Show that the function
(x,y) = (0,0)
f(x, y) = x² +5y
0
(x,y)=(0,0)
is not continuous at (0,0).
Answers
Lectures 26-27: Functions of Several Variables
(Continuity, Differentiability, Increment Theorem and Chain Rule)
The rest of the course is devoted to calculus of several variables in which we study continuity,
differentiability and integration of functions from R
n
to R, and their applications.
In calculus of single variable, we had seen that the concept of convergence of sequence played
an important role, especially, in defining limit and continuity of a function, and deriving some
properties of R and properties of continuous functions. This motivates us to start with the notion
of convergence of a sequence in R
n
. For simplicity, we consider only R
2 or R
3
. General case is
entirely analogous.
Convergence of a sequence : Let Xn = (x1,n, x2,n, x3,n) ∈ R
3
. We say that the sequence (Xn)
is convergent if there exists X0 ∈ R
3
such that kXn − X0k → 0 as n → ∞. In this case we say that
Xn converges to X0 and we write Xn → X0.
Note that corresponding to a sequence (Xn), Xn = (x1,n, x2,n, x3,n), there are three sequences
(x1,n)(x2,n) and (x3,n) in R, and vice-versa. Thus the properties of (Xn) can be completely understood in terms of the properties of the corresponding sequences (x1,n)(x2,n) and (x3,n) in R. For
example,
(i) Xn → X0 in R
3 ⇔ the coordinates xi,n → xi,0 for every i = 1, 2, 3 in R.
(ii) (Xn) is bounded (i.e., ∃ M such that kXnk ≤ M ∀ n) ⇔ each sequence (xi,n), i = 1, 2, 3, is
bounded.
Using the previous idea, we can prove the following results.
Problem 1: Every convergent sequence R
3
is bounded.
Problem 2 (Bolzano-Weierstrass Theorem): Every bounded sequence in R
3 has a convergent
subsequence.
In case of a sequence in R, to define the notion of convergence or boundedness, we use | | in
place of k k, hence it is clear how we generalized the concept of convergence or boundedness of
a sequence in R
1
to R
3
. Moreover, it is also now clear how to define the concepts of limit and
continuity of a function f : R
3 → R at some point X0 ∈ R
3
.
Limit and Continuity : (i) We say that L is the limit of a function f : R
3 → R at X0 ∈ R
3
(and
we write limX→X0
f(X) = L) if f(Xn) → L whenever a sequence (Xn) in R
3
, Xn 6= X0, converges
to X0.
(ii) A function f : R
3 → R is continuous at X0 ∈ R
3
if limX→X0
f(X) = f(X0).
Examples 1: (i) Consider the function f : R
2 → R, where f(x, y) = sin2
(x−y)
|x|+|y| when (x, y) 6= (0, 0)
and f(0, 0) = 0. We will show that this function is continuous at (0,0). Note that
| f(x, y) − f(0, 0) | ≤ | x − y |
2
| x | + | y |
≤ | x | + | y | (or | x − y | )
Therefore, whenever a sequence (xn, yn) → (0, 0), i.e, xn → 0 and yn → 0, we have f(xn, yn) →
f(0, 0). Hence f is continuous at (0, 0). In fact, this function is continuous on the entire R
2
.
(ii) Consider the function f : R
2 → R, where f(x, y) = √
xy
x2+y
2
when (x, y) 6= (0, 0) and f(0, 0) = 0.
This function is continuous at (0, 0), because, | √
xy
x2+y
2
| ≤ |x
2+y
2
√
|
x2+y
2
=
p
x
2 + y
2 → 0, as (x, y) → 0.
2
(iii) Let f(x, y) = 2xy
x2+y
2 ,(x, y) 6= (0, 0). We will show that this function does not have a limit at
(0, 0). Note that f(x, mx) → 2m
1+m2 as x → 0 for any m. This shows that the function does not
have a limit at (0, 0).
(iv) Let f(x, y) = x
2y
x4+y
2 when (x, y) 6= (0, 0) and f(0, 0) = 0. Note that f(x, mx) → 0 as x → 0.
But the function is not continuous at (0, 0) because f(x, x2
) → 1
2
as x → 0. Similarly we can show
that the function f(x, y) defined by f(x, y) = x
4−y
2
x4+y
2 when (x, y) 6= (0, 0) and f(0, 0) = 0 is not
continuous at (0, 0) by taking y = mx2 and allowing x → 0.
Partial derivatives : The partial derivative of f with respect to the first variable at X0 =
(x0, y0, z0) is defined by
∂f
∂x |X0= lim
h→0
f(x0 + h, y0, z0) − f(x0, y0, z0)
h
provided the limit exists. Similarly we define ∂f
∂y |X0
and ∂f
∂z |X0
.
Example 2: The function f defined by f(x, y) = 2xy
x2+y
2 at (x, y) 6= (0, 0) and f(0, 0) = 0 is not
continuous at (0, 0), however, the partial derivatives exist at (0, 0).
Problem 3: Let f(x, y) be defined in S = {(x, y) ∈ R
2
: a < x < b, c < y < d}. Suppose that the
partial derivatives of f exist and are bounded in S. Then show that f is continuous in S.
Solution : Let |fx(x, y)| ≤ M and |fy(x, y)| ≤ M for all (x, y) ∈ S. Then
f(x + h, y + k) − f(x, y) = f(x + h, y + k) − f(x + h, y) + f(x + h, y) − f(x, y)
= kfy(x+h, y +θ1k)+hfx(x+θ2h, y), (for some θ1, θ2 ∈ R, by the MVT).
Hence, |f(x + h, y + k) − f(x, y) |≤ M(|h| + |k|) ≤ 2M
√
h
2 + k
2.
Hence, for ² > 0, choose δ =
²