10.2, #12. If f(x) = x^{−1}, then f′(x) = −x^{−2}, f′′(x) = +2!x^{−3}, and, in general, f^{(n)}(x) = (−1)^{n}n!x^{−(n+1)}. Evaluating these derivatives at 1 we have f(1) = 1, f′(1) = −1, f′′(1) = 2!, and f^{(n)}(1) = (−1)^{n}n!. The taylor series is

1 − (x − 1) + (x − 1)^{2} − (x − 1)^{3} + − ⋯ + (−1)^{n}(x − 1)^{n} + · · · .

10.2, #19. sin θ = θ − θ^{3}/3! + θ^{5}/5! − θ^{7}/7! + · · ·. Thus

\frac{θ − sin θ}{θ^{3}} = \frac{θ^{3}/3! − θ^{5}/5! + θ^{7}/7! − ⋯}{θ^{3}} = 1/3! − θ^{2}/5! + θ^{4}/7! − · · · .

This has limit 1/6 as θ → 0.

10.2, #20. tan α = α + α^{3}/3 + · · ·. Thus

\frac{tan α}{α} = 1 + α^{2}/3 + · · · .

This has limit 1 as α → 0.

10.2, #21. \sqrt{1 + x} has Taylor series 1 + x/2 − x^{2}/8 + x^{3}/16 − 5x^{4}/128 + ⋯ using 10.2, #7 or the binomial series with p = 1/2. Thus

lim_{x→0} \frac{\sqrt{1 + x} − 1}{x} = lim_{x→0} \frac{x/2 − x^{2}/8 + x^{3}/16 − 5x^{4}/128 + ⋯}{x}

= lim_{x→0}(1/2 − x/8 + x^{2}/16 − 5x^{3}/128 + ⋯) = 1/2.

10.2, #28. The polynomials P_{n}(x) for n ≤ 4 appproximate \sqrt{1 + x} well for |x| ≤ 1/4. This estimate was made using the following parameters for the window: −1.4 ≤ x ≤ 3, −.5 ≤ y ≤ 2. So the interval of convergence based on this observation would be (−1/4, 1/4).

10.2, #30. f′(0), f′′(0), f′′′(0), and f^{(10)}(0) are related to the coefficients C_{1}, C_{2}, C_{3}, C_{10} of the Taylor series by the formula f^{(j)}(0) = j! C_{j}. Thus

f′(0) = 1, f′′(0) = 1, f′′′(0) = 2, f^{(10)}(0) = 9! = 362880

10.2, #31. The derivatives asked for are f′(0) = and f^{(6)}(0). They are related to the coefficients C_{1} and C_{6}, in the Taylor series by the formula in the previous problem, f′(0) = 0, since the coefficient of x is 0, and f^{(6)}(0) = 6!/2! = 6 · 5 · 4 · 3 = 360.