When Nearby Values Keep Varying
A function can stay within a fixed range near an input and still fail to have a limit there. The difficulty is not necessarily that its values become large. Instead, the values may continue to swing between different levels, no matter how close the input gets to the point.
The function \(f(x)=\sin(1/x)\), considered for \(x\ne 0\), is a standard example near zero. As positive \(x\) approaches zero, \(1/x\) grows without bound. The sine function therefore keeps cycling through its full range rather than settling at one output value.
Recall the intuitive meaning of a limit: if \(\lim_{x\to a}f(x)=L\), then by taking \(x\) sufficiently close to \(a\), the function values can be made as close as desired to \(L\). In particular, for any fixed small tolerance, all function values sufficiently near \(a\) must lie within that tolerance of \(L\). Persistent oscillation can contradict this requirement.
Definition: Persistent Oscillation Near an Input
“Bounded” means that the function’s values stay between some fixed lower and upper bounds. “Approaches a limit” means that the values eventually gather as close as desired to one number. These are different properties: bounded values can keep oscillating across a range and never gather near one number.
Worked Example: The Limit of \(\sin(1/x)\) at Zero
Worked Example: The Limit of \(\sin(1/x)\) at Zero
Determine whether \(\lim_{x\to 0}\sin(1/x)\) exists.
For positive integers \(n\), consider the inputs
As \(n\) increases, both denominators grow without bound, so both inputs become arbitrarily close to zero from the right. At the first input, the reciprocal is \(\frac{\pi}{2}+2\pi n\), an angle where sine equals \(1\):
At the second input, the reciprocal is \(\frac{3\pi}{2}+2\pi n\), an angle where sine equals \(-1\):
Thus values arbitrarily close to the input zero include both \(1\) and \(-1\). These outputs remain two units apart. They cannot both be arbitrarily close to one proposed limit. Therefore \(\lim_{x\to 0}\sin(1/x)\) does not exist.
Check: The inputs are positive and approach zero, so this argument already rules out a two-sided limit. It also shows that the right-hand limit fails. The function is bounded, since \(-1\leq \sin(1/x)\leq 1\), but boundedness does not make its values settle at one number.
Why Two Persistent Values Rule Out a Limit
The reasoning in the example is general. Suppose there are inputs arbitrarily close to \(a\) where the function takes the value \(A\), and also inputs arbitrarily close to \(a\) where it takes the value \(B\), with \(A\ne B\). If a limit \(L\) existed, function values near \(a\) would all have to become close to \(L\). But \(A\) and \(B\) cannot both be arbitrarily close to the same \(L\) while remaining a fixed distance apart.
For the sine example, the separation is \(|1-(-1)|=2\). If both values were within \(1/2\) of \(L\), the triangle inequality would give
which is impossible. Any proposed limit would require all sufficiently nearby function values to be within \(1/2\) of it. Yet arbitrarily close to zero there are inputs producing both \(1\) and \(-1\). This contradiction proves that no limit exists.
This argument explains the essential feature: the outputs do not merely vary; their variation does not die away as the input approaches the point. A graph may help reveal the repeated swings, but the proof rests on the function’s values at inputs as close as needed, not on a picture’s apparent shape.
Worked Example: A Shifted Oscillation
Worked Example: A Shifted Oscillation
For \(x\ne 0\), let \(g(x)=2+\sin(1/x)\). Determine whether \(g\) has a limit as \(x\to 0\).
Use the same kinds of inputs as before. At \(x_n=1/(\frac{\pi}{2}+2\pi n)\), the sine value is \(1\), so
At \(y_n=1/(\frac{3\pi}{2}+2\pi n)\), the sine value is \(-1\), so
Both kinds of input occur arbitrarily close to zero, while the corresponding outputs remain \(3\) and \(1\), a distance of \(2\) apart. If there were a limit \(L\), both values would need to become close to \(L\) near zero. Taking a tolerance of \(1/2\) would imply \(2=|3-1|\leq |3-L|+|L-1|<1\), a contradiction. Therefore \(\lim_{x\to 0}g(x)\) does not exist.
Check: Adding \(2\) shifts every output upward by \(2\), but does not reduce the separation between the recurring values. The range stays bounded between \(1\) and \(3\), and the limit still fails to exist.
Oscillation Does Not Always Prevent a Limit
Oscillation is a warning sign, but it does not automatically rule out a limit. What matters is whether the size of the oscillation remains substantial near the input. If the oscillation is multiplied by a factor that shrinks to zero, the whole function may still approach one value.
Worked Example: Oscillation with Shrinking Size
Define \(h(0)=0\), and for \(x\ne 0\) define \(h(x)=x\sin(1/x)\). Determine the limit as \(x\to 0\).
The sine function always satisfies \(-1\leq \sin(1/x)\leq 1\), so \(|\sin(1/x)|\leq 1\). Multiplying by \(|x|\) gives
As \(x\) approaches zero, the upper bound \(|x|\) becomes arbitrarily small. Consequently the distance from \(h(x)\) to \(0\) also becomes arbitrarily small, and \(\lim_{x\to 0}x\sin(1/x)=0\).
Check: For example, if \(|x|<0.01\), then \(|h(x)-0|\leq |x|<0.01\), regardless of the value of the sine factor. The oscillation remains in the factor \(\sin(1/x)\), but its effect on the product is reduced by \(x\).
A Common Mistake: Bounded Does Not Mean Convergent
A common incorrect argument is: “\(\sin(1/x)\) is always between \(-1\) and \(1\), so it must have a limit.” This confuses staying within bounds with approaching a single number. A bounded function might converge, but boundedness alone does not establish convergence.
For this particular function, the outputs \(1\) and \(-1\) recur arbitrarily close to zero. Their fixed separation shows that they do not gather near one number. The correction is to ask whether nearby outputs become close to one another and to a candidate limit—not merely whether they remain within some overall range.
Another tempting but invalid calculation is to write \(\sin(1/x)\to\sin(\infty)\). Infinity is not a real input at which the sine function can be evaluated. More importantly, \(1/x\) does not approach a particular angle: it passes through angles giving many different sine values repeatedly. The recurring values, rather than a substitution involving infinity, explain the failure.
How to Recognize This Type of Failure
When investigating a suspected oscillatory failure, look for two output levels that stay separated and keep occurring arbitrarily near the input in question. For a trigonometric expression, identify angles where the function takes convenient extreme values, then solve for inputs that produce those angles. Check that the resulting inputs can be made as close as needed to the point.
Find two distinct values that the function attains near the input, preferably values with a clear fixed separation.
Show that inputs producing each value can be taken arbitrarily close to the point of interest.
A limit would require all sufficiently nearby function values to be close to one number, which the separated recurring outputs prevent.
This approach is useful because a graph can conceal rapid oscillations when its scale is too coarse. Exact input choices can establish that the variation persists even when the graph is difficult to read. Conversely, an inequality such as \(|x\sin(1/x)|\leq |x|\) can show that oscillation is becoming too small to prevent a limit.
Check Your Understanding
Use the idea of recurring, separated output values to answer these questions.
- Why does the fact that \(\sin(1/x)\) stays between \(-1\) and \(1\) fail to prove that it has a limit at zero?
- Which inputs make \(\sin(1/x)=1\) in the first worked example, and why can they be made arbitrarily close to zero?
- What contradiction follows if a proposed limit for \(\sin(1/x)\) requires both \(1\) and \(-1\) to be within \(1/2\) of it?
- Why does adding \(2\) to \(\sin(1/x)\) leave the limit failure unchanged?
- How does the bound \(|x\sin(1/x)|\leq |x|\) explain why the product has limit zero?
