Section 14.2 Limits and Continuity
SECTION 14.2 LIMITS AND CONTINUITY ¤ 397 6. ( ) = 2 + 2
2− 2 is a rational function and hence continuous on its domain.
(2 −1) is in the domain of , so is continuous there and lim
()→(2−1) ( ) = (2 −1) =(2)2(−1) + (2)(−1)2 (2)2− (−1)2 = −2
3. 7. − is a polynomial and therefore continuous. Since sin is a continuous function, the composition sin( − ) is also
continuous. The function is a polynomial, and hence continuous, and the product of continuous functions is continuous, so
( ) = sin( − ) is a continuous function. Then lim
()→(2) ( ) =
2
=2sin
−2
=2sin2 = 2.
8.2 − is a polynomial and therefore continuous. Since√
is continuous for ≥ 0, the composition√
2 − is continuous where 2 − ≥ 0. The function is continuous everywhere, so the composition ( ) = √2−is a continuous function for 2 − ≥ 0. If = 3 and = 2 then 2 − ≥ 0, so lim
()→(32) ( ) = (3 2) = √
2(3)−2= 2.
9. ( ) = (4− 42)(2+ 22). First approach (0 0) along the -axis. Then ( 0) = 42 = 2for 6= 0, so
( ) → 0. Now approach (0 0) along the -axis. For 6= 0, (0 ) = −4222= −2, so ( ) → −2. Since has two different limits along two different lines, the limit does not exist.
10. ( ) = (54cos2)(4+ 4). First approach (0 0) along the -axis. Then ( 0) = 04= 0for 6= 0, so
( ) → 0. Next approach (0 0) along the -axis. For 6= 0, (0 ) = 544= 5, so ( ) → 5. Since has two different limits along two different lines, the limit does not exist.
11. ( ) = (2sin2)(4+ 4). On the -axis, ( 0) = 0 for 6= 0, so ( ) → 0 as ( ) → (0 0) along the
-axis. Approaching (0 0) along the line = , ( ) =2sin2
4+ 4 = sin2 22 = 1
2
sin
2
for 6= 0 and
lim→0
sin
= 1, so ( ) → 12. Since has two different limits along two different lines, the limit does not exist.
12. ( ) = −
( − 1)2+ 2. On the -axis, ( 0) = 0( − 1)2= 0for 6= 1, so ( ) → 0 as ( ) → (1 0) along the -axis. Approaching (1 0) along the line = − 1, ( − 1) = ( − 1) − ( − 1)
( − 1)2+ ( − 1)2 = ( − 1)2 2( − 1)2 = 1
2for 6= 1, so ( ) → 12 along this line. Thus the limit does not exist.
13. ( ) =
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.
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SECTION 14.2 LIMITS AND CONTINUITY ¤ 397 6. ( ) = 2 + 2
2− 2 is a rational function and hence continuous on its domain.
(2 −1) is in the domain of , so is continuous there and lim
()→(2−1) ( ) = (2 −1) =(2)2(−1) + (2)(−1)2 (2)2− (−1)2 = −2
3. 7. − is a polynomial and therefore continuous. Since sin is a continuous function, the composition sin( − ) is also
continuous. The function is a polynomial, and hence continuous, and the product of continuous functions is continuous, so
( ) = sin( − ) is a continuous function. Then lim
()→(2) ( ) =
2
=2sin
−2
=2sin2 = 2.
8.2 − is a polynomial and therefore continuous. Since√
is continuous for ≥ 0, the composition√
2 − is continuous where 2 − ≥ 0. The function is continuous everywhere, so the composition ( ) = √2−is a continuous function for 2 − ≥ 0. If = 3 and = 2 then 2 − ≥ 0, so lim
()→(32) ( ) = (3 2) = √
2(3)−2= 2.
9. ( ) = (4− 42)(2+ 22). First approach (0 0) along the -axis. Then ( 0) = 42 = 2for 6= 0, so
( ) → 0. Now approach (0 0) along the -axis. For 6= 0, (0 ) = −4222= −2, so ( ) → −2. Since has two different limits along two different lines, the limit does not exist.
10. ( ) = (54cos2)(4+ 4). First approach (0 0) along the -axis. Then ( 0) = 04= 0for 6= 0, so
( ) → 0. Next approach (0 0) along the -axis. For 6= 0, (0 ) = 544= 5, so ( ) → 5. Since has two different limits along two different lines, the limit does not exist.
11. ( ) = (2sin2)(4+ 4). On the -axis, ( 0) = 0 for 6= 0, so ( ) → 0 as ( ) → (0 0) along the
-axis. Approaching (0 0) along the line = , ( ) =2sin2
4+ 4 = sin2 22 = 1
2
sin
2
for 6= 0 and
lim→0
sin
= 1, so ( ) → 12. Since has two different limits along two different lines, the limit does not exist.
12. ( ) = −
( − 1)2+ 2. On the -axis, ( 0) = 0( − 1)2= 0for 6= 1, so ( ) → 0 as ( ) → (1 0) along the -axis. Approaching (1 0) along the line = − 1, ( − 1) = ( − 1) − ( − 1)
( − 1)2+ ( − 1)2 = ( − 1)2 2( − 1)2 = 1
2for 6= 1, so ( ) → 12 along this line. Thus the limit does not exist.
13. ( ) =
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.
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SECTION 14.2 LIMITS AND CONTINUITY ¤ 397 6. ( ) =2 + 2
2− 2 is a rational function and hence continuous on its domain.
(2 −1) is in the domain of , so is continuous there and lim
()→(2−1) ( ) = (2 −1) = (2)2(−1) + (2)(−1)2 (2)2− (−1)2 = −2
3. 7. − is a polynomial and therefore continuous. Since sin is a continuous function, the composition sin( − ) is also
continuous. The function is a polynomial, and hence continuous, and the product of continuous functions is continuous, so
( ) = sin( − ) is a continuous function. Then lim
()→(2) ( ) =
2
= 2sin
−2
=2sin2 =2.
8. 2 − is a polynomial and therefore continuous. Since√
is continuous for ≥ 0, the composition√
2 − is continuous where 2 − ≥ 0. The function is continuous everywhere, so the composition ( ) = √2−is a continuous function for 2 − ≥ 0. If = 3 and = 2 then 2 − ≥ 0, so lim
()→(32) ( ) = (3 2) = √
2(3)−2= 2.
9. ( ) = (4− 42)(2+ 22). First approach (0 0) along the -axis. Then ( 0) = 42= 2for 6= 0, so
( ) → 0. Now approach (0 0) along the -axis. For 6= 0, (0 ) = −4222= −2, so ( ) → −2. Since has two different limits along two different lines, the limit does not exist.
10. ( ) = (54cos2)(4+ 4). First approach (0 0) along the -axis. Then ( 0) = 04= 0for 6= 0, so
( ) → 0. Next approach (0 0) along the -axis. For 6= 0, (0 ) = 544 = 5, so ( ) → 5. Since has two different limits along two different lines, the limit does not exist.
11. ( ) = (2sin2)(4+ 4). On the -axis, ( 0) = 0 for 6= 0, so ( ) → 0 as ( ) → (0 0) along the
-axis. Approaching (0 0) along the line = , ( ) = 2sin2
4+ 4 = sin2 22 = 1
2
sin
2
for 6= 0 and
lim→0
sin
= 1, so ( ) →12. Since has two different limits along two different lines, the limit does not exist.
12. ( ) = −
( − 1)2+ 2. On the -axis, ( 0) = 0( − 1)2= 0for 6= 1, so ( ) → 0 as ( ) → (1 0) along the -axis. Approaching (1 0) along the line = − 1, ( − 1) = ( − 1) − ( − 1)
( − 1)2+ ( − 1)2 = ( − 1)2 2( − 1)2 =1
2for 6= 1, so ( ) →12 along this line. Thus the limit does not exist.
13. ( ) =
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.
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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. lim
()→(00)
2+ 2
2+ 2+ 1 − 1= lim
()→(00)
2+ 2
2+ 2+ 1 − 1·
2+ 2+ 1 + 1
2+ 2+ 1 + 1
= lim
()→(00)
2+ 2
2+ 2+ 1 + 1
2+ 2 = lim
()→(00)
2+ 2+ 1 + 1
= 2
18. ( ) = 4(2+ 8). On the -axis, ( 0) = 0 for 6= 0, so ( ) → 0 as ( ) → (0 0) along the -axis.
Approaching (0 0) along the curve = 4gives (4 ) = 828=12 for 6= 0, so along this path ( ) → 12as ( ) → (0 0). Thus the limit does not exist.
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 = 13then 6= 2 + , so
()→(013)lim ( ) = ( 0 13) = 02tan( · 13) = 1 · tan(3) =√3.
20. ( ) = +
2+ 2+ 2. Then ( 0 0) = 02= 0for 6= 0, so as ( ) → (0 0 0) along the -axis,
( ) → 0. But ( 0) = 2(22) = 12for 6= 0, so as ( ) → (0 0 0) along the line = , = 0,
( ) → 12. 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) = 12for 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
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) = 13for 6= 0, so ( ) → 13as ( ) → (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 R2andsin()
is continuous on R2
[continued]
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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) = 13for 6= 0, so ( ) → 13as ( ) → (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 R2andsin()
is continuous on R2
[continued]
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1
402 ¤ CHAPTER 14 PARTIAL DERIVATIVES
except possibly where = 0. To show that is continuous at those points, consider any point ( ) in R2where = 0.
Because is continuous, → = 0 as ( ) → ( ). If we let = , then → 0 as ( ) → ( ) and
()lim→()
sin()
= lim
→0
sin()
= 1by Equation 2.4.2 [ET 3.3.2]. Thus lim
()→() ( ) = ( )and is continuous on R2.
44. (a) ( ) =
0 if ≤ 0 or ≥ 4
1 if 0 4 Consider the path = , 0 4. [The path does not pass through (0 0)if ≤ 0 except for the trivial case where = 0.] If ≤ 0 then ( ) = 0. If 0then
= || = || || and ≥ 4 ⇔ || || ≥ 4 ⇔ 4
||≤ || ⇔ ||4−≤ || whenever is defined. Then ≥ 4 ⇔ || ≤ ||1(4−)so ( ) = 0for || ≤ ||1(4−)and ( ) → 0 as
( ) → (0 0) along this path.
(b) If we approach (0 0) along the path = 5, 0 then we have ( 5) = 1for 0 1 because 0 5 4there.
Thus ( ) → 1 as ( ) → (0 0) along this path, but in part (a) we found a limit of 0 along other paths, so
()lim→(00) ( )doesn’t exist and is discontinuous at (0 0).
(c) First we show that is discontinuous at any point ( 0) on the -axis. If we approach ( 0) along the path = , 0 then ( ) = 1 for 0 4, so ( ) → 1 as ( ) → ( 0) along this path. If we approach ( 0) along the path
= , 0 then ( ) = 0 since 0 and ( ) → 0 as ( ) → ( 0). Thus the limit does not exist and is discontinuous on the line = 0. is also discontinuous on the curve = 4: For any point ( 4)on this curve, approaching the point along the path = , 4gives ( ) = 0 since 4, so ( ) → 0 as ( ) → ( 4).
But approaching the point along the path = , 4gives ( ) = 1 for 0, so ( ) → 1 as ( ) → ( 4) and the limit does not exist there.
45. Since |x − a|2= |x|2+ |a|2− 2 |x| |a| cos ≥ |x|2+ |a|2− 2 |x| |a| = (|x| − |a|)2, we have
|x| − |a|
≤ |x − a|. Let
0be given and set = . Then if 0 |x − a| ,
|x| − |a|
≤ |x − a| = . Hence limx→a|x| = |a| and
(x) = |x| is continuous on R.
46. Let 0 be given. We need to find 0 such that if 0 |x − a| then | (x) − (a)| = |c · x − c · a| .
But |c · x − c · a| = |c · (x − a)| and |c · (x − a)| ≤ |c| |x − a| by Exercise 12.3.61 (the Cauchy-Schwartz Inequality). Set
= |c|. Then if 0 |x − a| , | (x) − (a)| = |c · x − c · a| ≤ |c| |x − a| |c| = |c| ( |c|) = . So is continuous on R.
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