1 votes 1 votes The Taylor series expansion of $\dfrac{\sin x}{x-\pi}$ at $x=\pi$ is given by $1+\frac{(x-\pi)^{2}}{3 !}+\ldots$ $-1-\frac{(x-\pi)^{2}}{3 !}+\ldots$ $1-\frac{(x-\pi)^{2}}{3 !}+\ldots$ $-1+\frac{(x-\pi)^{2}}{3 !}+\ldots$ Others gate2009-ec + – admin asked Sep 15, 2022 edited Jan 30, 2023 by Lakshman Bhaiya admin 46.4k points 49 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.