Oscillation Does Not Prevent a Limit
The function \(\sin(1/x^q)\) oscillates increasingly rapidly as \(x\) approaches zero. Its values keep moving between \(-1\) and \(1\), so the sine factor by itself does not settle toward a single value. But in \[ |x|^p\sin(1/x^q), \] the oscillation is multiplied by \(|x|^p\), which becomes small when \(p>0\). The key question is whether this shrinking factor controls the oscillation. It does: the product approaches zero for every \(p>0\) and \(q>0\).
There is a domain detail when \(p\) and \(q\) are arbitrary positive real numbers. The expression \(x^q\) is always real for \(x>0\), so the result below always gives the right-hand limit. If \(x^q\) is also real for negative \(x\), the same argument gives the two-sided limit. In particular, this applies when \(q\) is a positive integer.
The Power Envelope
A bound that contains an oscillating function but not its detailed behavior is called an envelope for the function. Here the sine factor stays between \(-1\) and \(1\), so multiplying by the nonnegative quantity \(|x|^p\) gives
The two outer expressions approach zero when \(p>0\). Thus, by the Squeeze Theorem, the middle expression also approaches zero. Notice what does—and does not—matter: \(p\) determines whether the envelope shrinks to zero, while \(q\) changes how quickly the sine oscillates but does not change the bound.
Why the Squeeze Works
The Squeeze Theorem says that if \(g(x)\leq f(x)\leq h(x)\) for all \(x\) sufficiently close to (but not equal to) zero, and if \(g\) and \(h\) both approach the same number \(L\), then \(f(x)\) also approaches \(L\). Here the relevant bounds are \(-|x|^p\) and \(|x|^p\), and both approach zero. The argument does not require the middle function to be monotone or to become less oscillatory.
For every real input \(u\), \(-1\leq\sin u\leq 1\).
Because \(|x|^p\geq0\), multiplication preserves the inequalities: \(-|x|^p\leq |x|^p\sin(1/x^q)\leq |x|^p\).
For \(p>0\), \(|x|^p\to0\), so both outer expressions tend to zero.
The bounded middle expression must tend to zero as well.
This is a complete justification: the bound holds throughout the punctured domain, and both bounds have the same limit. It also explains why no special estimate for the rapidly changing input \(1/x^q\) is needed.
Worked Examples
Worked Example: A Quadratic Envelope
Find \(\displaystyle\lim_{x\to0} |x|^2\sin(1/x^3)\). The expression is real on both sides of zero because the exponent \(3\) is an integer.
For every \(x\neq0\), \(-1\leq\sin(1/x^3)\leq1\). Since \(|x|^2\geq0\), multiply all three parts by \(|x|^2\):
Because \(|x|^2\to0\), the left bound tends to \(-0=0\), and the right bound tends to \(0\). The Squeeze Theorem therefore gives \[ \lim_{x\to0}|x|^2\sin(1/x^3)=0. \] Check: the original expression lies between two quantities that both approach zero, so its limit must equal their common limit, \(0\).
Worked Example: A Fractional Power with a Faster Oscillation
Find \(\displaystyle\lim_{x\to0^+}x^{1/2}\sin(1/x^4)\). The fractional power is real for \(x>0\), and the fourth power in the denominator makes the sine oscillate rapidly.
Use \(-1\leq\sin(1/x^4)\leq1\). Since \(x^{1/2}>0\) for \(x>0\), multiplication gives
As \(x\to0^+\), the two bounds approach zero: \[ \lim_{x\to0^+}(-x^{1/2})=0, \qquad \lim_{x\to0^+}x^{1/2}=0. \] Hence \[ \lim_{x\to0^+}x^{1/2}\sin(1/x^4)=0. \] Check: the factor \(x^{1/2}\) shrinks to zero, while the sine remains between \(-1\) and \(1\); the displayed bounds verify that their product is squeezed to zero.
Worked Example: A Different Positive Power and Oscillation Rate
Find \(\displaystyle\lim_{x\to0}|x|^{3/2}\sin(1/x^2)\). Both powers are chosen so the expression is real for \(x\neq0\).
For all such \(x\), \(-1\leq\sin(1/x^2)\leq1\). Multiplying by the nonnegative factor \(|x|^{3/2}\) yields
Since \(3/2>0\), \(|x|^{3/2}\to0\). Therefore the lower and upper bounds both approach zero, and \[ \lim_{x\to0}|x|^{3/2}\sin(1/x^2)=0. \] Check: the sine cannot have magnitude greater than one, so the absolute value of the entire expression is at most \(|x|^{3/2}\), which tends to zero. This confirms the result directly from the same envelope.
Which Exponent Controls the Limit?
It is useful to separate the two jobs played by the exponents. The positive exponent \(p\) makes the envelope vanish. The exponent \(q\) controls the input to the sine: a larger \(q\) generally makes the oscillations occur more rapidly near zero, but it cannot make the sine's magnitude exceed one. Consequently, no comparison such as \(p>q\) is required.
| Part of the expression | Its role in the argument |
|---|---|
| \(|x|^p\), with \(p>0\) | Provides an envelope that approaches zero. |
| \(\sin(1/x^q)\) | Oscillates, but always stays between \(-1\) and \(1\). |
| \(q>0\) | Determines the oscillation rate; it does not affect the envelope bound. |
The condition \(p>0\) is essential to this particular conclusion. If \(p=0\), the envelope no longer shrinks. For example, for \(q>0\), take \[ x_n=(\pi/2+2\pi n)^{-1/q} \quad\hbox{and}\quad y_n=(3\pi/2+2\pi n)^{-1/q}, \] where \(n\) is a positive integer. Both sequences are positive and approach zero, while \[ \sin(1/x_n^q)=1 \quad\hbox{and}\quad \sin(1/y_n^q)=-1. \] Thus \(\sin(1/x^q)\) has no right-hand limit at zero. This comparison emphasizes why the shrinking power, rather than the oscillation rate, is decisive.
A Common Mistake: Using a Local Bound at a Large Input
A tempting but incorrect move is to use the local sine estimate \(|\sin u|\leq |u|\) with \(u=1/x^q\), and then infer that the expression is bounded by a power like \(|x|^{p-q}\). The local sine estimate is useful when its input is close to zero. Here, as \(x\to0\), the input \(1/x^q\) grows in magnitude rather than approaching zero. More importantly, \(|\sin u|\leq|u|\) is a poor bound for large \(|u|\); it does not show that the sine factor is small.
Even if the displayed inequality happens to hold, the resulting estimate may fail to approach zero when \(p\leq q\). That failure is a failure of the chosen estimate, not evidence that the limit is nonzero. The reliable bound is \(|\sin u|\leq1\), which is valid for every real \(u\):
Since \(p>0\), this upper bound tends to zero. The correction is to choose a bound suited to the actual input: it is rapidly varying and large near zero, so use the global bound on sine rather than a local estimate near input zero.
Using the Result Reliably
For an expression of this form, keep the reasoning short and precise. First confirm that the function is defined on the side or sides from which the limit is taken. Next use \(|\sin u|\leq1\) to bound the absolute value of the product. Finally, use \(p>0\) to conclude that the power envelope tends to zero. The value of \(q\) need not be compared with \(p\).
Check Your Understanding
For each question, identify the bound that controls the oscillation and check the domain before deciding whether the limit is one-sided or two-sided.
- Use a squeeze argument to find \(\displaystyle\lim_{x\to0}|x|^{1/3}\sin(1/x^2)\).
- For \(\displaystyle x^4\sin(1/x^7)\), which exponent controls whether the power envelope vanishes, and which controls the oscillation rate?
- Why does the estimate \(|\sin(1/x^q)|\leq |1/x^q|\) not provide a useful vanishing bound near zero?
- For arbitrary positive real \(q\), why is it safest to state the result as a right-hand limit unless the expression is known to be real-valued for negative \(x\)?
- What changes in the envelope argument if \(p=0\)?
