Why the Choice of Bounds Matters
An expression can keep oscillating as \(x\) approaches a point, even when the whole expression has a limit. The key is often to separate the oscillating factor from the part that shrinks.
Since sine and cosine always stay between \(-1\) and \(1\), they cannot grow without bound. If an oscillating factor is multiplied by a quantity that approaches zero, a suitable bound can show that the product is forced toward zero. The important work is choosing a bound that is both valid and narrow enough to be useful.
Definitions and the Squeeze Theorem
For every real number \(u\), the trigonometric bounds are
Equivalently, \(|\sin u|\leq 1\) and \(|\cos u|\leq 1\). A second useful inequality, established in the earlier material on the local sine bound, is \(|\sin u|\leq |u|\). The first pair of bounds works for any argument, including an argument that grows without bound. The local sine bound can be sharper when the argument itself approaches zero.
The bounds must be in the correct order and must approach the same value. Merely knowing that an expression stays within a fixed interval does not establish its limit: the interval may remain too wide to force a unique value.
A Practical Method for Choosing Bounds
Begin by identifying which factors can oscillate and which factors can become small. Then choose a bound that preserves the order of the expression and has endpoint limits that agree. For an expression whose absolute value is easy to estimate, it is often simplest to find a nonnegative function \(B(x)\) such that \(|f(x)|\leq B(x)\). This immediately gives \(-B(x)\leq f(x)\leq B(x)\).
Use \(|\sin u|\leq 1\) or \(|\cos u|\leq 1\) when the argument may oscillate rapidly.
If a sine factor has an argument approaching zero, consider \(|\sin u|\leq |u|\).
When multiplying or dividing inequalities, check whether the factor is positive or negative. Absolute values can avoid reversing an inequality incorrectly.
Confirm that the lower and upper bounds approach the same finite number before applying the Squeeze Theorem.
Worked Examples
Worked Example: A Constant Plus a Bounded Oscillation
Find \(\lim_{x\to 0}\left(3+x^2\sin(1/x)\right)\).
The expression is defined for \(x\neq 0\). The argument \(1/x\) may oscillate rapidly, so use the bound \(-1\leq \sin(1/x)\leq 1\). Because \(x^2\geq 0\), multiplying throughout by \(x^2\) preserves the inequality:
Adding \(3\) to all three parts gives explicit bounds for the original expression:
As \(x\to 0\), \(x^2\to 0\), so both bounds approach \(3\):
The Squeeze Theorem therefore gives \(\lim_{x\to 0}\left(3+x^2\sin(1/x)\right)=3\). Checking the squeeze: the lower and upper bounds meet at the same limiting value, so the oscillatory term cannot prevent the expression from approaching \(3\).
Worked Example: Use a Local Sine Bound to Obtain a Power Bound
Find \(\lim_{x\to 0}\frac{\sin(x^2)}{x}\cos(1/x)\).
The expression is defined for \(x\neq 0\). The cosine factor has absolute value at most \(1\), and the sine factor has an argument \(x^2\) approaching zero. Apply the local sine bound and then the cosine bound:
The last equality uses \(|x^2|=x^2\) and \(x^2/|x|=|x|\) for \(x\neq 0\). The absolute-value estimate is equivalent to
Both bounds approach \(0\) as \(x\to 0\). Thus the limit is \(0\). As a check, the entire expression is trapped between two quantities whose distance from zero is \(|x|\), and that distance tends to zero.
Worked Example: Bound Each Term Before Combining Them
Find \(\lim_{x\to 0}\left(x^2\cos(1/x)+\frac{\sin(x^3)}{x}\right)\).
Both terms are defined for \(x\neq 0\). For the first term, use \(|\cos(1/x)|\leq 1\):
For the second term, apply \(|\sin u|\leq |u|\) with \(u=x^3\):
The triangle inequality now bounds the absolute value of the sum:
Consequently,
Since both \(-2x^2\) and \(2x^2\) approach \(0\), the limit is \(0\). The check is direct: the absolute value of the full expression is at most \(2x^2\), which tends to zero.
Why the Squeeze Works
The logic behind the Squeeze Theorem can be seen directly from the definition of a limit. Suppose \(g(x)\leq f(x)\leq h(x)\) near \(a\), and both bounds approach \(L\). Given any \(\varepsilon>0\), the limits of \(g\) and \(h\) ensure that sufficiently close to \(a\),
Combining these inequalities with \(g(x)\leq f(x)\leq h(x)\) gives
Therefore \(|f(x)-L|<\varepsilon\) for all sufficiently close \(x\neq a\). This is precisely what it means for \(f(x)\) to approach \(L\). The bounds leave no room for \(f(x)\) to approach a different value.
Choosing Between Constant and Power Bounds
A constant bound is useful for controlling an oscillating factor, but often it is not enough by itself. For example, \(|\cos(1/x)|\leq 1\) only traps \(\cos(1/x)\) between \(-1\) and \(1\); those endpoints do not approach the same number. In a product such as \(x^2\cos(1/x)\), the extra factor \(x^2\) turns that constant bound into the shrinking bounds \(-x^2\) and \(x^2\).
Power bounds are especially helpful near zero. If the estimate reduces to a positive power such as \(|x|^p\), where \(p>0\), then \(|x|^p\to 0\) as \(x\to 0\). A larger positive power may give a narrower bound, but it is not necessary to find the narrowest possible estimate. The bound only needs to be valid and have the required endpoint limit.
Common Pitfall: Multiplying by a Negative Quantity
Suppose you try to bound \(x\sin(1/x)\) by multiplying \(-1\leq \sin(1/x)\leq 1\) by \(x\). For \(x>0\), this gives \(-x\leq x\sin(1/x)\leq x\). But when \(x<0\), multiplying by \(x\) reverses the inequality signs; it is incorrect to use the same ordered bounds \(-x\leq x\sin(1/x)\leq x\) on both sides of zero.
Instead, use absolute values:
This gives the correctly ordered bounds for positive and negative \(x\):
Both bounds tend to \(0\), so the limit is \(0\). The correction works because \(|x|\geq 0\) regardless of the sign of \(x\), and \(|x|\to 0\) as \(x\to 0\).
Check Your Understanding
For each problem, identify bounds that are valid near the stated limit and explain whether their limits determine the result.
- Find \(\lim_{x\to 0}\left(5+x^4\cos(1/x^2)\right)\).
- Find \(\lim_{x\to 0}\frac{\sin(x^3)}{x^2}\sin(1/x)\).
- For \(\lim_{x\to 0}x\cos(1/x)\), write lower and upper bounds that work for both positive and negative \(x\).
- Why do the bounds \(-1\leq \sin(1/x)\leq 1\) alone fail to establish a limit for \(\sin(1/x)\) as \(x\to 0\)?
- In a proposed squeeze, what must be true of the limits of the lower and upper bounds to force a unique value?
