Section 2.6: Putting the Squeeze on a Limit © 2001 by Karl Hahn

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An Arabian Night

"Long ago a merchant's son set off for Cairo to trade for goods at the bazaar. When after a week he did not return, his sister grew worried, and when after two weeks he did not return, she became distraught and could not sleep. So on the fifteenth day she sought advice from the old hermit who lived far outside of town.

"'What has become of my brother,' she asked him.

"The old man threw a handful of seeds on his divining board, examined them carefully, then spoke. 'Your brother has met with a bad end,' he said. 'On the road to Cairo he encountered an evil sorcerer, disguised as another traveling merchant, who lured your brother far from the road under the ruse that your brother might trade for goods the sorcerer had in his treasure-house. When your brother became thirsty, the sorcerer offered him drink, but with the first swallow your brother was transformed into a single grain of black sand lying in the middle of the wide desert, whereupon the thieving sorcerer made off with all your brother's gold.'

"The young woman burst into tears, and try as he might, the old man was unable to console her. When she paused for a moment in her crying he attempted to offer her hope by saying, 'All is not lost, my child. If you can recover that black sand-grain from the desert and cast it into the fountain at the caliph's palace, your brother will be restored and the evil-doer will be revealed.'

"But this only renewed her torrent of tears. At last she caught her breath and sobbed, 'And how am I to find a single grain of sand in the vastness of a trackless desert? There is no hope at all.'

"The hermit reached into his basket and extracted two porcelain saucers. 'Take these,' he said. 'Ride into the desert and throw them before you, one to your right, the other to your left. They will roll, and you are to ride with them, always keeping the one to your right and the other to your left. Watch only the saucers and fear nothing, for they will lead you to that single grain of sand which is your brother.'

"The young woman rose the next morning hours before the sun. While her father still slept, she dressed herself in her brother's clothes, mounted her father's horse, and rode to the edge of town. There she grasped the saucers, one in each hand, and threw them to the ground, one to the right and one to the left, just as the hermit had instructed. At first they rolled so far apart she could barely keep track of them both. But determined as she was, she kept the horse between them by riding at once far to the right to spot the one saucer, and later far to the left to spot the other. Back and forth she rode, always riding between them. Hour after hour the two saucers continued to roll, leading her deeper and deeper into the desert. As the day wore on, though, the saucers came closer and closer together. By noon they were only a few strides apart. By late afternoon there was barely enough room between them for the horse. As the sun sank low, she had to dismount and continue on foot, for the horse would no longer fit between the two saucers. When the sun touched the horizon, the saucers were too close together to walk between, and their rolling slowed. Now she had to follow just behind them. She found she was in a broad expanse of white sand. Soon the saucers were only a palm-width apart. She dropped to her hands and knees and crawled after them, always staring at the strip of white sand between them. As the sun disappeared, they were only a finger's width apart. In another minute only a blade of grass would have fit between them. And just when only the width of a hair would fit between them, there in the last light of day, with her eye just above the two saucers, she saw the single grain of black sand alone in the sea of white grains she had come through.

"Eagerly she drew a spoon from under her cloak, scooped up the bit sand that contained the black grain, and poured it into a tiny urn that hung from a gold chain around her neck."

Shaherezade smiled. "The sun has risen my love," she said. "You will have to wait until tomorrow evening to hear how this resourceful girl survived the perils of the desert that night and then found her way back to town, and of what befell the evil sorcerer."

Neither saucer ever touched the grain of black sand. Yet because they both tended to the same limit, the young woman traveling always between them was inexorably led to the same limit they were going to. Her wandering path was gradually squeezed between the two saucers until, in the end, both they and she converged on that one grain of sand among all the others in the great desert. This is the essence of the squeeze theorem.

So if I tell you, for example, that some  f(x)  always lies between  -1/x  and  1/x,  what can you say about the limit of  f(x)  as x goes to infinity? For positive x then, it is always true that

     1              1
   -    <  f(x)  <                                                eq. 2.6-1
     x              x

    lim   -
   x → ∞

 1
    =  0                                               eq. 2.6-2a
 x

    lim
   x → ∞

  1
     =  0                                                eq. 2.6-2b
  x

As close as you'd like  -1/x  and  1/x  to be to zero, you can always find an x large enough to make them that way. That is what going to a limit means. Of course  -1/x  will always be less than zero and  1/x  always greater than zero. And still  f(x)  must remain between them all of the way. Under these circumstances it ought to be clear that wherever the two saucers,  -1/x  and  1/x,  are going,  f(x)  must go there too.

So here is the squeeze theorem in math-speak:

If  g(x) ≤ f(x) ≤ h(x),  and both  g(x)  and  h(x)  go to the same limit, L, as x approaches a, then the limit of  f(x)  as x approaches a must be L also.

Here are some notes to go with that:

1. a can be any real number, or it can be +∞ or -∞.
2. If a is a real number (that is, it's finite) then the region over which  g(x) ≤ f(x) ≤ h(x)  must hold is throughout some open interval that contains a (remember that an open interval is one that does not contain its endpoints). That interval can be as small as you like, just as long as the condition holds over some open interval with a in it.
3. If a is -∞ then the region over which  g(x) ≤ f(x) ≤ h(x)  must hold is from some arbitrary place on the real number line going to the left forever. That is on some half-open interval that extends to -∞.
4. If a is +∞ then the region over which  g(x) ≤ f(x) ≤ h(x)  must hold is from some arbitrary place on the real number line going to the right forever. That is on some half-open interval that extends to +∞.

Let's take a popular example. Let  f(x) = x² sin(1/x).  Clearly you can't evaluate this function at  x = 0  because you'd get a division by zero. But what is

lim x2 sin x → 0

1   x

eq. 2.6-3

The graph on the right shows a closeup of this  f(x),  enveloped above and below by  x2  and  -x2  respectively.

Observe how, like the Arabian woman's ride, this function oscillates between the edges of the envelope, but is confined by that same envelope to be closer and closer to zero as x goes toward zero.

This behavior should not surprise you. Recall from previous math classes that you've had that it is always the case that  -1 ≤ sin(x) ≤ 1.  So it must also be true that  -x² ≤ x²sin(x) ≤ x².  Observe that both  -x²  and  x²  go to zero as x goes to zero. So the squeeze theorem clearly indicates that  x²sin(x)  must do so also. It has to. It is being squeezed from both sides by functions that also go to zero. It has to stay between them. Where else can it go but to the same place  -x²  and  x² go?

Can we prove the squeeze theorem using a delta-epsilon contract? We start with an  f(x)  that always lies between  g(x)  and  h(x)  in our region of interest around  x = a.  We also stipulate that both  g(x)  and  h(x)  go to some limit, L, as x goes to a. We have to prove that  f(x)  must also go to L in the limit as x goes to a. So if I gave you some  ε > 0,  and told you to find the δ so that whenever  |x - a| < δ,  it would always be the case that  |f(x) - L| < ε,  what recipe would you use to find a δ that would guarantee that contract?

The key here is that both  g(x)  and  h(x)  already have such recipes. Why? They have to because we stipulated that they both go to L in the limit, and that is what going to the limit means. So we choose a recipe from one of the two, and the one we choose is the one that gives the smaller δ. We know that  f(x)  can't possibly go lower than  g(x)  (the lower edge of the envelope), and with x within our chosen δ of a,  g(x)  can't be any farther than ε from the limit. So  L - f(x)  can't be any more than ε. And you can make a complementary argument using  h(x)  (the upper edge of the envelope) to show that  f(x) - L  can't be any larger than ε. That clinches it. Bounded from both side as it is, that means that  |f(x) - L|  can't be any more than ε either.

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