Ex 12.1, 22 - 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