Chapter 12 - Sequences and Series - 12.3 Taylor Polynomials at 0 - 12.3 Exercises - Page 632: 45

See proof

$$P(k)=1-e^{-2k}$$ Using the Taylor series for $e^{x}$ centered at $0$ because $k$ is small it follows: $$P(k)=1-(1-2k+2k^{2}+\dots)$$ $$P(k)=1-1+2k-2k^{2}+\dots$$ $$P(k)=2k-2k^{2}+\dots$$ $$P(k)=2k(1-k+\dots$$ Since $k \to 0$ it follows that $1 \gt \gt -k+\dots $ therefore $1-k+\dots \approx 1$ $$P(k)\approx 2k\cdot 1$$ $$P(k)\approx 2k$$