Why Rescaling Matters
Knowing that \(\sin x/x\) approaches \(1\) as \(x\) approaches \(0\) gives us a way to evaluate many nearby limits. The key is to rewrite a new expression so that its sine argument and its denominator match. A constant factor can then remain outside the familiar limit.
This detail matters: the limit of \(\sin(kx)/x\) is generally not \(1\), and the limit of \(\sin(kx)/\sin(mx)\) generally depends on both constants. Negative constants affect the sign, while zero constants require separate checks because some of the usual algebraic rewrites would involve division by zero.
We will use this established limit and rescale its input. The constants \(k\) and \(m\) below are fixed real numbers. The limits in this tutorial are two-sided limits as \(x\) approaches \(0\).
Rescaling \(\sin(kx)/x\)
First suppose \(k\ne 0\). For \(x\ne 0\), multiply and divide by \(k\), arranging the expression so the sine argument appears in the denominator:
As \(x\to 0\), the input \(kx\) also approaches \(0\). Thus the fundamental sine limit applies to the quotient on the right. If \(k=0\), then \(\sin(kx)/x=\sin 0/x=0\) whenever \(x\ne 0\), so its limit is \(0\). Together these cases give the result for every real \(k\).
The limit equals the constant multiplying \(x\) inside the sine, not \(1\) in general. In particular, a negative \(k\) gives a negative limit, and \(k=0\) gives zero.
Worked Example: A Positive Constant Multiple
Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(5x)}{x}\).
Because \(5\ne 0\), multiply and divide by \(5\). For \(x\ne 0\),
As \(x\to 0\), \(5x\to 0\). Therefore,
Check: This agrees with the theorem, which gives the limit \(k=5\).
Worked Example: A Negative Constant Multiple
Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(-3x)}{x}\).
For \(x\ne 0\), rewrite the denominator using the same constant as the sine argument:
Since \(-3x\to 0\), the quotient tends to \(1\). Hence,
Check: The constant is \(k=-3\), so the theorem also gives \(-3\). The negative sign must not be discarded.
Rescaling a Ratio of Sines
Now consider \(\sin(kx)/\sin(mx)\). For the main case, assume both \(k\ne 0\) and \(m\ne 0\). When the terms are defined and \(x\ne 0\), factor each sine into a familiar quotient and its matching input:
The identity follows by multiplying the factors on the right: the constants \(k/m\), the factor \(kx\) in the denominator, and the factor \(mx\) in the numerator combine to \(1\), leaving \(\sin(kx)/\sin(mx)\). As \(x\to 0\), both \(kx\) and \(mx\) tend to \(0\). The first sine quotient tends to \(1\), and the reciprocal of the second sine quotient tends to \(1\). Therefore the limit is \(k/m\).
There is a small domain issue to check. Since \(\sin u/u\to 1\), that quotient is nonzero for all sufficiently small nonzero \(u\): for example, it is greater than \(1/2\) there. When \(m\ne 0\), a sufficiently small nonzero \(x\) gives a nonzero \(u=mx\). Consequently \(\sin(mx)\ne 0\) for all sufficiently small nonzero \(x\), so the original ratio is defined in a punctured interval around \(0\).
When \(k=0\), the quotient is zero wherever it is defined: the numerator is \(\sin(0x)=0\), while the denominator is nonzero for sufficiently small nonzero \(x\) if \(m\ne 0\). Its limit is therefore \(0=0/m\). This case should be handled directly rather than by writing \(\sin(kx)/(kx)\), which would divide by zero.
Worked Example: A Ratio with a Negative Constant
Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(4x)}{\sin(-2x)}\).
Here \(k=4\) and \(m=-2\), so neither constant is zero. For sufficiently small nonzero \(x\), the quotient can be rearranged as
The factors have limits \(-2\), \(1\), and \(1\), respectively. Thus,
Check: The ratio theorem gives \(k/m=4/(-2)=-2\), matching the rescaling calculation.
Worked Example: A Zero Constant in the Numerator
Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(0x)}{\sin(5x)}\).
The numerator is \(\sin(0x)=0\). Because \(m=5\ne 0\), the denominator is nonzero for sufficiently small nonzero \(x\), as established by the sine limit. Thus the quotient equals \(0\) throughout that punctured neighborhood:
Therefore,
Check: The ratio theorem gives \(k/m=0/5=0\), in agreement with the direct evaluation.
Why the Rescaling Works
The method does not depend on a special trick for each value of \(k\) or \(m\). It works because multiplying \(x\) by any fixed constant still produces an input approaching \(0\). The fundamental sine limit can therefore be applied to \(\sin(kx)/(kx)\) whenever \(k\ne 0\), and to \(\sin(mx)/(mx)\) whenever \(m\ne 0\).
For a sine of \(kx\), place \(kx\) in the denominator to form \(\sin(kx)/(kx)\).
For \(\sin(kx)/x\), the factor is \(k\); for a sine ratio, the constants leave the factor \(k/m\).
Each matched sine quotient tends to \(1\) as its input tends to \(0\).
Do not divide by a constant that is zero, and confirm that the original denominator is nonzero near the point when required.
This is a complete justification of the scaling formulas: each rewrite is an identity on the relevant punctured neighborhood, and each remaining factor has a limit supplied by the fundamental sine limit. The zero cases follow from the original expressions rather than from an invalid division by zero.
Common Pitfall: Losing the Constant Factor
A frequent error is to treat every sine quotient as though its limit were \(1\), even when the arguments have different constant multiples. For example, for \(\sin(3x)/\sin(5x)\), one might incorrectly cancel the \(x\)-dependence and claim the limit is \(1\). That ignores the constants needed to match each sine argument.
The correction is to preserve both scaling factors. The same care prevents a sign error when a constant is negative. Also, if the denominator constant \(m\) is zero, \(\sin(mx)\) is identically zero; a formula such as \(k/m\) cannot be used to assign a limit to an undefined quotient.
Why These Limits Are Useful
These scaling results provide basic tools for evaluating trigonometric limits near zero. They let you reduce expressions with constant multiples of \(x\) to the fundamental sine limit, while making the effect of each constant explicit. In more complicated expressions, correctly identifying this factor is often what separates a valid limit calculation from an incorrect one.
For this topic, keep three checks in view: angles must be in radians, constants outside the standard sine quotient must be retained, and any zero constant in a denominator must be examined before applying a quotient formula.
Check Your Understanding
Use the fundamental sine limit and show the constant factors in each calculation.
- Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(7x)}{x}\).
- Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(-2x)}{x}\).
- Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(3x)}{\sin(8x)}\).
- Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(0x)}{\sin(-4x)}\).
- Why can the formula for \(\sin(kx)/\sin(mx)\) not be used when \(m=0\)?
