Ex 13.1, 31 - Chapter 13 Class 11 Limits and Derivatives
Last updated at Sept. 17, 2019 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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