Ex 12.1, 22 - Chapter 12 Class 11 Limits and Derivatives

Last updated at Dec. 16, 2024 by

Ex 12.1, 22 - lim x->π/2 tan 2x/x-π/2 - Limits Class 11 - Ex 12.1 - Ex 12.1

part 2 - Ex 12.1, 22 - Ex 12.1 - Serial order wise - Chapter 12 Class 11 Limits and Derivatives
part 3 - Ex 12.1, 22 - Ex 12.1 - Serial order wise - Chapter 12 Class 11 Limits and Derivatives

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Ex 12.1, 22 lim┬(x → π/2) tan⁡2x/(x − π/2) lim┬(x → π/2) tan⁡2x/(x − π/2) Putting y = x – π/2 When x → 𝜋/2 y → 𝜋/2 – 𝜋/2 y → 0 So, our equation becomes lim┬(x→π/2) tan⁡2x/(x − π/2) = lim┬(y→0) (tan⁡2(𝜋/2 + 𝑦)/𝑦) = lim┬(y→0) ((〖tan 〗⁡〖(𝜋 + 2𝑦〗))/𝑦) = lim┬(y→0) (tan⁡2𝑦/𝑦) = lim┬(y→0) (1/𝑦 . sin⁡2𝑦/cos⁡2𝑦 ) = lim┬(y→0) (sin⁡2𝑦/𝑦 . 1/cos⁡2𝑦 ) = lim┬(y→0) sin⁡2𝑦/𝑦 ×lim┬(y→0) 1/cos⁡2𝑦 Multiply & Divide by 2 (As tan⁡〖(𝜋+𝑥〗)=tan x) = lim┬(y→0) (sin⁡2𝑦/𝑦 "× " 2/2).lim┬(y→0) 1/cos⁡2𝑦 = 2 lim┬(y→0) (𝒔𝒊𝒏⁡𝟐𝒚/𝟐𝒚).lim┬(y→0) 1/cos⁡2𝑦 Using ( lim)┬(x→0) (sin⁡x )/x = 1 Replacing x by 2y. lim┬(x→0) sin⁡2𝑦/2y = 1 = 2 × 1 × lim┬(y→0) 1/cos⁡2𝑦 = 2 × 1/cos⁡〖2(0)〗 = 2/cos⁡0 = 2/1 = 2 (As cos 0 = 1)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo