# Ex 13.1, 31 - Chapter 13 Class 11 Limits and Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 13.1, 31 If the function f(x) satisfies limx → 1 𝑓 𝑥 − 2𝑥2 − 1 = π , evaluate limx→1 f(x) . Given limx→1 𝑓 𝑥 − 2 𝑥2 − 1 = π limx→1 𝑓 𝑥−2 limx→1 (𝑥2−1) = π limx→1 (f(x) – 2) = π × limx→1 (x2 – 1) limx→1 f(x) – limx→1 2 = π ( limx→1 x2 – limx→1 1) Finding limits, putting x = 1 limx→1 f(x) – 2 = π × ((1)2 – 1) limx→1 f(x) – 2 = π × 0 limx→1 f(x) – 2 = π × 0 limx→1 f(x) – 2 = 0 limx→1 f(x) = 2 Thus 𝒍𝒊𝒎𝐱→𝟏 f (x) = 2

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.