# Ex 13.1, 15 - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

Last updated at Nov. 30, 2019 by Teachoo

Last updated at Nov. 30, 2019 by Teachoo

Transcript

Ex 13.1, 15 Evaluate the Given limit: lim┬(x→π) (sin (π − x))/(π(π − x)) lim┬(x→π) (sin (π − x))/(π(π − x)) Let y = π – x So, when x → π y → π – π y → 0 So, our equation becomes (𝒍𝒊𝒎)┬(𝒙→𝝅) (𝒔𝒊𝒏 (𝝅 − 𝒙))/(𝝅(𝝅 − 𝒙)) = (𝒍𝒊𝒎)┬(𝒚→𝟎) (𝒔𝒊𝒏 𝒚)/(𝝅 𝒚) = 1/𝜋 lim┬(y→0) (sin 𝑦)/y = 1/𝜋 × 1 = 𝟏/𝝅 Using lim┬(x→0) sin𝑥/𝑥 = 1 Replace x with y

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Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.