Ex 13.1, 31 - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
Last updated at Sept. 6, 2021 by Teachoo
Limits - Limit exists
Last updated at Sept. 6, 2021 by Teachoo
Ex 13.1, 31 If the function f(x) satisfies lim┬(x → 1) (𝑓(𝑥) − 2)/(𝑥2 − 1) = π , evaluate lim┬(x→1) f(x) . Given lim┬(x→1) (𝑓(𝑥) − 2)/(𝑥^2 − 1) = π (lim┬(x→1) 𝑓(𝑥) − 2)/(lim┬(x→1) 〖(𝑥〗^2 − 1) ) = π lim┬(x→1) (f(x) – 2) = π × lim┬(x→1) (x2 – 1) lim┬(x→1) f(x) – lim┬(x→1) 2 = π (lim┬(x→1) x2 – lim┬(x→1) 1) By Algebra of limits (𝑙𝑖𝑚)┬(𝑥→𝑎) (𝑓(𝑥))/(𝑔(𝑥)) = ((𝑙𝑖𝑚)┬(𝑥→𝑎) 𝑓(𝑥))/((𝑙𝑖𝑚)┬(𝑥→𝑎) 𝑔(𝑥)) (lim┬(x→1) 𝑓(𝑥) − 2)/(lim┬(x→1) 〖(𝑥〗^2 − 1) ) = π lim┬(x→1) (f(x) – 2) = π × lim┬(x→1) (x2 – 1) lim┬(x→1) f(x) – lim┬(x→1) 2 = π (lim┬(x→1) x2 – lim┬(x→1) 1) Finding limits, putting x = 1 lim┬(x→1) f(x) – 2 = π × ((1)2 – 1) lim┬(x→1) f(x) – 2 = π × 0 lim┬(x→1) f(x) – 2 = π × 0 lim┬(x→1) f(x) – 2 = 0 lim┬(x→1) f(x) = 2 Thus (𝒍𝒊𝒎)┬(𝐱→𝟏) f (x) = 2