− 1
2+ 2 = lim
→0+
−2− 1
2 = lim
→0+
−2(−2)
2 [using l’Hospital’s Rule]
= lim
→0+−−2 = −0= −1 42. lim
()→(00)
sin(2+ 2)
2+ 2 = lim
→0+
sin(2)
2 , which is an indeterminate form of type 00. Using l’Hospital’s Rule, we get
lim
→0+
sin(2)
2
= limH
→0+
2 cos(2)
2 = lim
→0+
cos(2) = 1.
Or: Use the fact that lim
→0
sin
= 1.
43. ( ) =
sin()
if ( ) 6= (0 0) 1 if ( ) = (0 0)
From the graph, it appears that is continuous everywhere. We know
is continuous on R2and sin is continuous everywhere, so sin()is continuous on R2and sin()
is continuous on R2
[continued]
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