Introduction
Near zero, \(\sin x\) behaves like \(x\) when angles are measured in radians. The standard sine quotient limit makes this precise. A small change in the argument or denominator introduces a constant factor, and careful rescaling keeps that factor visible.
For example, \(\sin(5x)/x\) does not have the same limit as \(\sin x/x\). The argument changes five times as quickly as \(x\), so the limit is multiplied by \(5\). If the denominator is \(2x\), it is also divided by \(2\).
The standard sine quotient
The result used throughout this tutorial is the standard sine quotient limit, proved by squeezing in the earlier material on the sine quotient:
The variable in this limit is \(u\). To use it in an expression involving \(x\), we rewrite that expression so that its sine argument also appears in the denominator. This is the essential rescaling step.
Angles must be in radians for this standard limit to equal \(1\). If angles are measured in degrees, the corresponding quotient has a different limiting value. The formulas below therefore use radian measure.
Rescaling \(\sin(ax)/x\)
Let \(a\) be a nonzero real constant. For every \(x\ne 0\), the product \(ax\) is also nonzero, and we can multiply and divide by \(a\) to obtain an exact equality:
As \(x\to 0\), the quantity \(ax\to 0\). Thus the quotient on the right has the form of the standard sine quotient, with \(u=ax\). Taking limits gives the result.
Rewrite \(\sin(ax)/x\) as \(a\bigl(\sin(ax)/(ax)\bigr)\).
Because \(ax\to 0\), the quotient \(\sin(ax)/(ax)\) tends to \(1\).
The remaining constant \(a\) gives the limit \(a\).
This derivation is an application of the standard sine quotient limit and the limit law for a constant multiple. It does not require substituting \(x=0\): the original quotient is undefined at \(x=0\), but a limit depends on values at nearby \(x\ne 0\).
Including a constant in the denominator
Now let \(a\ne 0\) and \(b\ne 0\). Factor the denominator as \(bx=b\cdot x\), and use the same rescaling:
For \(x\ne 0\), all divisions in this identity are valid because \(a\) and \(b\) are nonzero. Since \(ax\to 0\), the quotient \(\sin(ax)/(ax)\) tends to \(1\). Multiplication by the fixed factor \(a/b\) therefore gives the limit \(a/b\).
The numerator’s scale contributes the factor \(a\), while the denominator’s scale contributes division by \(b\). Keeping both contributions separate is a reliable way to avoid sign and coefficient errors.
Worked examples
Worked Example: A scaled numerator and denominator
Evaluate \(\displaystyle\lim_{x\to 0}\frac{\sin(5x)}{2x}\).
Here \(a=5\) and \(b=2\), so both constants are nonzero. Rewrite the expression using the argument \(5x\) in the denominator:
As \(x\to 0\), \(5x\to 0\), and therefore \(\sin(5x)/(5x)\to 1\). It follows that
Check: the exact rewrite has outside factor \(5/2\), and the remaining quotient tends to \(1\). Thus the computed limit retains both the numerator factor \(5\) and denominator factor \(2\).
Worked Example: A negative argument scale
Evaluate \(\displaystyle\lim_{x\to 0}\frac{\sin(-3x)}{4x}\).
Both constants are nonzero. Rewriting with \(-3x\) as the denominator of the sine quotient gives
Since \(-3x\to 0\), the quotient \(\sin(-3x)/(-3x)\) tends to \(1\). Hence
Check: the factor outside the standard quotient is negative, and the standard quotient tends to positive \(1\). The resulting limit is therefore negative, as calculated.
Worked Example: A negative denominator scale
Evaluate \(\displaystyle\lim_{x\to 0}\frac{\sin(7x)}{-2x}\).
Here \(a=7\) and \(b=-2\). Rescale by the argument \(7x\):
As \(x\to 0\), the quotient \(\sin(7x)/(7x)\to 1\), so
Check: \(a/b=7/(-2)=-7/2\), which agrees with the rescaled expression and the limit.
What the conditions mean
The formula for \(\sin(ax)/(bx)\) requires \(b\ne 0\), because \(b=0\) makes the denominator zero for every \(x\), so the expression is not defined on a punctured interval around zero. For the rescaling proof as written, \(a\ne 0\) is also required: it allows division by \(ax\).
If \(a=0\) and \(b\ne 0\), the original numerator is \(\sin(0)=0\). For every \(x\ne 0\), the quotient is \(0/(bx)=0\), so its limit is \(0\), agreeing with \(a/b=0\). Thus the formula \(\lim_{x\to0}\sin(ax)/(bx)=a/b\) also holds for \(a=0\), provided \(b\ne0\), although the rescaling proof must be replaced by this direct check.
Similarly, if \(a=0\), then \(\sin(ax)/x=0\) for every \(x\ne0\), and the limit is \(0=a\). The nonzero-constant condition in the rescaling argument is what permits the division used there; checking zero constants separately prevents an invalid division by zero.
Common pitfall: losing a scale factor
A frequent error is to recognize the standard sine quotient but forget the constant created when the argument is placed in the denominator. For example, the following calculation is wrong:
The equality is false: changing the denominator from \(x\) to \(5x\) divides the quotient by \(5\). The correct equality restores that factor:
A practical check is to compare the original expression with the standard quotient. If the sine argument is \(ax\), the quotient \(\sin(ax)/(ax)\) needs a factor \(a\) to recover a denominator of \(x\). If the denominator is \(bx\), the outside factor is \(a/b\).
Why rescaling matters
The standard sine quotient handles one particular form, \(\sin u/u\). Many nearby limits become accessible by rewriting them exactly into that form and retaining the resulting constant. The method depends on identifying the full argument of the sine, not merely noticing that both numerator and denominator approach zero.
These evaluations also provide a useful test of algebraic bookkeeping. A negative argument scale makes the outside factor negative; a negative denominator scale does the same through division by \(b\). The quotient \(\sin(ax)/(ax)\) still tends to \(1\) in either case, so all sign changes belong in the outside factor \(a/b\).
Check Your Understanding
For each limit, identify the scale factor before evaluating. Assume angles are measured in radians.
- Evaluate \(\displaystyle\lim_{x\to0}\frac{\sin(4x)}{x}\).
- Evaluate \(\displaystyle\lim_{x\to0}\frac{\sin(3x)}{6x}\).
- Evaluate \(\displaystyle\lim_{x\to0}\frac{\sin(-2x)}{5x}\).
- Explain why \(\sin(4x)/(4x)\) alone is not equal to \(\sin(4x)/x\).
- What happens to \(\lim_{x\to0}\sin(ax)/(bx)\) when \(a=0\) and \(b\ne0\)?
