Why Constants Matter in Limits
A limit describes the value a function approaches as its input approaches a specified point. A fixed multiplier changes the size of the function’s values, so it also changes the value they approach. If a function approaches \(L\), multiplying all its values by a constant \(c\) makes them approach \(cL\).
This rule is simple to use, but it is important to keep track of what is fixed and what is changing. The multiplier must be a constant, not a quantity that varies with \(x\). We will also distinguish this rule from the constant limit law, which says that a function whose values are constantly \(C\) has limit \(C\).
Limit Notation and the Two Laws
For a function \(f\) defined near \(a\), the statement \(\lim_{x\to a} f(x)=L\) means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that, whenever \(x\) is in the domain of \(f\) and \(0<|x-a|<\delta\), we have \(|f(x)-L|<\varepsilon\). The condition \(0<|x-a|\) means that the value of \(f\) at \(a\), if defined, is not required to equal \(L\).
These laws address different situations. The constant multiple law starts with a function that has a known limit and scales that limit. The constant limit law evaluates the limit of a function whose nearby values are already fixed at one constant.
Worked Example: A Negative Constant Multiplier
Worked Example: A Negative Constant Multiplier
Suppose \(\lim_{x\to 2} g(x)=4\). Find \(\lim_{x\to 2} -3g(x)\).
The multiplier is the fixed constant \(-3\), and the known limit of \(g\) is \(4\). By the constant multiple law, multiply the limit by \(-3\):
Check the result from the definition. Given any \(\varepsilon>0\), the assumption \(\lim_{x\to 2}g(x)=4\) provides a \(\delta>0\) such that \(0<|x-2|<\delta\) implies \(|g(x)-4|<\varepsilon/3\). For those \(x\),
Thus the values of \(-3g(x)\) approach \(-12\), as the law predicts. The negative multiplier reverses the sign of the limiting value.
Worked Example: Scaling a Linear Function
Worked Example: Scaling a Linear Function
Evaluate \(\lim_{x\to 0} -4(x+2)\), using the constant multiple law.
First, as \(x\) approaches \(0\), \(x+2\) approaches \(2\). This can be seen directly: \(|(x+2)-2|=|x|\), so for every \(\varepsilon>0\), choosing \(\delta=\varepsilon\) ensures \(0<|x|<\delta\) gives \(|(x+2)-2|<\varepsilon\). Therefore,
Now multiply the known limit by the fixed constant \(-4\):
To check directly, let \(\varepsilon>0\) and choose \(\delta=\varepsilon/4\). If \(0<|x|<\delta\), then
This verifies that the limit is \(-8\).
Why the Constant Multiple Law Works
The definition of a limit measures the error between a function’s value and its proposed limit. If the values of \(f(x)\) are within a small distance of \(L\), then the values of \(cf(x)\) are within \(|c|\) times that distance of \(cL\). The absolute value is essential: a negative multiplier reverses signs, but distances remain nonnegative.
Let \(\varepsilon>0\) be arbitrary. We must ensure that \(|cf(x)-cL|<\varepsilon\) whenever \(x\) is sufficiently close to \(a\), but not equal to \(a\).
If \(c=0\), then \(cf(x)=0\) and \(cL=0\) wherever \(f\) is defined. Thus \(|cf(x)-cL|=0<\varepsilon\), and any positive \(\delta\) works.
Since \(|c|>0\), the number \(\varepsilon/|c|\) is positive. The assumption \(\lim_{x\to a}f(x)=L\) gives a \(\delta>0\) such that \(0<|x-a|<\delta\) implies \(|f(x)-L|<\varepsilon/|c|\).
For those \(x\), factor out the fixed multiplier and use the absolute-value identity \(|cu|=|c||u|\):
\(|cf(x)-cL|=|c(f(x)-L)|=|c||f(x)-L|<|c|(\varepsilon/|c|)=\varepsilon\).
In both cases the definition is satisfied, so \(\lim_{x\to a}cf(x)=cL\). This proof also explains why the input error may need to be smaller when \(|c|\) is large: multiplication magnifies the error by the factor \(|c|\).
Worked Example: A Constant Function with a Different Point Value
Worked Example: A Constant Function with a Different Point Value
Define \(q(x)=9\) for \(x\ne 4\), and set \(q(4)=-2\). Find \(\lim_{x\to 4}q(x)\).
Every domain point different from \(4\) has \(q(x)=9\). The value at \(4\) is not used in the limit definition, which only considers \(0<|x-4|<\delta\). Therefore the candidate limit is \(9\).
To check it, let any \(\varepsilon>0\) be given and choose \(\delta=1\). Whenever \(x\) is in the domain and \(0<|x-4|<1\), we have \(x\ne4\), so \(q(x)=9\). Consequently,
Thus \(\lim_{x\to4}q(x)=9\), even though \(q(4)=-2\). This is an application of the constant limit law, not a claim that a function’s limit must equal its value at the approach point.
Common Mistake: Scaling in the Wrong Direction
A frequent error is to divide the limiting value by the multiplier. For example, if \(\lim_{x\to a} f(x)=6\), a mistaken calculation might say:
This is incorrect because the function’s values are being multiplied by \(5\), so their limiting value is also multiplied by \(5\). The correct calculation is
A related error occurs with negative constants: replacing \(c\) by \(|c|\) in the answer would lose the sign. The absolute value \(|c|\) appears when measuring the size of an error in the proof; the limiting value itself is \(cL\), with its sign preserved.
When to Use Each Law
| What is known | Law to use | Result |
|---|---|---|
| \(\lim_{x\to a}f(x)=L\), and \(c\) is fixed | Constant multiple law | \(\lim_{x\to a}cf(x)=cL\) |
| \(f(x)=C\) at all nearby domain points except possibly \(a\) | Constant limit law | \(\lim_{x\to a}f(x)=C\) |
The distinction is useful when simplifying limit calculations one step at a time. First identify the limiting value of the function being scaled; then multiply that value by the fixed coefficient. For a function that is already constant near the approach point, its nearby values directly identify the limit.
Summary
The constant multiple law says that multiplying a function by a fixed real number \(c\) multiplies its finite limit by \(c\). Its epsilon-delta justification follows by controlling the original function’s error to within \(\varepsilon/|c|\) when \(c\ne0\); when \(c=0\), the resulting error is exactly zero. The constant limit law says that values fixed at \(C\) near the approach point have limit \(C\), regardless of a possible different value at the point itself.
Check Your Understanding
Use the two laws and the definition of a limit where appropriate.
- If \(\lim_{x\to a}f(x)=-4\), what is \(\lim_{x\to a}3f(x)\)?
- If \(\lim_{x\to b}h(x)=5\), what is \(\lim_{x\to b}-\frac{1}{2}h(x)\)?
- Why does the proof of the constant multiple law use \(\varepsilon/|c|\) when \(c\ne0\)?
- If \(r(x)=12\) for all \(x\ne 1\) near \(1\), but \(r(1)=0\), what is \(\lim_{x\to1}r(x)\)?
- Identify the error in the claim \(\lim_{x\to a}(-2)f(x)=L/2\) when \(\lim_{x\to a}f(x)=L\).
