Check sibling questions

Ex 13.1, 22 - lim x->pi/2 tan 2x/x-pi/2 - Chapter 13 Class 11

Ex 13.1, 22 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Ex 13.1, 22 - Chapter 13 Class 11 Limits and Derivatives - Part 3

This video is only available for Teachoo black users

 

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

Ex 13.1, 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)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.