MATH 209, Lab 1
Richard M. Slevinsky∗
Problems
1. Find the domain and range of the following functions and sketch their graphs:
p
(a) z = f (x, y) = x2 + y 2 ;
The domain is {(x, y) : x ∈ R, y ∈ R}.
The range is {z : z ∈ R, z ≥ 0}.
(b) z = f (x, y) = 3 x2 ;
The domain is {(x, y) : x ∈ R, y ∈ R}.
The range is {z : z ∈ R, z ≥ 0}.
p
(c) x = f (y, z) = 4 − y 2 − z 2 .
The domain is {(y, z) : y ∈ R, z ∈ R, y 2 + z 2 ≤ 4}.
The range is {x : x ∈ R, 0 ≤ x ≤ 2}.
2. Find each of the following limits, or show that the limit does not exist:
2 x2 y
;
2 x4 + 3 y 2
Let y = m x2 . Then we have:
(a) lim(x,y)→(0,0)
2 m x4
2 x2 y
=
lim
,
x→0 2 x4 + 3 m2 x4
(x,y)→(0,0) 2 x4 + 3 y 2
2m
2m
= lim
=
.
2
x→0 2 + 3 m
2 + 3 m2
lim
But this is different for every value m we choose, so the limit DNE.
∗
Contact: [email protected]
1
x2 y 3
;
2 x2 + 3 y 2
Let y = m x. Then we have:
(b) lim(x,y)→(0,0)
x2 y 3
m 3 x5
=
lim
,
x→0 2 x2 + 3 m2 x2
(x,y)→(0,0) 2 x2 + 3 y 2
lim
m3
x3 = 0.
x→0 2 + 3 m2
= lim
Clearly, it works for any line y = m x. However, how can we be sure? Switch to polar coordinates:
x2 y 3
r2 cos2 θr3 sin3 θ
=
lim
,
r→0 2 r 2 cos2 θ + 3 r 2 sin2 θ
(x,y)→(0,0) 2 x2 + 3 y 2
lim
cos2 θ sin3 θ
r3 ,
r→0 2 cos2 θ + 3 sin2 θ
cos2 θ sin3 θ 3
= lim
r = 0.
r→0 2 + sin2 θ
= lim
2
3 ey
.
(c) lim(x,y)→(e,10)
10 ln x
Sometimes it is as easy as plugging in the values!
2
3 ey
3 e100
lim
=
= 0.3e100 .
10 ln e
(x,y)→(e,10) 10 ln x
Exercises
1. Find the domain and range and sketch the graph for each of the following functions:
(a) z = f (x, y) = 2 y;
The domain is {(x, y) : x ∈ R, y ∈ R}.
The range is {z : z ∈ R}.
(b) y = f (x, z) = 3 x2 + z;
The domain is {(x, y) : x ∈ R, z ∈ R}.
The range is {y : y ∈ R}.
p
(c) z = f (x, y) = 12 − x2 − y 2 .
The domain is {(x, y) : x ∈ R, y ∈ R, x2 + y 2 ≤ 12}.
√
The range is {z : z ∈ R, 0 ≤ z ≤ 12}.
2. Find the limit, or show that the limit does not exist:
x2 − y 2
;
x2 + y 2
Let y = m x, then:
(a) lim(x,y)→(0,0)
x2 − y 2
x2 − m 2 x2
=
lim
,
x→0 x2 + m2 x2
(x,y)→(0,0) x2 + y 2
lim
=
1 − m2
.
1 + m2
But this is different for every value m we choose, so the limit DNE.
2
6 x3
(b) lim(x,y)→(0,0) p
;
x2 + y 2
In polar coordinates:
6 x3
6 r3 cos3 θ
p
,
= lim p
(x,y)→(0,0)
x2 + y 2 r→0 r2 cos2 θ + r2 sin2 θ
6 r3 cos3 θ
√
= lim
= lim 6 r2 cos3 θ = 0.
r→0
r→0
r2
lim
x2 − 2 y 3 − 4 z
.
x + y3 + 2 z
Substituting values, we find:
(c) lim(x,y,z)→(2,1,0)
x2 − 2 y 3 − 4 z
22 − 2 13 − 4 · 0
2
=
= .
3
3
x
+
y
+
2
z
2
+
1
+
2
·
0
3
(x,y,z)→(2,1,0)
lim
3. Show:
sin(x + y)
= 1;
x+y
In polar coordinates:
(a) lim(x,y)→(0,0)
sin(x + y)
sin(r(cos θ + sin θ))
= lim
,
r→0
x+y
r(cos θ + sin θ)
(x,y)→(0,0)
√
sin(r 2 sin(θ + π/4))
√
= lim
,
r→0
r 2 sin(θ + π/4)
lim
cos θ + sin θ =
√
2 sin(θ + π/4)
sin(rα)
= 1.
r→0
rα
Recall : lim
sin x + sin y
α−β
= 1. [Hint: sin α + sin β = 2 sin α+β
cos
.]
2
2
x+y
Using the hint, we obtain:
2 sin x+y
cos x−y
sin x + sin y
2
2
lim
=
lim
,
Recall : cos(0) = 1
x+y
x+y
(x,y)→(0,0)
(x,y)→(0,0)
sin x+y
·1
2
=
lim
= 1,
from part (a).
x+y
(b) lim(x,y)→(0,0)
(x,y)→(0,0)
x4
4. Find the value of k such that f (x, y) =
x2 + y 2

k
continuous everywhere? Why?


3
2
, (x, y) 6= (0, 0)
, (x, y) = (0, 0)
is continuous at (0, 0). Is f (x, y)
Taking the limit, we find:
x4
r4 cos4 θ
,
=
lim
r→0 r 2 cos2 θ + r 2 sin2 θ
(x,y)→(0,0) x2 + y 2
lim
= lim r2 cos4 θ = 0.
r→0
Therefore, k = 0. It is then continuous everywhere, since f (x, y) is otherwise a rational function and
there are no other singularities.
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