When an Oscillating Factor Is Multiplied by a Small Factor
As \(x\) approaches zero, the input \(1/x\) does not approach a finite real number. Instead, it becomes arbitrarily large in magnitude, and \(\sin(1/x)\) and \(\cos(1/x)\) continue to oscillate between \(-1\) and \(1\). Their values do not settle toward a single number.
A product can nevertheless have a limit. In expressions such as \(x\sin(1/x)\), the factor \(x\) becomes small while the sine factor remains bounded. The Squeeze Theorem lets us turn that observation into a proof.
The Bounds We Need
For every real number \(u\), sine and cosine satisfy
Equivalently, \(|\sin u|\leq 1\) and \(|\cos u|\leq 1\). These inequalities hold no matter how large or small \(u\) is. In particular, they still hold when \(u=1/x\), as long as \(x\neq 0\).
We will use the Squeeze Theorem in the following form: if \(g(x)\leq f(x)\leq h(x)\) for all \(x\) sufficiently close to \(a\), except possibly at \(a\), and if both \(g(x)\) and \(h(x)\) tend to the same finite number \(L\) as \(x\) tends to \(a\), then \(f(x)\) also tends to \(L\).
For oscillatory products at zero, absolute-value bounds often give the most direct squeeze. If we establish \(|f(x)|\leq b(x)\), where \(b(x)\geq 0\) and \(b(x)\to 0\), then
Since the lower and upper bounds both tend to zero, the Squeeze Theorem gives \(\lim_{x\to 0}f(x)=0\).
Worked Example: \(x\sin(1/x)\)
Worked Example: \(x\sin(1/x)\)
Evaluate \(\lim_{x\to 0}x\sin(1/x)\). The expression is defined for \(x\neq 0\), which is enough to consider its limit at zero.
First use \(|\sin u|\leq 1\) with \(u=1/x\):
Multiply both sides by the nonnegative quantity \(|x|\). This gives
Writing the absolute-value inequality as a two-sided bound,
As \(x\to 0\), both \(-|x|\) and \(|x|\) tend to zero. Therefore, the Squeeze Theorem gives
Check: The bound also verifies the limit directly in epsilon language. Given any \(\varepsilon>0\), choose \(\delta=\varepsilon\). If \(0<|x|<\delta\), then
This is exactly the condition for the product to tend to zero.
Why the Squeeze Works
The key step is not that the sine factor approaches a particular value; it does not. The key step is that its magnitude never exceeds \(1\). Consequently, the magnitude of the product is at most \(|x|\), and \(|x|\) can be made as small as desired by taking \(x\) close enough to zero.
More explicitly, for any \(\varepsilon>0\), the condition \(0<|x|<\varepsilon\) ensures
Thus the product lies within \(\varepsilon\) of zero. This explains why the Squeeze Theorem applies: the product is trapped between bounds whose distance from zero shrinks with \(x\).
Worked Example: \(x^2\cos(1/x)\)
Worked Example: \(x^2\cos(1/x)\)
Evaluate \(\lim_{x\to 0}x^2\cos(1/x)\). The expression is again considered only for \(x\neq 0\).
The cosine bound gives \(\left|\cos(1/x)\right|\leq 1\). Since \(x^2\geq 0\), multiplication by \(x^2\) preserves the inequality
The lower bound tends to zero because \(-x^2\to 0\), and the upper bound tends to zero because \(x^2\to 0\). Hence
Check: The inequality gives
Given \(\varepsilon>0\), take \(\delta=\sqrt{\varepsilon}\). If \(0<|x|<\delta\), then \(x^2<\delta^2=\varepsilon\), so the absolute difference from the claimed limit is less than \(\varepsilon\). This verifies the result.
Worked Example: A Product of Sine and Cosine
Worked Example: \(x\sin(1/x)\cos(1/x)\)
Evaluate \(\lim_{x\to 0}x\sin(1/x)\cos(1/x)\). Both trigonometric factors are bounded, so their product is bounded as well.
Using \(|\sin u|\leq 1\) and \(|\cos u|\leq 1\), we obtain
Multiplying by \(|x|\) gives
Equivalently,
Both bounds tend to zero as \(x\to 0\), so the Squeeze Theorem gives
Check: If \(0<|x|<\varepsilon\), then the absolute value of the product is at most \(|x|<\varepsilon\). Thus its distance from zero is less than \(\varepsilon\), as required.
Scaled Inputs and Coefficients
The same reasoning applies when the oscillating input is scaled. For example, consider \(3x\cos(5/x)\). The value \(5/x\) still gives a real input whenever \(x\neq 0\), so \(|\cos(5/x)|\leq 1\). Therefore,
Since \(3|x|\to 0\), the product tends to zero. The constant multiplier changes the size of the bound but does not prevent it from shrinking to zero.
The essential ingredients in these examples are simple: the trigonometric factor has a fixed bound, and the other factor makes the resulting bound tend to zero. A more complicated expression may require additional algebra to expose these ingredients, but the squeeze argument itself remains the same.
A Common Mistake: Substituting into the Oscillating Factor
A tempting but incorrect calculation is to substitute \(x=0\) into \(x\sin(1/x)\), write \(1/0=\infty\), and then try to evaluate \(0\sin(\infty)\). This is not valid: \(1/0\) is undefined, and \(\infty\) is not a real input at which sine can be evaluated. Direct substitution does not determine the limit.
Another incorrect conclusion is that the product has no limit because \(\sin(1/x)\) has no limit. The factor \(\sin(1/x)\) does oscillate without settling. For instance, as positive integers \(n\) grow, the inputs
both approach zero, while \(\sin(1/x_n)=1\) and \(\sin(1/y_n)=-1\). But for the product, the accompanying factor \(x\) shrinks: the absolute-value bound \(\left|x\sin(1/x)\right|\leq |x|\) forces the product toward zero.
A Reliable Procedure
For products of the kind considered here, organize the argument in three steps:
Use \(|\sin u|\leq 1\) or \(|\cos u|\leq 1\), including when \(u\) depends on \(x\).
Use absolute values and multiply by the magnitude of the other factor, taking care that the multiplier used in an inequality is nonnegative.
If the resulting nonnegative bound tends to zero, place the product between its negative and positive versions and apply the Squeeze Theorem.
This procedure is useful whenever a factor oscillates but stays within known bounds. It separates two questions that should not be confused: whether the oscillating factor itself has a limit, and whether the entire product has a limit.
Summary
The bounds \(|\sin u|\leq 1\) and \(|\cos u|\leq 1\) hold for every real input, including inputs such as \(1/x\) when \(x\neq 0\). They imply that \(x\sin(1/x)\), \(x^2\cos(1/x)\), and \(x\sin(1/x)\cos(1/x)\) are trapped between bounds that tend to zero. By the Squeeze Theorem, each product has limit zero at the origin.
Check Your Understanding
For each question, focus on a bound for the full product rather than trying to substitute into its oscillating factor.
- Use an absolute-value bound and the Squeeze Theorem to evaluate \(\lim_{x\to 0}x\cos(1/x)\).
- Why is it valid to use \(|\sin(1/x)|\leq 1\) even though \(1/x\) does not approach a finite value as \(x\to 0\)?
- Find a bound for \(\left|2x\sin(3/x)\right|\), and use it to evaluate its limit at zero.
- Explain why the fact that \(\sin(1/x)\) has no limit does not, by itself, show that \(x\sin(1/x)\) has no limit.
- For \(x^2\cos(1/x)\), state a lower and an upper bound that have the same limit as \(x\to 0\).
