Why Oscillation Needs Special Care
Some functions keep varying as the input grows, even though their values stay within a fixed range. The sine function is a familiar example: its values repeatedly move between \(-1\) and \(1\). This persistent variation can prevent a limit at infinity.
A bounded oscillating factor does not always prevent a limit, however. If it is multiplied by a factor that tends to zero, the product may tend to zero regardless of the oscillation. The key is to distinguish persistent variation from variation whose overall size is being reduced.
Limits at Infinity and Subsequences
A function \(f\) has limit \(L\) as \(x\) tends to positive infinity if, for every \(\varepsilon>0\), there is a number \(X\) such that \(x>X\) implies \(|f(x)-L|<\varepsilon\). In other words, all sufficiently far-out function values must be as close to \(L\) as desired.
One useful consequence of this definition is a test using sequences. If \(f(x)\) tends to \(L\), then along every sequence of inputs \(x_n\) that tends to positive infinity, the values \(f(x_n)\) must tend to \(L\). Therefore, to show that a proposed limit does not exist, it is enough to find two such sequences along which the function values approach different limits.
The test follows directly from the definition: if \(f(x)\to L\), then for each \(\varepsilon>0\), every sufficiently large input has function value within \(\varepsilon\) of \(L\). The inputs \(x_n\) and \(y_n\) eventually exceed that threshold, so both sequences of function values must tend to \(L\). They cannot instead tend to two different numbers.
Persistent Oscillation Can Prevent a Limit
Worked Example: Showing That Sine Has No Limit at Infinity
Consider \(f(x)=\sin x\). Choose two sequences of inputs:
Both sequences tend to positive infinity, since adding \(2\pi n\) makes each input grow without bound. Periodicity of sine gives
Thus the function values along the first sequence tend to \(1\), while those along the second tend to \(-1\). Since these limits differ, the subsequence test shows that \(\sin x\) has no limit at positive infinity. The calculations also check the result at every selected input: the two value sequences are exactly \(1,1,1,\ldots\) and \(-1,-1,-1,\ldots\).
This reasoning applies to any function for which one can find two sequences of arbitrarily large inputs whose function values approach different numbers. The important evidence is not simply that the function varies. It is that the variation persists by a fixed amount arbitrarily far out.
Worked Example: A Growing Factor Does Not Suppress Sine's Oscillation
Consider \(f(x)=x\sin x\). Use the same input sequences:
On the first sequence, \(\sin x_n=1\), so
On the second sequence, \(\sin y_n=-1\), so
The first sequence of function values grows without bound, and the second decreases without bound. In particular, they cannot both approach the same finite number. Hence \(x\sin x\) has no finite limit at positive infinity. Substitution into the original expression confirms both values and their opposite directions of growth.
When Decay Suppresses Oscillation
Now suppose that an oscillating factor remains bounded while another factor tends to zero. The product must tend to zero. The reason is that bounded oscillation cannot make the product larger than a fixed multiple of the decaying factor.
Here is a direct proof from the definition of a limit. If \(M=0\), then \(g(x)=0\) for all \(x>X_0\), and the product is eventually zero. If \(M>0\), let \(\varepsilon>0\). Since \(h(x)\to0\), there is a threshold \(X_1\) such that
For \(x\) larger than both \(X_0\) and \(X_1\), the bound on \(g\) and the product rule for absolute values give
This is precisely the definition of \(g(x)h(x)\to0\). Notice that \(g\) does not need to have a limit. Only its eventual boundedness is required.
Worked Example: A Sine Factor Divided by a Growing Denominator
Evaluate the limit of \(\frac{\sin x}{x+2}\) as \(x\to+\infty\). For \(x>0\), the standard bound \(|\sin x|\le1\) and the positive denominator give
As \(x\to+\infty\), the right-hand side tends to zero. Equivalently,
The squeeze theorem therefore yields
The estimate checks the whole tail of the function, not just selected values: the absolute value of every output is at most \(1/(x+2)\), and that bound becomes arbitrarily small.
Worked Example: A Bounded Oscillating Numerator Over a Radical
Find the limit of \(\frac{5+2\cos x}{\sqrt{x^2+9}}\) as \(x\to+\infty\). Since \(-1\le\cos x\le1\), adding \(5\) and multiplying the inequalities by \(2\) gives
For \(x>0\), the numerator is positive and \(\sqrt{x^2+9}\ge x\). Consequently,
The last expression tends to zero as \(x\to+\infty\), so the squeeze theorem gives
The bounds verify the conclusion for every sufficiently large \(x\): the numerator never exceeds \(7\), while the denominator grows at least as fast as \(x\).
Decay Is the Deciding Feature
The same oscillating factor can behave differently depending on what multiplies it. A factor tending to zero can suppress bounded oscillation, but a factor tending to a nonzero number generally does not. For example, multiplying \(\sin x\) by \(1+1/x\) does not remove its oscillation.
Worked Example: A Multiplier Near One Does Not Remove Oscillation
For \(x>0\), consider \(f(x)=(1+1/x)\sin x\). Along \(x_n=\frac{\pi}{2}+2\pi n\), the sine factor equals \(1\), so
Along \(y_n=\frac{3\pi}{2}+2\pi n\), the sine factor equals \(-1\), and therefore
The two subsequential limits are different, so \(f\) has no limit at positive infinity. The factor \(1+1/x\) tends to \(1\), not to zero; the computed subsequences confirm that it does not damp the persistent oscillation.
More generally, if \(g\) has subsequences tending to distinct finite values \(A\) and \(B\), and \(h(x)\to c\ne0\), then along those same subsequences the products \(h(x)g(x)\) tend to \(cA\) and \(cB\). Since \(c\ne0\) and \(A\ne B\), these limits remain distinct. A multiplier tending to a nonzero constant therefore preserves that particular separation between subsequential values.
A Common Mistake: Bounded Does Not Mean Convergent
A tempting but incorrect argument is: “Since \(-1\le\sin x\le1\), the sine function is bounded, so it must have a limit.” Boundedness only says the values stay within a fixed interval. It does not say they settle near one number. The sequences for which \(\sin x_n=1\) and \(\sin y_n=-1\) directly contradict convergence.
A related error is to replace an oscillating factor by its average value when evaluating a limit. For instance, it is not valid to treat \(\sin x\) as zero in \(\sin x+1/x\). Along the two sine subsequences, the full expression has limits \(1\) and \(-1\), respectively, because \(1/x\to0\). Thus the expression has no limit.
Check Your Understanding
For each question, use bounds or sequences as appropriate, and explain what your reasoning establishes.
- Find two sequences tending to positive infinity that show \(\cos x\) has no limit.
- Evaluate \(\lim_{x\to+\infty}\frac{3\sin x}{x}\), and state the bound that justifies your answer.
- Does boundedness of \(g(x)\) alone imply that \(g(x)\) has a limit? Give a reason.
- Use subsequences to determine whether \((2+1/x)\sin x\) has a limit at positive infinity.
- State the condition on \(g\) and \(h\) that guarantees \(g(x)h(x)\to0\) by the bounded-factor theorem.
