Why Integer Powers Have Their Own Limit Rule
When a function approaches a finite value, its powers often approach the corresponding powers of that value. The reason is not a separate calculation rule: a positive integer power is a repeated product, so its limit follows by applying the product law repeatedly.
The same idea does not extend to every integer exponent without conditions. A zeroth power requires a nonzero base wherever it is evaluated, while a negative power involves a reciprocal. In particular, a negative-power limit rule requires the base to approach a nonzero value.
The Positive Integer Power Rule
The hypotheses ensure that the product law can be used at each step: every factor has the same finite limit \(L\). The approach may be from either side when the corresponding limit exists, and the same reasoning applies to a one-sided limit.
For example, when \(n=4\), the expression is \(f(x)f(x)f(x)f(x)\). The product law first gives a limit of \(L^2\) for the product of two factors. Applying it to that product and another factor gives \(L^3\), and one more application gives \(L^4\). The general argument is induction.
Why Repeated Products Give the Rule
We prove the theorem for all positive integers \(n\). The proof uses the product law for limits: if two functions have finite limits as \(x\to a\), then their product has a limit equal to the product of those limits.
\([f(x)]^1=f(x)\), so the claim is exactly the assumption \(\lim_{x\to a}f(x)=L=L^1\).
For some positive integer \(k\), assume \(\lim_{x\to a}[f(x)]^k=L^k\).
Since \([f(x)]^{k+1}=[f(x)]^k f(x)\), and the two factors have finite limits \(L^k\) and \(L\), their product has limit \(L^kL=L^{k+1}\).
Thus the statement holds for \(n=1\), and whenever it holds for \(k\), it holds for \(k+1\). By mathematical induction, it holds for every positive integer \(n\). The proof explains the mechanism: each additional factor contributes one additional factor of \(L\) to the limit.
Worked Example: A Fourth Power Limit
Worked Example: A Fourth Power Limit
Find \(\lim_{x\to 3}(2x+1)^4\) using the integer power rule.
First, the constant-multiple and sum laws give the limit of the base:
The exponent \(4\) is a positive integer, so the theorem applies. Repeated multiplication gives
Check the arithmetic by grouping the four factors into two pairs: \(7\cdot7=49\), and \(49\cdot49=2401\). Thus the computed fourth power is \(2401\), as claimed.
The Zeroth Power
For a nonzero number \(u\), the usual integer-power convention is \(u^0=1\). The restriction matters: \(0^0\) is not defined by this convention. Consequently, if the expression is \(f(x)^0\), it is defined only at points where \(f(x)\ne 0\).
A punctured neighborhood means values of \(x\) sufficiently close to \(a\), with \(x=a\) excluded. The nonzero condition is included because otherwise the expression \(f(x)^0\) would be undefined at points in the neighborhood. When the condition holds, the function whose limit is being taken is simply the constant function \(1\).
Worked Example: A Zeroth Power Limit
Worked Example: A Zeroth Power Limit
Find \(\lim_{x\to 1}(x+2)^0\).
For \(x\) close to \(1\), the base is nonzero. For instance, if \(|x-1|<1\), then \(0<x<2\), so \(2<x+2<4\). Thus \(x+2\ne0\) throughout this neighborhood, and the zeroth power is defined there.
At every such \(x\), \((x+2)^0=1\). Therefore
Check: the expression equals \(1\) at every point sufficiently close to \(1\) (other than the point \(1\) itself if excluded), so its limit is indeed \(1\).
Negative Integer Powers
A negative integer exponent is defined using reciprocals. If \(u\ne0\) and \(m\) is a positive integer, then \(u^{-m}=1/u^m\). This definition turns a negative-power limit into a reciprocal limit.
To see why the base is nonzero close enough to \(a\), use the definition of a nonzero limit. Since \(L\ne0\), values of \(f(x)\) sufficiently close to \(L\) satisfy \(|f(x)-L|<|L|/2\). The triangle inequality then gives \(|f(x)|>|L|/2>0\). Thus the reciprocal is defined near \(a\). The positive integer power rule gives \(\lim_{x\to a} f(x)^m=L^m\), which is nonzero. The reciprocal law for limits now gives the stated result.
This argument uses the quotient law and its nonzero-denominator condition, established in the earlier material on the quotient law. The condition \(L\ne0\) is essential: without it, the reciprocal law cannot be applied to the limit of \(f(x)^m\).
Worked Example: A Negative Power Limit
Worked Example: A Negative Power Limit
Find \(\lim_{x\to 2}\left(\frac{x+1}{2}\right)^{-2}\).
First find the limit of the base. By the sum and constant-multiple laws,
This limit is nonzero, so the negative integer power rule applies. With \(m=2\), it gives
Check: the positive square of the limiting base is \(9/4\), and \( (9/4)(4/9)=1\). Hence its reciprocal is \(4/9\), confirming the result.
Common Pitfall: Taking a Reciprocal of a Zero Limit
A tempting but invalid calculation is to apply the negative-power rule when the base limit is zero. For example, one might write
The final expression is undefined. The problem is not just a missing arithmetic step: the negative-power theorem requires a nonzero limit, but here \(\lim_{x\to0}x=0\). In fact, \(x^{-2}=1/x^2\) is defined for \(x\ne0\), and its values grow without bound as \(x\) approaches \(0\). For any \(M>0\), if \(0<|x|<1/\sqrt{M}\), then \(x^{-2}>M\). Thus there is no finite real limit, and the reciprocal rule cannot be used to produce one.
Check Your Understanding
For each question, identify the relevant exponent rule and check its conditions before calculating.
- If \(\lim_{x\to a}f(x)=-2\), what is \(\lim_{x\to a}f(x)^3\)? Briefly explain why the rule applies.
- If \(\lim_{x\to a}g(x)=0\), can the negative-power rule determine \(\lim_{x\to a}g(x)^{-2}\)? Why or why not?
- What additional domain condition is needed to conclude \(\lim_{x\to a}h(x)^0=1\)?
- If \(\lim_{x\to a}p(x)=4\), find \(\lim_{x\to a}p(x)^{-2}\), showing the reciprocal step.
- In one sentence, explain why the positive integer power rule follows from the product law.
