Introduction
A familiar trigonometric limit describes what happens near zero: as an angle \(h\) approaches zero, \(\sin h\) is approximately \(h\). That fact can also reveal what happens near other zeros of the sine function. The sign is not always the same, however: at successive integer multiples of \(\pi\), the sine function alternates direction.
To evaluate a limit near \(n\pi\), where \(n\) is any fixed integer, measure the input’s displacement from \(n\pi\). After that shift, the standard sine limit applies, while the parity of \(n\) determines whether the result is positive or negative.
Shifting the Input to the Nearby Zero
When \(x\) approaches \(n\pi\), the difference \(x-n\pi\) approaches zero. Set \(h=x-n\pi\), or equivalently \(x=n\pi+h\). This substitution describes the same approach using a variable that tends to zero.
The standard sine limit, established in the earlier material on the geometric derivation of the sine limit, is \(\lim_{h\to 0}\frac{\sin h}{h}=1\), for \(h\) in radians. We will use it after rewriting \(\sin x\) in terms of \(h\).
Why the Sign Depends on \(n\)
The sine addition identity gives \(\sin(a+b)=\sin a\cos b+\cos a\sin b\). Use it with \(a=n\pi\), \(b=h\). For any integer \(n\), \(\sin(n\pi)=0\) and \(\cos(n\pi)=(-1)^n\). Thus the sine at the shifted input is a signed copy of \(\sin h\).
Since \(x-n\pi=h\), for \(x\ne n\pi\) the quotient can be rewritten exactly: \(\frac{\sin x}{x-n\pi}=(-1)^n\frac{\sin h}{h}\). The point \(x=n\pi\) itself is excluded from the quotient because its denominator is zero, but a limit depends on values arbitrarily near the point, not on the value at the point.
Complete Derivation
The substitution is legitimate because \(x\to n\pi\) if and only if \(h=x-n\pi\to0\). For every \(x\ne n\pi\), the denominator \(h\) is nonzero, so the exact quotient identity holds in a punctured neighborhood of the limit point. Taking limits of that identity gives the result.
Set \(h=x-n\pi\). Then \(x=n\pi+h\), and \(x\to n\pi\) is equivalent to \(h\to0\).
\(\sin(n\pi+h)=\sin(n\pi)\cos h+\cos(n\pi)\sin h=(-1)^n\sin h\).
Because \(x-n\pi=h\), \(\frac{\sin x}{x-n\pi}=(-1)^n\frac{\sin h}{h}\) whenever \(h\ne0\).
The constant factor \((-1)^n\) is unchanged, and \(\lim_{h\to0}\frac{\sin h}{h}=1\). Therefore the limit is \((-1)^n\cdot1=(-1)^n\).
This also explains why the limit exists from both sides. The shifted quotient \(\sin h/h\) approaches \(1\) whether \(h\) is positive or negative, so multiplying it by the fixed sign \((-1)^n\) gives the same limit from either side.
Worked Examples
Worked Example: A Limit Near \(4\pi\)
Evaluate \(\displaystyle \lim_{x\to4\pi}\frac{\sin x}{x-4\pi}\).
Here \(n=4\). Let \(h=x-4\pi\), so \(x=4\pi+h\) and \(h\to0\). The angle addition identity gives
The denominator is \(x-4\pi=h\). Therefore, for \(h\ne0\), the quotient is \(\frac{\sin h}{h}\), and its limit is \(1\). The parity check agrees: \(4\) is even, so \((-1)^4=1\).
Check: The exact rewrite is \(\frac{\sin x}{x-4\pi}=\frac{\sin h}{h}\). Applying the standard sine limit gives \(1\), matching the theorem’s value \((-1)^4=1\).
Worked Example: A Limit Near \(3\pi\)
Evaluate \(\displaystyle \lim_{x\to3\pi}\frac{\sin x}{x-3\pi}\).
Set \(h=x-3\pi\), so \(x=3\pi+h\) and \(h\to0\). Since \(\sin(3\pi)=0\) and \(\cos(3\pi)=-1\),
Also \(x-3\pi=h\). For \(h\ne0\), the quotient is \(-\frac{\sin h}{h}\), so the limit is \(-1\). The sign follows from \(3\) being odd: \((-1)^3=-1\).
Check: Substituting the shifted expression into the original quotient gives \(\frac{-\sin h}{h}=-\frac{\sin h}{h}\). Its limit is \(-1\cdot1=-1\), as required.
Worked Example: A Limit Near \(-\pi\)
Evaluate \(\displaystyle \lim_{x\to-\pi}\frac{\sin x}{x+\pi}\).
This time \(n=-1\). Let \(h=x+\pi=x-(-\pi)\), so \(x=-\pi+h\) and \(h\to0\). Using \(\sin(-\pi)=0\) and \(\cos(-\pi)=-1\),
The denominator \(x+\pi\) equals \(h\), so the quotient becomes \(-\frac{\sin h}{h}\). Its limit is \(-1\). Although the integer is negative, the parity rule still applies: \((-1)^{-1}=-1\).
Check: The shifted quotient is exactly \(-\sin h/h\), and the standard sine limit makes its limit \(-1\). This matches the theorem’s value for \(n=-1\).
Parity and the Pattern of Signs
The result can be read directly from whether \(n\) is even or odd. If \(n\) is even, \((-1)^n=1\); if \(n\) is odd, \((-1)^n=-1\). This includes negative integers: for example, \(-2\) is even and \(-3\) is odd. There is no separate rule for approaching a negative multiple.
| Integer \(n\) | Sign factor | Limit |
|---|---|---|
| Even, such as \(2\) or \(-2\) | \((-1)^n=1\) | \(1\) |
| Odd, such as \(3\) or \(-1\) | \((-1)^n=-1\) | \(-1\) |
The sign is about the ratio, not about the numerator alone. For an odd \(n\), the sine near \(n\pi\) has the opposite sign from the displacement \(h\). Consequently, numerator and denominator have opposite signs on both sides, and the quotient approaches a negative value from either side.
Common Mistake: Dropping the Alternating Sign
A tempting but incorrect shortcut is to write \(\sin(n\pi+h)=\sin h\) for every integer \(n\). This silently ignores that shifting by an odd multiple of \(\pi\) reverses the sign of sine.
Worked Example: Correcting a Sign Error Near \(\pi\)
Consider \(\displaystyle \lim_{x\to\pi}\frac{\sin x}{x-\pi}\). The incorrect shortcut \(\sin(\pi+h)=\sin h\) would produce \(\lim_{h\to0}\sin h/h=1\).
But the addition identity gives the actual numerator:
Since \(x-\pi=h\), the correct quotient is \(-\sin h/h\), whose limit is \(-1\). The parity check confirms this: \(n=1\) is odd, so \((-1)^1=-1\).
Check: The corrected expression uses the exact numerator \(-\sin h\); dividing by the exact denominator \(h\) and applying the standard sine limit gives \(-1\), not \(1\).
Another way to avoid the same error is to keep both parts of the addition identity in view. The \(\sin(n\pi)\cos h\) term vanishes, but \(\cos(n\pi)\sin h\) remains, and its coefficient is precisely the alternating factor \((-1)^n\).
When This Result Applies
The integer condition matters because the sine zeros occur at integer multiples of \(\pi\), and the identity \(\cos(n\pi)=(-1)^n\) supplies the sign. The limit concerns \(x\to n\pi\) for a fixed integer \(n\), and the standard limit used in the proof assumes radian measure. Degrees would change the scaling in the corresponding small-angle limit.
At the point itself, both \(\sin x\) and \(x-n\pi\) are zero, so direct substitution does not evaluate the quotient. The shift resolves that indeterminate form by expressing the nearby values as a constant sign times \(\sin h/h\). No value needs to be assigned to the original quotient at \(x=n\pi\) in order to find its limit.
Check Your Understanding
For each question, use a shifted variable and state the sign factor before taking the limit.
- Evaluate \(\displaystyle \lim_{x\to2\pi}\frac{\sin x}{x-2\pi}\).
- Evaluate \(\displaystyle \lim_{x\to-3\pi}\frac{\sin x}{x+3\pi}\).
- If \(n\) is an odd integer, what is the value of \(\displaystyle \lim_{x\to n\pi}\frac{\sin x}{x-n\pi}\)?
- Why is the expression \(\sin(n\pi+h)=\sin h\) not valid for every integer \(n\)?
- Why does the shifted sine limit determine a two-sided limit, even when \(n\) is odd?
