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Example: Find the limit if it exists.
lim ln [ ( 1+ y2 ) / (x2
+ xy ) ]
(x,y)
→ (1/2,0)
Note: ( 1+ y2 ) / (x2 +
xy) is a rational function and is
therefore continuous.
For continuous
functions lim f(x,y)
= f(a,b)
(x,y) → (a,b)
So lim [ ( 1+ y2 ) / (x2 +
xy ) ] = 4
(x,y) → (1/2,0)
Also ln ( s ) is a continuous function as
long as s >
0.
Thus lim ln [ (1+ y2 ) / (x2
+ xy ) ] = ln 4
.
(x,y) → (1/2,0)
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Example: lim [ ( x2 +
2 sin2y) / ( 2x2 + 2y2 ) ]
(x,y) →
(0,0)
First evaluate the limit
along y = 0.
lim [ x2 / 2x2
]
= 1/2
(x,y) → (x,0)
Next evaluate along x = 0.
for y ≠ 0 lim [ 2 sin2y / 2y2] =
lim [sin y / y] 2
(0,y) → (0,y) (0,y) → (0,y)
if y = 0
lim sin y / y =
1
y
→ 0
Since the values are
different for each direction, the limit does not exist.
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