Oscillation Near a Finite Input
A function can remain bounded near an input and still fail to approach any single value. One way this happens is when its input to a periodic function changes without bound as \(x\) approaches a finite number. In expressions such as \(\sin\!\left(\frac{1}{x-a}\right)\), the distance \(x-a\) shrinks toward zero while its reciprocal grows in magnitude. The sine function then passes repeatedly through its full range.
The key question is not whether the function is bounded, but whether its values settle near one number. To answer that rigorously, we can find input values approaching \(a\) from one side for which the function repeatedly takes different exact values.
One-Sided Limits and Sequences
A finite right-hand limit \(L\) at \(a\) means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that whenever \(a<x<a+\delta\), the function is defined and \(|f(x)-L|<\varepsilon\). A finite left-hand limit uses \(a-\delta<x<a\) instead. In either case, the requirement is that all sufficiently nearby inputs from the specified side give outputs close to the same \(L\).
Here “from that side” means \(x_n>a\) for a right-hand limit or \(x_n<a\) for a left-hand limit. The theorem follows directly from the definition: given \(\varepsilon>0\), choose its corresponding \(\delta\). Since \(x_n\to a\), eventually \(0<|x_n-a|<\delta\), with \(x_n\) on the required side. The one-sided limit condition then gives \(|f(x_n)-L|<\varepsilon\) for all sufficiently large \(n\). That is exactly \(f(x_n)\to L\).
The contrapositive is especially useful: if we can exhibit two sequences approaching \(a\) from the same side whose function values approach two different numbers, there cannot be a one-sided limit. Both sequences would have to approach \(L\) if the limit existed, which is impossible when their limiting values differ.
Why the Reciprocal Sine Oscillates
The sine function repeats every \(2\pi\). In particular, \(\sin(\frac{\pi}{2}+2\pi n)=1\) and \(\sin(\frac{3\pi}{2}+2\pi n)=-1\) for every integer \(n\). These angles have magnitudes that grow as \(n\) grows. Taking their reciprocals produces positive distances that shrink toward zero.
For example, to make \(\sin\!\left(\frac{1}{x-a}\right)=1\) at inputs to the right of \(a\), choose \(x-a=\frac{1}{\frac{\pi}{2}+2\pi n}\). To make the sine equal \(-1\), choose \(x-a=\frac{1}{\frac{3\pi}{2}+2\pi n}\). Both distances tend to zero, so both sequences approach the same finite input \(a\) from the right, even though the function values stay separated.
Worked Example: A Right-Hand Limit That Does Not Exist
Consider \(f(x)=\sin\!\left(\frac{1}{x-2}\right)\) as \(x\to2^+\). The function is defined for \(x\ne2\), and the right-hand limit concerns inputs \(x>2\).
For each \(n\in\mathbb{N}\), define
Each denominator is positive, so \(x_n>2\) and \(y_n>2\). As \(n\to\infty\), both denominators grow without bound, so both reciprocal terms tend to zero. Therefore \(x_n\to2^+\) and \(y_n\to2^+\).
Substituting \(x_n\) into the function gives
Similarly,
Thus \(f(x_n)\to1\) while \(f(y_n)\to-1\). These are different limits for sequences approaching 2 from the same side. By the sequence test, \(\lim_{x\to2^+}f(x)\) does not exist. The exact substitutions also check that both sequences give the claimed sine values.
Oscillation From the Left
The same reasoning works on the left, but the reciprocal \(1/(x-a)\) is negative there. The angles used to obtain \(1\) and \(-1\) must therefore be negative. Sine is odd, so \(\sin(-\theta)=-\sin(\theta)\); keeping track of this sign prevents an otherwise easy error.
Worked Example: A Left-Hand Limit With an Affine Output
Let \(g(x)=4+2\sin\!\left(\frac{1}{x+1}\right)\), and consider \(x\to-1^-\). For \(n\in\mathbb{N}\), set
Both subtracted fractions are positive, so \(u_n<-1\) and \(v_n<-1\). Their denominators grow without bound, so \(u_n\to-1^-\) and \(v_n\to-1^-\).
For the first sequence, \(u_n+1=-\frac{1}{\frac{\pi}{2}+2\pi n}\), and hence
For the second, \(v_n+1=-\frac{1}{\frac{3\pi}{2}+2\pi n}\), giving
The sine values are \(-1\) and \(1\), respectively, so the corresponding outputs are exactly 2 and 6. Since these sequences approach \(-1\) from the left but their function values approach different numbers, \(\lim_{x\to-1^-}g(x)\) does not exist.
Bounded Does Not Mean Convergent
The sine function always lies between \(-1\) and \(1\). That bound does not imply that \(\sin\!\left(\frac{1}{x-a}\right)\) approaches a number: the examples show that it repeatedly reaches both endpoints arbitrarily close to \(a\). A function can be bounded without settling toward a limit.
There is an important contrast. If the oscillating factor is multiplied by a quantity that shrinks to zero, the whole expression may have a limit. The size of the oscillations then shrinks even though their frequency increases.
Worked Example: A Shrinking Amplitude Allows a Limit
Consider \(h(x)=(x-1)\sin\!\left(\frac{1}{x-1}\right)\) as \(x\to1^+\). The expression is defined for \(x\ne1\). Since \(-1\le\sin(t)\le1\) for every real \(t\), and \(x-1>0\) when \(x>1\), multiplying the inequality by \(x-1\) gives
As \(x\to1^+\), both the left and right bounds tend to zero. More directly, for any \(\varepsilon>0\), choose \(\delta=\varepsilon\). If \(1<x<1+\delta\), then
This verifies the definition of a right-hand limit with \(L=0\), so \(\lim_{x\to1^+}h(x)=0\). The bound checks the conclusion even though the sine factor itself continues to oscillate.
A Reliable Test for Oscillation
For an expression involving a reciprocal distance, the following approach keeps the reasoning organized:
For a right-hand limit, construct inputs greater than \(a\); for a left-hand limit, construct inputs less than \(a\).
For sine, convenient targets are \(1\) and \(-1\), obtained at angles \(\frac{\pi}{2}+2\pi n\) and \(\frac{3\pi}{2}+2\pi n\), with signs adjusted on the left.
Solve for \(x\) so that the reciprocal distance equals the chosen angle. Check that the resulting inputs are on the required side and approach \(a\).
Substitute each input into the full function. Different limiting output values prove that the one-sided limit does not exist.
Common Mistake: Treating Sine at Infinity as a Limit
A tempting but incorrect argument is: “As \(x\to a\), \(1/(x-a)\to\infty\), so \(\sin(1/(x-a))\to0\).” This is not a valid limit calculation. The notation \(\sin(\infty)\) does not represent a sine value, and sine does not approach zero as its input grows.
The correction is to inspect values the function actually takes. For instance, as \(x\to2^+\), the sequences in the first example make the sine equal to \(1\) and \(-1\) repeatedly. Those exact values contradict the requirement that all nearby outputs approach one common number. Also note that \(1/(x-a)\) changes sign across \(a\), so a left-hand argument must account for negative reciprocal values rather than copying a right-hand sequence unchanged.
Why the Method Matters
A graph can suggest rapid oscillation, but a sequence argument gives a precise reason that a one-sided limit fails. The method is especially useful when the oscillations become too frequent to assess from a sketch. It also separates two distinct questions: does the oscillating factor itself settle, and does the complete expression settle after any shrinking multiplier is included?
The decisive test is whether every input sequence approaching the point from the chosen side must produce function values approaching the same number. Two sequences with different limiting outputs are enough to rule out a limit. If instead the oscillation is multiplied by a factor whose magnitude shrinks, a direct bound may show that the complete function still approaches a finite value.
Check Your Understanding
Use exact sequences or a direct estimate to justify each answer.
- For \(f(x)=\sin\!\left(\frac{1}{x-3}\right)\), construct two sequences approaching 3 from the right that give function values 1 and \(-1\).
- Why do two sequences approaching \(a\) from the same side with different limiting function values rule out a one-sided limit?
- When \(x\to a^-\), what sign does \(1/(x-a)\) have, and why does that matter when choosing sine angles?
- Does boundedness of \(\sin\!\left(\frac{1}{x-a}\right)\) establish that it has a limit? Explain.
- Use the inequality \(|\sin(t)|\le1\) to show that \((x-4)\sin\!\left(\frac{1}{x-4}\right)\) tends to zero as \(x\to4^+\).
