MATH280 Tutorial 3: A bit more on limits and the derivative
Exercise 1 (2.2.48 pg. 108). Establish rigorously that
lim
(x,y)→(0,0)
x3 + y 3
= 0.
x2 + y 2
Suggestions:
(a) Show that |x| ≤ k(x, y)k and |y| ≤ k(x, y)k.
(b) Show that x3 + y 3 ≤ 2(x2 + y 2 )3/2 .
3 3
(c) Show that if 0 ≤ k(x, y)k < δ, then xx2 +y
< 2δ.
+y 2 (d) Now prove that lim(x,y)→(0,0)
x3 +y 3
x2 +y 2
= 0.
Solution. (a) If x = 0, then |x| = 0 ≤ k(x, y)k and our first inequality holds. Otherwise, we can
write
r
r
2
p
√
y
y2
k(x, y)k = x2 + y 2 = x2 1 + 2 = |x| 1 + 2 .
x
x
q
2
2
Because 1 ≤ 1 + xy 2 , we have 1 ≤ 1 + xy 2 , and so, multiplying both sides of the last inequality
by |x|,
r
y2
|x| ≤ |x| 1 + 2 = k(x, y)k .
x
The proof that |y| ≤ k(x, y)k is similar.
Remark. We can also see part (a) by applying the Cauchy-Schwarz inequality. Let u := (x, y),
v := (1, 0) and w := (0, 1). Writing these out in coordinates, we have kuk = k(x, y)k , kvk =
1, kwk = 1 and
|(u · v)| = |x · 1 + y · 0| = |x| ,
|(u · w)| = |x · 0 + y · 1| = |y| .
By Cauchy-Schwarz, we have the inequalities
|(u · v)| ≤ kuk kvk
and
|(u · w)| ≤ kuk kwk
that become, when written out in coordinates, our desired inequalities
|x| ≤ k(x, y)k · 1
and
|y| ≤ 1 · k(x, y)k ,
respectively.
(b) Applying the triangle inequality and part (a), we have
3
x + y 3 ≤ x3 + y 3 = |x|3 + |y|3
≤ k(x, y)k3 + k(x, y)k3
(part(a))
3
= 2 k(x, y)k
= 2(x2 + y 2 )3/2 .
(c) We have, applying part (b),
3
x + y 3 x3 + y 3 2(x2 + y 2 )3/2
=
≤
= 2(x2 + y 2 )1/2 = 2 k(x, y)k < 2δ,
x2 + y 2 |x2 + y 2 |
x2 + y 2
as k(x, y)k < δ by hypothesis.
1
MATH280 Tutorial 3: A bit more on limits and the derivative
(d) Let > 0 be given.
We wish to find δ > 0 such that for each (x, y) satisfying k(x, y)k < δ
x3 +y3 we have x2 +y2 < . But from part (c), we see that choosing δ = 2 works. Indeed, if
k(x, y)k < δ = 2 , we have
3
x + y3 x2 + y 2 < 2δ = 2 2 = ,
as desired.
Exercise 2 (2.3.33, pg. 124). Find an equation for the hyperplane tangent to the 4-dimensional
paraboloid x5 = 10 − (x21 + 3x22 + 2x23 + x24 ) at (2, −1, 1, 3, −8).
Solution. By formula (8) on pg. 117 of the textbook, the tangent hyperplane to a hypersurface
defined by the equation xn+1 = f (x1 , . . . , xn ) at the point (a1 , . . . , an , f (a1 , . . . , an )) ∈ Rn+1 is
given by the equation
xn+1 = f (a1 , . . . , an ) + Df (a1 , . . . , an )(x1 − a1 , . . . , xn − an ),
where Df (a1 , . . . , an ) is the (total) derivative of f at the point (a1 , . . . , an ). Intuitively, the map
Df (a1 , . . . , an ) is the linear transformation Rn → R that ‘best approximates’ f : Rn → R at
(a1 , . . . , an ).
Write g(x1 , . . . , x4 ) = 10−(x21 +3x22 +2x23 +x24 ). First, let us check that the point (2, −1, 1, 3, −8)
is on the hypersurface defined by x5 = g(x1 , . . . , x4 ). Indeed, g(2, −1, 1, 3) = 10 − (4 + 3 + 2 + 9) =
10 − 18 = −8 (this is not logically a part of the solution, but it is a nice ‘reality check’).
The partials of g at an arbitrary point (x1 , . . . , x4 ) are
∂g
= −2x1 ,
∂x1
∂g
= −6x2 ,
∂x2
∂g
= −4x3 ,
∂x3
∂g
= −2x4 .
∂x4
Hence,
Dg(2, −1, 1, 3) = [−4, 6, −4, −6]
and
Dg(2, −1, 1, 3)(x1 − 2, x2 + 1, x3 − 1, x4 − 3) = −4(x1 − 2) + 6(x2 + 1) − 4(x3 − 1) − 6(x4 − 3)
= −4x1 + 6x2 − 4x3 − 6x4 + 36.
Our tangent hyperplane at (2, −1, 1, 3, −8) is then described by the equation
x5 = −8 + (−4x1 + 6x2 − 4x3 − 6x4 + 36)
or, with somewhat more symmetry in the variables,
4x1 − 6x2 + 4x3 + 6x4 + x5 − 28 = 0.
Exercise 3 (2.3.37, pg. 124). Let f (x, y, z) = x2 + xyz + y 3 z.
(a) Find a linear approximation for f (x, y, z) at (1.01, 1.95, 2.2).
(b) How accurate is the approximation in part (a)?
Solution. (a) To begin, note that our function is polynomial in x, y and z, so that partial derivatives of all orders clearly exist and are continuous. Therefore, by theorem 3.10 of the textbook
(pg. 119), f is differentiable.
By definition of the derivative, for any function f : Rn → Rm and point a in the domain of f ,
the linear function ha : Rn → Rm given by ha (x) := f (a) + Df (a)(x − a) approximates f ‘well
to the first degree’ at a. More precisely, we have
kf (x) − ha (x)k
= 0.
x→a
kx − ak
lim
2
(1)
MATH280 Tutorial 3: A bit more on limits and the derivative
So a natural choice of linear function to approximate f at (1.01, 1.95, 2.2) is hb for some
point b in a neighbourhood of (1.01, 1.95, 2.2). For simplicity of computations, let’s choose
b = (1, 2, 2).
The partials of f with respect to x, y and z are
∂f
= 2x + yz,
∂x
∂f
= xz + 3y 2 z,
∂y
∂f
= xy + y 3 .
∂z
Then
Df (1, 2, 2) = [6, 26, 10] ,
and f (1, 2, 2) = 21, so that our linear approximation is
h(1,2,2) (x, y, z) = 21 + 6(x − 1) + 26(y − 2) + 10(z − 2).
Then f at (1.01, 1.95, 2.2) is approximately
h(1,2,2) (1.01, 1.95, 2.2) = 21+6(1.01−1)+26(1.95−2)+10(2.2−2) = 21+0.06−1.3+2 = 21.76.
(b) Evaluating f at (1.01, 1.95, 2.2), we find
f (1.01, 1.95, 2.2) = (1.01)2 + (1.01)(1.95)(2.2) + (1.95)3 (2.2) = 21.6657.
Out of
find the ratio (??) in our case. Writing a = (1.01, 1.95, 2.2), we
curiosity, we can
have f (a) − h(1,2,2) (a) = 21.76 − 21.6657 = 0.0943. On the other hand, k(1, 2, 2) − ak =
p
(0.01)2 + (0.05)2 + (0.2)2 ∼ 0.2064, so
f (a) − h(1,2,2) (a)
1
∼ .
k(1, 2, 2) − ak
2
3