The LNM Institute of Information Technology
Department of Mathematics
Mathematics-1 & MTH102
Quiz-II
Time: 30 minutes
Date: 01/11/2018
Maximum Marks: 10
1 Examine the existence of the limit of following function at (0, 0)

2x

, if (x, y) 6= (0, 0)
2
f (x, y) =
x + x + y2
 0,
if (x, y) = (0, 0).
[3]
Solution: Along x axis, that is y = 0
2x
2x
= lim 2
x→0 x + x
+ x + y2
lim
(x,y)→(0,0) x2
lim
(x,y)→(0,0) x2
2x
2
= lim
= 2.
x→0 x + 1
+ x + y2
[1.5]
Along y axis, that is 0 = 0
lim
(x,y)→(0,0) x2
lim
(x,y)→(0,0) x2
0
2x
= lim 2
2
y→0
+x+y
y
2x
= lim 0 = 0.
y→0
+ x + y2
Limit is not unique or path dependent so Limit does not exist.
[1.5]
p
3 For the function f (x, y) = x2 + y 2 , find directional derivatives in all possible directions at point (0, 0) (if it exists). [2]
Solution: Let u = (u1 , u2 ) be unit vector.
f (tu1 , tu2 ) − f (0, 0)
t
p
2
t (u21 + u22 )
Du f (0, 0) = lim
t→0
t
Du f (0, 0) = lim
t→0
[1]
Du f (0, 0) = lim
t→0
|t|
.
t
Since
u21 + u22 = 1
Limit does not exist. Hence directional derivative does not exist.
1
[1]
h
i
2 Find iterated(repeated) limit, lim lim f (x, y) where f (x, y) is defined as
y→0 x→0

 x + y , if x 6= y
f (x, y) =
x−y
 0,
if x = y.
[2]
Solution: For y 6= 0, we compute lim f (x, y).
x→0
Since we are interested in computing limit as x → 0, for given any y 6= 0, our x will be such that |x| < |y|. Hence
f (x, y) =
x+y
, for |x| < |y|.
x−y
This means
x+y
0+y
=
= −1
x→0 x − y
0−y
lim
h
i
[1] Since for each y 6= 0, lim f (x, y) = −1, therefore lim lim f (x, y) = −1
x→0
[1]
y→0 x→0
4 Find all the points of local maxima and local minima for the function f (x, y) = x3 + y 3 − 3xy + 15.
[3]
Solution: Since f is a polynomial function, hence the critical points of f are exactly those points where ∇f (x, y) = (0, 0).
fx = 3x2 − 3y = 0 =⇒ y = x2
fy = 3y 2 − 3x = 0 =⇒ y 2 = x
=⇒ (x2 )2 = x =⇒ x(x3 − 1) = 0 =⇒ x = 0, 1
So (0, 0) and (1, 1) are the only critical points of f .
[1]
fxx = 6x, fyy = 6y, fxy = −3
∆f (x, y) = 36xy − 9
∆f (0, 0) = −9 < 0 =⇒ (0, 0) is a saddle point.
[1]
∆f (1, 1) = 36 − 9 > 0, fxx (1, 1) = 6 > 0 =⇒ (1, 1) is a point of local minimum.
[1]
2