Lesson 6 of 12 · 7 min
A standard limit for powers
NCERT §12.3.2, Theorem 2, Example 3
Back at the start: the average from 2 s to t was 4.9(t² − 2²)/(t − 2), and it tended to 4.9 × 4. Where does that 4 come from, and is there a quick rule for such limits?
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The lesson in notes
In short
For any positive integer n, lim_{x→a} (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹.
Reason: xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + xⁿ⁻²a + … + aⁿ⁻¹). After cancelling x − a, each of the n terms tends to aⁿ⁻¹.
The result also holds when n is any rational number, provided a is positive.
Worked: lim_{x→1} (x¹⁵ − 1)/(x¹⁰ − 1). Divide top and bottom by x − 1 and use the theorem on each: 15 ÷ 10 = 3/2.
Worked: lim_{x→0} (√(1 + x) − 1)/x. Put y = 1 + x, so y → 1; the limit is lim (y^(1/2) − 1^(1/2))/(y − 1) = ½ · 1^(−1/2) = 1/2.