When Trigonometric Bounds Reveal a Limit
Direct substitution is useful when a function is continuous at the value being approached and substitution gives a definite value. But expressions involving trigonometric functions can instead produce an indeterminate form, or contain an oscillating factor whose values do not settle down. In such cases, a useful strategy is to bound the expression by simpler quantities that do have known limits.
The basic estimates come from the fact that sine cannot exceed the size of its angle near zero, and that cosine differs from \(1\) by at most a quantity proportional to the square of its angle. A local estimate for tangent follows from the sine bound and the fact that cosine stays away from zero near \(0\).
The Basic Trigonometric Bounds
Throughout this tutorial, angles are measured in radians. For every real number \(u\), the sine function satisfies
This estimate is especially useful when the angle \(u\) itself depends on \(x\), such as \(u=x^2\) or \(u=5x\). It then controls the sine by the size of that inner expression.
A related estimate controls the difference between cosine and \(1\). The identity \(1-\cos u=2\sin^2(u/2)\), together with the sine bound, gives
The lower bound holds because \(\cos u\le 1\). The upper bound shows why this difference is small near zero: it is no larger than half the square of the angle.
Tangent requires extra care because \(\tan u=\sin u/\cos u\), and division by cosine is safe only where cosine is nonzero. For \(|u|\le \pi/4\), cosine is positive and at least \(\sqrt{2}/2\). Therefore,
Thus \(|\tan u|\le 2|u|\) whenever \(|u|\le\pi/4\). This is a local estimate: it is guaranteed only when the angle lies in that interval. For a limit as \(x\to0\), an angle such as \(x^2\) eventually satisfies this condition.
From Absolute Bounds to Limits
The Squeeze Theorem says that if \(g(x)\le f(x)\le h(x)\) for all \(x\) sufficiently close to \(a\), except possibly at \(a\), and both \(g(x)\) and \(h(x)\) tend to the same number \(L\), then \(f(x)\to L\). For limits of zero, it is often easiest to prove an absolute-value estimate. If
then \(-B(x)\le f(x)\le B(x)\), and both bounds tend to \(0\). The theorem therefore gives \(f(x)\to0\). The estimate must hold on a punctured neighborhood of the point: the function need not be defined at the point itself.
Worked Examples
Worked Example: A Cosine Difference Divided by \(x\)
Evaluate \[ \lim_{x\to0}\frac{1-\cos(4x)}{x}. \] Direct substitution gives \(0/0\), so it does not determine the limit. For \(x\ne0\), apply the cosine bound with \(u=4x\):
Taking absolute values of the quotient gives
Consequently,
As \(x\to0\), both \(-8|x|\) and \(8|x|\) tend to \(0\). The Squeeze Theorem gives \[ \lim_{x\to0}\frac{1-\cos(4x)}{x}=0. \] The absolute-value estimate also checks the sign issue: the denominator changes sign across zero, so the quotient was not assumed to be nonnegative.
Worked Example: A Sine with a Shrinking Angle and an Oscillating Factor
Evaluate \[ \lim_{x\to0}\frac{\sin(x^2)\cos(1/x)}{x}. \] The expression is defined for every \(x\ne0\). Direct substitution cannot be used on \(\cos(1/x)\), because its angle does not approach a finite value. Instead, use \(|\cos v|\le1\) for every real \(v\), and use the sine bound with angle \(x^2\):
Thus the expression lies between \(-|x|\) and \(|x|\). Both bounds tend to \(0\) as \(x\to0\), so \[ \lim_{x\to0}\frac{\sin(x^2)\cos(1/x)}{x}=0. \] The check is precisely the bound above: the full expression has magnitude at most \(|x|\), which vanishes at the required point, regardless of the oscillations of \(\cos(1/x)\).
Worked Example: A Tangent with a Quadratic Angle
Evaluate \[ \lim_{x\to0}\frac{\tan(3x^2)}{x}. \] The tangent estimate requires \(|3x^2|\le\pi/4\). This condition holds whenever \(x^2\le\pi/12\), so it holds for every \(x\) sufficiently close to \(0\). For nonzero \(x\) in that neighborhood,
Therefore,
The lower and upper bounds both tend to \(0\). Hence the limit is \(0\). This verifies both parts of the argument: the local tangent estimate applies near zero, and its resulting bound tends to zero.
Why the Estimates Work
The cosine estimate is not an unrelated formula to memorize. It follows directly from a trigonometric identity and the sine estimate. Replacing \(u\) by \(u/2\) in \(|\sin v|\le|v|\) gives
Squaring both nonnegative sides and multiplying by \(2\) yields
The identity \(1-\cos u=2\sin^2(u/2)\) then gives \(1-\cos u\le u^2/2\). Also \(1-\cos u\ge0\), since \(\cos u\le1\). This proves the two-sided estimate used above.
The tangent bound has a similarly direct explanation. Near zero, \(\cos u\) is positive and bounded below, so dividing \(|\sin u|\le|u|\) by \(|\cos u|\) increases the bound by a fixed factor, but not without limit. On \(|u|\le\pi/4\), that factor is at most \(\sqrt{2}\), which is less than \(2\). The resulting estimate is enough for squeezing because a fixed multiple of a quantity tending to zero still tends to zero.
Common Pitfall: Using the Sine Bound Directly for Tangent
A tempting but incorrect step is to claim \(|\tan u|\le|u|\) simply because \(|\sin u|\le|u|\). The sine estimate does not bound the quotient \(\sin u/\cos u\) by \(|u|\); the denominator must also be accounted for. For example, at \(u=\pi/6\),
So \(\tan u\le u\) is already false at this positive angle. The correction is to control cosine on a specified neighborhood, then divide by its positive lower bound. For \(|u|\le\pi/4\), the valid conclusion is \(|\tan u|\le2|u|\).
A related sign error occurs when dividing an inequality by \(x\) without considering whether \(x\) is positive or negative. When the sign can change, first bound the absolute value of the whole expression. The resulting pair of inequalities, \(-B(x)\le f(x)\le B(x)\), works on both sides of the point.
A Reliable Squeezing Procedure
When a trigonometric expression resists direct substitution, organize the reasoning in a few deliberate steps:
Separate the outer function from its argument. For example, in \(\sin(x^2)\), the angle to estimate is \(x^2\).
Use \(|\sin u|\le|u|\), \(0\le1-\cos u\le u^2/2\), or the local estimate \(|\tan u|\le2|u|\) when its condition holds.
Ensure denominators are nonzero in the punctured neighborhood and check any local restriction on the angle.
Reduce the estimate to a quantity with a known limit, then apply the Squeeze Theorem.
Why These Bounds Matter
Trigonometric inequalities let us prove limits without needing to know the exact behavior of every factor. In particular, an oscillating factor may fail to have a limit by itself, yet a different factor can make the entire product small. The bound shows this quantitatively: if the absolute value of the full expression is no greater than a quantity tending to zero, then no oscillation can prevent the expression from tending to zero.
The essential checks are that each inequality is valid for the angle being used, any local condition is satisfied near the limit point, and the final bounds really do approach the same number. With these checks in place, squeezing converts a difficult-looking trigonometric expression into a comparison of simpler functions.
Check Your Understanding
For each question, identify which trigonometric bound applies and what domain or sign condition must be checked.
- What upper bound does \(0\le1-\cos(5x)\le u^2/2\) give when the angle \(u\) is \(5x\)?
- Why does \(|\sin(x^3)|\le|x|^3\) hold for every real \(x\)?
- What condition on \(u\) allows the estimate \(|\tan u|\le2|u|\) used here?
- How can an absolute-value bound give two inequalities suitable for the Squeeze Theorem?
- Why is it unsafe to divide an inequality by \(x\) without checking whether \(x\) is positive or negative?
