Why Absolute Values Help with Oscillating Expressions
Trigonometric expressions can keep changing as their inputs vary. For example, \(\sin(1/x)\) oscillates as \(x\) approaches zero, so its values do not settle toward one number. Taking the absolute value changes the signs but does not stop the oscillation: \(|\sin(1/x)|\) still varies between \(0\) and \(1\).
A limit can nevertheless exist when an oscillating term is multiplied by a factor that shrinks to zero. The key is to use bounds that are valid regardless of the oscillation. Absolute values are especially useful because they give a nonnegative lower bound, while familiar trigonometric inequalities provide upper bounds.
The Bounds and When to Use Them
The first bound, \(|\sin u|\leq1\), holds for every real \(u\). It is useful when the sine input may oscillate or may not approach a finite value. If a nonnegative factor multiplying \(|\sin u|\) approaches zero, the product is trapped between zero and that factor.
The second bound, \(|\sin u|\leq|u|\), is useful when the sine input approaches zero. It follows from the standard local sine bound; in particular, for \(u\) near zero, \(|\sin u|\leq|u|\). The absolute values on both sides matter: the inequality must remain valid whether \(u\) is positive or negative.
The Squeeze Theorem says that if \(L(x)\leq F(x)\leq U(x)\) for all \(x\) sufficiently close to \(a\), except possibly at \(a\), and if both \(L(x)\) and \(U(x)\) approach the same finite number \(M\) as \(x\to a\), then \(F(x)\to M\). In the examples below, nonnegativity often supplies the lower bound \(0\).
A General Squeezing Pattern
Suppose \(A(x)\geq0\) near \(a\), and \(g(x)\to0\) as \(x\to a\). For \(x\) close enough to \(a\), \(g(x)\) is close enough to zero for the local sine bound to apply. Multiplying that bound by \(A(x)\) preserves the inequality because \(A(x)\) is nonnegative:
Thus, if \(A(x)|g(x)|\to0\), the Squeeze Theorem gives \(A(x)|\sin(g(x))|\to0\). This conclusion depends on the product \(A(x)|g(x)|\) tending to zero; the bound alone does not guarantee that every expression containing \(|\sin(g(x))|\) has limit zero.
Worked Examples
Worked Example: A Shrinking Factor Times a Rapid Oscillation
Evaluate \(\displaystyle\lim_{x\to0}x^2\left|\sin\left(\frac1x\right)\right|\).
The sine input \(1/x\) does not approach a finite value as \(x\to0\), so a bound that requires the input to be near zero is not appropriate. Instead, use the global bound \(|\sin u|\leq1\). Since \(x^2\geq0\), multiplication preserves the inequality:
As \(x\to0\), the lower bound \(0\) approaches \(0\), and the upper bound \(x^2\) also approaches \(0\). Therefore, by the Squeeze Theorem,
Check: the original expression is nonnegative, and its value never exceeds \(x^2\). Since \(x^2\) becomes arbitrarily small near zero, the expression is forced to approach zero even though the sine factor keeps oscillating.
Worked Example: A Sine Input That Approaches Zero
Evaluate \(\displaystyle\lim_{x\to0}\frac{|\sin(x^2)|}{|x|}\).
Here \(x^2\to0\), so the local bound \(|\sin u|\leq|u|\) applies with \(u=x^2\) for \(x\) sufficiently close to zero. For \(x\ne0\), \(|x|>0\), so dividing the nonnegative inequality by \(|x|\) preserves its order:
The equality \(x^2/|x|=|x|\) holds for every \(x\ne0\). Both the lower bound \(0\) and upper bound \(|x|\) approach zero as \(x\to0\). Therefore,
Check: for nonzero \(x\), the expression is nonnegative and at most \(|x|\). Values bounded above by a quantity approaching zero must themselves approach zero.
Worked Example: An Absolute-Value Sine Ratio
Evaluate \(\displaystyle\lim_{x\to0}\frac{|\sin(4x)|}{|x|}\).
For \(x\ne0\), rewrite the quotient so that the standard sine limit \(\lim_{t\to0}\sin(t)/t=1\) applies. Since \(|4x|=4|x|\) and \(4>0\),
As \(x\to0\), the input \(4x\to0\), so \(\sin(4x)/(4x)\to1\). The absolute value function is continuous, so the absolute value of this ratio also approaches \(|1|=1\). It follows that
Check: the equivalent expression is \(4\left|\sin(4x)/(4x)\right|\); the ratio inside the absolute value approaches \(1\), confirming the limit \(4\). The absolute values are essential for the expression to stay nonnegative on both sides of zero.
Why the Bounds Work
The Squeeze Theorem applies because the bounds control every possible value of the oscillating expression, not just selected values. Here is the argument for the general pattern in a limit at \(x=a\). Assume \(A(x)\geq0\), \(g(x)\to0\), and \(A(x)|g(x)|\to0\). By the local sine bound, there is a neighborhood of \(a\) in which
The left inequality follows from nonnegativity of absolute values and \(A(x)\). The right inequality follows from \(|\sin(g(x))|\leq|g(x)|\), multiplied by the same nonnegative \(A(x)\). Both outside expressions approach zero, so the Squeeze Theorem forces the middle expression to approach zero as well. No claim about the individual oscillating values is needed.
Use \(|\sin u|\leq1\) for arbitrary or rapidly oscillating inputs; use \(|\sin u|\leq|u|\) when the input approaches zero.
Check that any factor used to multiply or divide the bounds is nonnegative in the neighborhood under consideration.
Apply the Squeeze Theorem only when the lower and upper bounds approach the same value.
A Common Sign Error
A frequent mistake is to remove absolute values from the local sine bound and write \(|\sin x|\leq x\) for inputs that may be negative. This is false. For example, if \(x=-0.1\), then \(|\sin x|>0\), while \(x=-0.1<0\); a positive number cannot be less than or equal to a negative one.
The correct inequality is \(|\sin x|\leq|x|\) near zero. For a limit involving a quotient, check the sign of the denominator before multiplying or dividing an inequality. For instance, with \(|x|\) in the denominator, the denominator is positive whenever \(x\ne0\), so division preserves the inequality. Replacing \(|x|\) by \(x\) without considering the side of approach can change the sign and invalidate the argument.
Choosing Between the Two Sine Bounds
The bound \(|\sin u|\leq1\) and the local bound \(|\sin u|\leq|u|\) answer different questions. The first limits the size of a sine value no matter how large or erratic its input becomes. The second can give a sharper estimate, but it requires the input to be close to zero. Choosing the wrong one can leave an expression uncontrolled or use a bound outside the conditions where it is valid.
| Situation | Useful estimate | What to verify |
|---|---|---|
| The sine input may oscillate without approaching zero | \(0\leq|\sin u|\leq1\) | The factor multiplying \(|\sin u|\) is nonnegative and has a useful limit. |
| The sine input approaches zero | \(0\leq|\sin u|\leq|u|\), locally | The input really does approach zero, and any multiplying factor preserves the inequality. |
| An absolute-value ratio has a scaled sine input | Rewrite using \(\sin t/t\to1\) | The scaled input approaches zero and the absolute-value factors are handled consistently. |
Check Your Understanding
For each problem, decide which estimate or limit applies, and show how the bounds lead to the conclusion.
- Evaluate \(\displaystyle\lim_{x\to0}|x|\left|\sin\left(\frac{2}{x}\right)\right|\).
- Use a local sine bound to evaluate \(\displaystyle\lim_{x\to0}\frac{|\sin(3x^2)|}{|x|}\).
- Evaluate \(\displaystyle\lim_{x\to0}\frac{|\sin(5x)|}{|x|}\) using the standard sine limit.
- Explain why \(|\sin u|\leq u\) is not a valid bound for all \(u\) near zero, and state the corrected bound.
- In the general estimate \(0\leq A(x)|\sin(g(x))|\leq A(x)|g(x)|\), what conditions make the inequality valid and allow the Squeeze Theorem to establish a zero limit?
