Two limits of powers and the 1^infinity race. Self-test Problems 9 and 10.
Let them predict first.
Reveal: 0.8^100 is about 2e-10, so 0.8^n goes to 0. 0.8^(-n) = 1.25^n: 9.313 at n = 10, about 4.9e9 at n = 100; it diverges to +infinity.
a^n = e^(n ln a). On a log scale the exponent n ln a is a straight line through the origin; the sign of ln a decides where it goes.
n times 0.8^n rises for a few steps and then collapses. In logs, ln n grows slowly while n ln a falls in a straight line.
Let them predict first.
Reveal: neither wins. The values 2, 2.25, 2.370, 2.594, 2.705, 2.717 climb to e, about 2.718. The 2022 answer let the base reach 1 before the exponent moved.
Each step multiplies by 1 + a/n. More, smaller steps hug the smooth exponential curve.
Let them predict first.
Reveal: n ln(1+a/n) is about n times a/n, which is a. So the power tends to e^a. Check with a = 1.3: 1.222 at n = 10, 1.299 at n = 1000.
Route 2 in full: logs, a 0/0 form, L'Hopital, exponentiate back. Do not stop at a.
Let them predict first.
Reveal: A tends to e, B to +infinity, C to 1. The exponent times ln(base) decides: 1, about n, about 1/n.
Exponent times (base minus 1). If it tends to a finite L, the power tends to e^L; plus infinity gives plus infinity, minus infinity gives 0. On an exam, show the log step when asked to derive.
Let them predict first. One item at a time.
Reveal 1: n/2 times 4/n is 2, so the limit is e^2, about 7.389.
Reveal 2: n/(n+3) = 1 - 3/(n+3), so exponent times (base - 1) is n times -3/(n+3), which tends to -3. The limit is e^(-3), about 0.0498.
Reveal 3: sqrt(n) times 1/n is 1/sqrt(n), which tends to 0, so the limit is e^0 = 1.
Many trials, each rare. P(X=0) = (1 - lambda/n)^n tends to e^(-lambda): the same limit.
Five take-aways. Then do the drills by hand.