Why a Sum Can Settle Down When Its Terms Do Not
As \(x\) approaches zero, the values of \(\sin(1/x)\) and \(\cos(1/x)\) keep oscillating. Neither function approaches a single value. Yet multiplying their sum by a power of \(x\) that approaches zero can make the entire expression approach a definite limit.
The key is not to find limits for the oscillating terms. Instead, bound each term, combine those bounds to control the whole sum, and then show that the vanishing power forces the resulting bounds to zero.
Bounding Each Term Before Combining
For every real number \(t\), the basic trigonometric bounds are
These bounds remain true when \(t\) is replaced by an expression such as \(1/x\), \(3/(x-2)\), or any other real-valued angle. In particular, whenever the expressions are defined,
If two quantities satisfy \(|u(x)|\leq A\) and \(|v(x)|\leq B\), then the triangle inequality gives
Thus, even if neither \(u(x)\) nor \(v(x)\) has a limit, their sum has a fixed bound. For a sum of several terms, apply the same inequality repeatedly: if \(|u_i(x)|\leq M_i\) for each term, then
The constants \(M_i\) do not depend on \(x\). This is what allows the bounds to remain useful while the trigonometric arguments oscillate more and more rapidly.
A General Squeeze Result
A punctured neighborhood of \(a\) means values of \(x\) close to \(a\), but not equal to \(a\). The functions \(u_i\) do not need to have limits at \(a\); they only need to stay bounded near \(a\).
Why the Result Works
Set \(C=M_1+\cdots+M_n\). The triangle inequality and the assumed bounds give
For any real number \(y\), the inequality \(|y|\leq C\) is equivalent to \(-C\leq y\leq C\). Therefore,
Since \(|x-a|^p\geq 0\), multiplying each part of this inequality by \(|x-a|^p\) preserves the order:
Because \(p>0\), \(|x-a|^p\) approaches zero as \(x\) approaches \(a\). Both the left and right bounds therefore approach zero. The Squeeze Theorem now shows that the expression between them approaches zero as well. This proves the result.
Worked Examples
Worked Example: A Power Times the Sum of Sine and Cosine
Evaluate
For \(x\neq 0\), each trigonometric term is bounded:
Adding the lower bounds and adding the upper bounds gives
Since \(x^2\geq 0\), multiplication by \(x^2\) preserves the inequality:
As \(x\to0\), the lower bound \(-2x^2\) and the upper bound \(2x^2\) both approach zero. The Squeeze Theorem gives
Check: The derived interval for the expression has endpoints \(-2x^2\) and \(2x^2\), both of which tend to zero. Thus the bounds verify the stated limit without requiring either trigonometric term to converge.
Worked Example: Unequal Coefficients Near a Different Point
Evaluate
For \(x\neq 2\), the sine term is between \(-1\) and \(1\). The cosine term is also between \(-1\) and \(1\), so multiplying its bounds by \(2\) gives
Adding this inequality to the sine bound gives
The factor \((x-2)^2\) is nonnegative, so multiplying through yields
Both outer expressions approach zero as \(x\to2\). Hence
Check: The combined oscillating sum is bounded between \(-3\) and \(3\), so the product is trapped between \(-3(x-2)^2\) and \(3(x-2)^2\). Both bounds have limit zero at \(2\).
Worked Example: Combining Bounds for Three Oscillating Terms
Evaluate
First bound each term separately. Since sine and cosine lie between \(-1\) and \(1\),
Adding the three lower bounds gives \(-3-1-2=-6\). Adding the upper bounds gives \(3+1+2=6\). Therefore,
Because \(x^4\geq0\), multiplication by \(x^4\) preserves the order:
As \(x\to0\), both bounds approach zero. Thus the Squeeze Theorem gives
Check: The individual term bounds combine to the interval \([-6,6]\), and the product is trapped between \(-6x^4\) and \(6x^4\). Both endpoints tend to zero, confirming the limit.
A Sign Pitfall with Odd Powers
It is important to check whether the factor multiplying an inequality is nonnegative. For example, \(x^3\) is negative when \(x<0\). Multiplying an inequality by \(x^3\) on both sides of zero can reverse its order, so the bounding step must account for the sign.
A tempting but incorrect calculation is to start with \(-2\leq S(x)\leq2\) and write \(-2x^3\leq x^3S(x)\leq2x^3\) for all \(x\) near zero. At \(x=-1\), those proposed endpoints are \(2\) and \(-2\), in the wrong order. The corrected bound uses the absolute value:
Equivalently, \(-2|x|^3\leq x^3S(x)\leq2|x|^3\), and both bounds approach zero as \(x\to0\). This illustrates a useful general approach: bound the absolute value of the entire product, then use the equivalent two-sided inequality.
Why This Method Matters
The oscillating sum itself may have no limit. For instance, the sine and cosine terms in the first example continue to vary as \(x\) approaches zero. The limit exists because the power factor shrinks the size of the entire bounded sum, not because the oscillations stop.
When evaluating a limit of a vanishing power times a sum of oscillatory terms, use this sequence of steps:
Use the basic sine and cosine bounds, including any constant coefficients.
Add lower bounds and upper bounds, or use the triangle inequality to bound the absolute value of the sum.
Multiply inequalities directly only when the factor is nonnegative; otherwise, use an absolute-value bound.
Show that the resulting lower and upper bounds both approach the same limit.
The method applies to any finite sum whose terms are bounded near the point, not just to sine and cosine. The essential conditions are that the sum has a fixed bound in a punctured neighborhood and that the factor multiplying it approaches zero.
Check Your Understanding
For each question, focus on finding bounds for the whole expression before deciding whether the Squeeze Theorem applies.
- Find a constant \(C\) such that \(\left|\sin(1/x)+3\cos(1/x)\right|\leq C\) whenever \(x\neq0\).
- Use term-by-term bounds to evaluate \(\lim_{x\to0}x^2\left(2\sin(1/x)-\cos(1/x)\right)\).
- For \(x\to1\), what constant bound can you use for \(\left|\sin(1/(x-1))+4\cos(1/(x-1))\right|\)?
- Why is it unsafe to multiply a two-sided inequality by \(x^3\) without checking whether \(x\) is positive or negative?
- Does the theorem require each oscillating term to have a limit at the point? Explain what condition it requires instead.
