Squeeze Theorem Application

8/16/2019 Squeeze Theorem Application

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3 WAYS TO USE THE SQUEEZE THEOREM

Find the limit, or show the limit does not exist.

• lim(x, y)→(0,0)x2 y−x3

x2+ y2

For any limit problem, start by taking a limit along some “easy” paths. If the

limits along different paths disagree, you know straight away that the limit does

not exist; if the limits along different paths agree, then we still can’t tell whether

the limit exists or not, but we have a better idea of how we might try to prove

what the limit is.

There are many “easy” paths to approach the origin, namely straight lines, andwe can try them all at once:

Approach (0, 0) along y = mx (note that the denominator is nonzero along

these paths except at (0, 0), so the function is defined on all these paths and we

can take limits along these paths without problems):

limx→0

x2mx − x3

x2 + (mx)2 = lim

x→0

mx − x

1 + m2

= m(0) − 0

1 + m2

= 0

In the first step, we cancelled x2 from the top and the bottom because, near

x = 0 (but not actually at that point), x2 is nonzero. For the second step, mx−x1+m2

is a polynomial, which is continuous, so its limit at x = 0 is just this polynomial

evaluated at 0.

We’re trying to find the limit of a rational function, so next we should check

whether it’s homogeneous (in the sense that I defined). Set x = ta, y = tb. The

function becomes t2atb − t3a

t2a + t2b

If we want both terms in the denominator to have the same degree, we must

have a = b. Then the terms in the numerator will have degree 3a, whilst the

terms in the denominator have degree 2a ̸= 3a (since a, b are nonzero when we

test for homogeneity), so the function isn’t homogeneous.1


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MATH 51: 3 WAYS TO USE THE SQUEEZE THEOREM 2

Let’s try to prove that the limit is 0, using the squeeze theorem. We want a

“simple” function f (x, y) with

x2 y− x3

x2

+ y2

≤ f (x, y)

since that would mean

− f (x, y) ≤ x2 y− x3

x2 + y2 ≤ f (x, y)

(the two sides don’t necessarily have to be a function and its negative, the squeeze

theorem will work as long as the two “squeezing” functions have the same limit.

But, if you’re trying to show the limit is zero, it’s often useful to bound the

absolute value of the function, like I’m doing here.)

We want f (x, y) to be greater than a fraction. There are two ways to do this:

make the numerator bigger, or make the denominator smaller.

Solution 1: make the numerator bigger. I’m going to add some positive

terms to the numerator so it becomes a multiple of x2 + y2, and then I can cancel

with the denominator. The drawback of this method is that you have to think of

what to add.

Now,x2 y− x3

= x2| y− x| (0.1)

≤ x2 | y− x|+ y2 | y− x| (0.2)

(0.1) uses the fact that x2 ≥ 0

(0.2) is adding in y2 | y− x|, which is ≥ 0

So we have

−| y− x| ≤ x2 y− x3

x2 + y2 ≤ | y− x|

and | y− x|, − | y− x| are both continuous, so

lim(x, y)→(0,0)

| y− x| = |0− 0| = 0

and

lim(x, y)→(0,0)

− | y− x| = − |0− 0| = 0

. Then, by the squeeze theorem,

lim(x, y)→(0,0)

x2 y− x3

x2 + y2 = 0


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.

Solution 2: make the denominator smaller. (due to lecturer Joan). Again,

we want to make the numerator and denominator cancel. Observe that x2 + y2 ≥

x2 ≥ 0, so x2 y− x3

x2 + y2

≤x2 y− x3

x2

= | y− x|

and we’d like use the squeeze theorem as in solution 1, except the reasoning in

the line above only makes sense when x̸ = 0, since we have x2 in the denominator

at one point. (This is the main drawback of this method.) We need to check

separately that, when x = 0 (and y ̸= 0),

x2 y−x3

x2

+ y2

is still bounded by | y− x|.But this is easy

02 y− 03

02 + y2

= 0 ≤ | y− x|

so we indeed have

−| y− x| ≤ x2 y− x3

x2 + y2 ≤ | y− x|

for all (x, y) ̸= (0, 0) (ie the inequality holds at all points where the fraction is

defined), and we can finish the problem as in solution 1.

Solution 2a: Here’s a nifty rewrite from my fellow TA Rebecca, which bypasses

the above case-checking. Since x2 + y2 ≥ x2 ≥ 0,

0 ≤ x2

x2 + y2 ≤ 1

for all (x, y)̸ = (0, 0). Multiplying through by | y− x| gives

0 ≤

x2 y− x3

x2 + y2

≤ | y− x|

and now we continue as before.

Solution 3: use polar coordinates. (as suggested by Matthew, and with help

from Sharad). Polar coordinates is not a substitute for the squeeze theorem,

merely a way to find squeezing functions easily in the very special case where the


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denominator has the formx2 + y2

nfor some n.

x2 y− x3

x2 + y2

=

r3 cos2 θ sin θ − cos3 θ

r2 cos2 θ + sin2 θ

=

r cos2 θ (sin θ − cos θ)

Now | sin θ − cos θ| ≤ 2 and | cos θ| ≤ 1, so the right hand side ≤ 2|r|. Hence

−2r ≤ x2 y− x3

x2 + y2 ≤ 2r|

As limr→0 2r = 0, we can use the squeeze theorem to conclude that lim(x, y)→(0,0)x2 y−xy2

x2+ y2 =

0.

It’s important that your squeezing function has the same limit as r → 0 irre-

spective of how θ changes on this path. It’s not enough for this limit to exist forevery fixed value of θ.

Amy Pang, 2010

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Squeeze Theorem Application