Bounding a function you cannot simplify

Try

A quantity f(x) lies between −x² and x² whenever x is near zero. At x=0.1 it lies between −0.01 and 0.01. At x=0.01 it lies between −0.0001 and 0.0001. You do not know its exact value. Can you still know its limit?

Predict

Predict the difference between sin(1/x) and x² sin(1/x) near zero. The first oscillates increasingly rapidly. Does multiplying by x² remove oscillation, or does it shrink the range in which oscillation occurs? Those are different statements.

Experiment

Draw the two envelope curves y=x² and y=−x², marking the following vertical intervals. Any value of x² sin(1/x) must fit inside the interval because sine always lies between −1 and 1.

Distance	Upper envelope	Lower envelope	Maximum absolute output
1/2	1/4	−1/4	1/4
1/10	1/100	−1/100	1/100
1/100	1/10000	−1/10000	1/10000

The same bounds work for positive and negative x. To force the output distance from zero to be at most 0.0001, it suffices to make the input distance at most 0.01. This is a guarantee for every permitted input in that region, not just the sample rows.

Understand

The squeeze theorem says: if a lower function is no larger than f and f is no larger than an upper function throughout a punctured neighborhood of a, and both bounding functions approach the same number L, then f approaches L. A punctured neighborhood means all sufficiently nearby inputs except a itself. The bounds need not describe the value at a.

For x² sin(1/x), its absolute value is at most x². Both ±x² approach zero, so the limit is zero. For x sin(1/x), use the bound |x sin(1/x)|≤|x|. Writing −x≤x sin(1/x)≤x without checking the sign of x would fail for negative inputs.

Persistent oscillation alone has no limit. Along inputs 1/(π/2+2πn), sin(1/x)=1; along inputs 1/(3π/2+2πn), it equals −1. Here n is a positive integer growing without bound, so both input sequences approach zero. The conflicting outputs prevent a common limit. Multiplying by a shrinking amplitude changes this conclusion without making the underlying sine stop oscillating.

Bounds −1 and 1 do not squeeze a function to zero: their limits disagree. Nor do bounds valid only at a finite list of samples establish a neighborhood condition. Both mistakes substitute a suggestive picture for the theorem’s hypotheses.

Apply

In each practice argument, name the lower and upper bounds, identify the target input, and show that the two bounding limits agree. For a shifted bound such as |f(x)−5|≤3|x−2|, interpret the left side as output error relative to 5, not as the size of f itself.

Continue

Next, connect a limit to the actual function value through continuity. The squeeze theorem can establish a missing limit, but continuity additionally requires a defined value that matches it.