Why Reciprocal Trigonometric Inputs Have Limits
As \(n\) increases through the positive integers, \(1/n\) approaches zero. Consequently, expressions such as \(\sin(1/n)\) and \(\cos(1/n)\) involve angles that get closer and closer to zero. The trigonometric functions themselves are familiar, but scaling their values by \(n\), \(n^2\), or another power can produce different limits.
The key is to compare each expression with a standard limit at zero. In particular, the ratio \(\sin x/x\) approaches \(1\) as \(x\) approaches zero in radians. A related limit describes how quickly cosine approaches \(1\). These limits let us evaluate many sequences by rewriting them into a form where the known ratio appears.
The Sine Limit at Zero
For nonzero \(x\) near zero, the quotient \(\sin x/x\) compares the sine of an angle with the angle itself. Its limit is \(1\). One justification uses the familiar unit-circle inequalities. For \(0<x<\pi/2\), the area comparison between an inscribed triangle, a circular sector, and a containing triangle gives \(\sin x<x<\tan x\). Dividing these inequalities by the positive quantities involved yields \(\cos x<\sin x/x<1\). Since \(\cos x\) approaches \(1\) as \(x\) approaches zero, the squeeze theorem gives the result for positive \(x\); the quotient is even, so the same limit holds from the negative side.
The domain restriction matters: the quotient is not defined at \(x=0\), but a limit describes values at nearby nonzero inputs. In a sequence, \(x=1/n\) is positive and nonzero for every \(n\in\mathbb{N}\), so the ratio can be used directly.
A Related Limit for Cosine
The cosine difference \(1-\cos x\) also tends to zero as \(x\to0\), but when it is multiplied by a factor that grows, simply knowing that it tends to zero is not enough. The useful scale is \(x^2\). The identity \(1-\cos x=2\sin^2(x/2)\) converts the cosine difference into a square of a sine.
As \(x\to0\), the input \(x/2\) also tends to zero, so the sine ratio on the right tends to \(1\). Squaring and multiplying by \(1/2\) gives the quadratic cosine limit.
Work with the Ratio Before Taking the Limit
For reciprocal inputs, substituting \(x=1/n\) makes the standard ratios especially useful. The expressions \(n\sin(1/n)\) and \(n^2(1-\cos(1/n))\) can each be rewritten using a ratio that has a known limit. This is more informative than evaluating the unscaled function values alone.
Set \(x=1/n\), so \(x\to0\) as \(n\to\infty\).
Rewrite the expression to contain \(\sin x/x\) or \((1-\cos x)/x^2\).
Use the known ratio limit and the algebraic factors to find the sequence limit.
Worked Examples
Worked Example: A Scaled Sine Sequence
Evaluate \(a_n=n\sin(1/n)\).
Set \(x=1/n\). Since \(n=1/x\), rewrite the term as a ratio:
As \(n\to\infty\), \(1/n\to0\). The sine limit at zero therefore gives
Check: The rewritten expression is exactly the standard sine ratio with input \(1/n\), which tends to zero. Its limit is \(1\), confirming the result.
Worked Example: A Quadratically Scaled Cosine Difference
Evaluate \(b_n=n^2\bigl(1-\cos(1/n)\bigr)\).
Set \(x=1/n\), so \(n^2=1/x^2\). The term becomes
The quadratic cosine limit applies because \(1/n\) is nonzero and approaches zero in radians. Thus
Check: Substituting \(x=1/n\) into the ratio \((1-\cos x)/x^2\) gives precisely \(b_n\), and that ratio tends to \(1/2\). The positive sign also agrees with \(1-\cos x\geq0\).
Worked Example: A Scaled First-Order Sine Difference
Evaluate \(c_n=n\bigl(\sin(1/n)-1/n\bigr)\).
Let \(x=1/n\). Distribute the factor \(n=1/x\) and combine the terms over \(x\):
As \(n\to\infty\), \(x\to0\), and \(\sin x/x\to1\). Therefore
Check: Recombining \(\sin x/x-1\) gives \((\sin x-x)/x\), the original expression after setting \(x=1/n\). The limiting ratio is \(1\), so the difference from \(1\) has limit \(0\).
A More Strongly Scaled Sine Difference
The previous example does not by itself determine the limit of \(n^2(\sin(1/n)-1/n)\): an additional factor of \(n\) is present. To handle this stronger scaling, use a bound on the difference. For \(x\geq0\), the identity \(1-\cos x=2\sin^2(x/2)\), together with \(|\sin u|\leq |u|\), gives \(0\leq1-\cos x\leq x^2/2\). By the fundamental theorem of calculus, \(x-\sin x=\int_0^x(1-\cos u)\,du\). Integrating the bound gives \(0\leq x-\sin x\leq x^3/6\).
Worked Example: A Quadratically Scaled Sine Difference
Evaluate \(d_n=n^2\bigl(\sin(1/n)-1/n\bigr)\).
Put \(x=1/n>0\). The bound above implies \(-x^3/6\leq\sin x-x\leq0\). Multiplying by \(n^2=1/x^2\), which is positive, preserves the inequalities:
Thus
Both bounds tend to zero as \(n\to\infty\), so the squeeze theorem gives \(\lim_{n\to\infty}d_n=0\).
Check: The original term is \((\sin x-x)/x^2\) when \(x=1/n\). Its absolute value is at most \(x/6=1/(6n)\), which tends to zero, confirming the limit.
Common Pitfall: Ignoring the Scale
A frequent error is to conclude that \(n^2(1-\cos(1/n))\to0\) because \(\cos(1/n)\to1\). This only establishes that the parenthesized difference tends to zero; it does not settle what happens after multiplication by \(n^2\). In general, a sequence tending to zero multiplied by an unbounded sequence can tend to zero, a nonzero number, or fail to have a limit.
For example, the incorrect reasoning would be: \(1-\cos(1/n)\to0\), therefore \(n^2(1-\cos(1/n))\to0\). The correction is to factor using the scale that matches the cosine difference:
not zero. The scale \(x^2\) is essential here. Always rewrite the entire scaled expression before applying a limit law.
Why These Limits Are Useful
The sine ratio describes the leading behavior of \(\sin x\) near zero: its value is close to \(x\) when \(x\) is small. The cosine ratio describes the leading size of \(1-\cos x\): it is close to \(x^2/2\). With \(x=1/n\), these comparisons explain why \(\sin(1/n)\) is on the scale \(1/n\), while \(1-\cos(1/n)\) is on the scale \(1/n^2\).
These are not claims that the functions equal those approximations. They are precise statements about ratios and limits. In particular, before multiplying a small trigonometric quantity by a growing power of \(n\), determine its scale using the appropriate ratio or a valid bound. The standard limit laws for sums, products, and quotients can then be applied where their conditions hold; for quotients, the denominator sequence must have a nonzero limit.
Check Your Understanding
For each question, identify the relevant standard ratio or bound before evaluating the limit.
- Evaluate \(\lim_{n\to\infty} n\sin(1/n)\).
- Evaluate \(\lim_{n\to\infty} n^2(\cos(1/n)-1)\).
- Evaluate \(\lim_{n\to\infty} n\bigl(\sin(1/n)-1/n\bigr)\).
- Why does \(\cos(1/n)\to1\) alone not determine \(\lim_{n\to\infty}n^2(1-\cos(1/n))\)?
- What angle convention is required for the sine and cosine limits used here?
