Check sibling questions

Ex 13.1, 21 - Find lim x -> 0 (cosec x - cot x) - Teachoo

Ex 13.1, 21 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Ex 13.1, 21 - Chapter 13 Class 11 Limits and Derivatives - Part 3

This video is only available for Teachoo black users

Ex 13.1, 21 - Chapter 13 Class 11 Limits and Derivatives - Part 4 Ex 13.1, 21 - Chapter 13 Class 11 Limits and Derivatives - Part 5

 

This video is only available for Teachoo black users

Ex 13.1, 21 - Chapter 13 Class 11 Limits and Derivatives - Part 6 Ex 13.1, 21 - Chapter 13 Class 11 Limits and Derivatives - Part 7


Transcript

Ex 13.1, 21 (Method 1) Evaluate the Given limit: lim┬(x→0) (cosec x – cot x) lim┬(x→0) (cosec x – cot x) = lim┬(x→0) (1/sin⁡𝑥 −cos⁡𝑥/sin⁡𝑥 ) = lim┬(x→0) (1 −〖 cos〗⁡𝑥)/sin⁡〖𝑥 〗 Putting x = 0 = (1 − cos⁡〖0 〗)/sin⁡0 = (1 − 1)/0 = 0/0 Using cosec θ = (1 )/(sin θ) cot θ = (cos θ)/(sin θ) Since it is coming 0/0 Hence, we need to simplify lim┬(x→0) (1 −〖 cos〗⁡𝑥)/sin⁡〖𝑥 〗 = lim┬(x→0) (1 −〖 cos〗⁡𝑥)/sin⁡〖𝑥 〗 × (𝟏 + 𝐜𝐨𝐬⁡𝒙)/(𝟏 + 𝒄𝒐𝒔⁡𝒙 ) = lim┬(x→0) (1^2 −〖 cos^2〗⁡𝑥)/(sin⁡〖𝑥 〗 (1 + cos⁡𝑥 ) ) = lim┬(x→0) (𝟏 −〖 〖𝒄𝒐𝒔〗^𝟐〗⁡𝒙)/(sin⁡〖𝑥 〗 (1 + cos⁡𝑥 ) ) = lim┬(x→0) 〖 〖𝐬𝐢𝐧〗^𝟐〗⁡𝒙/(sin⁡〖𝑥 〗 (1 + cos⁡𝑥 ) ) = lim┬(x→0) 〖 𝑠𝑖𝑛〗⁡𝑥/((1 + cos⁡𝑥 ) ) Putting x = 0 = 〖 𝑠𝑖𝑛〗⁡0/((1 + cos⁡0 ) ) = 0/(1 + 1) = 0/2 = 0 Ex 13.1, 21 (Method 2) Evaluate the Given limit: lim┬(x→0) (cosec x – cot x) (𝑙𝑖𝑚)┬(𝑥→0) (cosec x – cot x) = (𝑙𝑖𝑚)┬(𝑥→0) ((𝑐𝑜𝑠𝑒𝑐 𝑥 − 𝑐𝑜𝑡⁡〖𝑥) 〗)/𝟏 = (𝑙𝑖𝑚)┬(𝑥→0) (𝑐𝑜𝑠𝑒𝑐 𝑥 −〖 𝑐𝑜𝑡〗⁡𝑥)/(𝒄𝒐𝒔𝒆𝒄𝟐 𝒙 − 𝒄𝒐𝒕𝟐 𝒙) = (𝑙𝑖𝑚)┬(𝑥→0) (𝑐𝑜𝑠𝑒𝑐 𝑥 −〖 𝑐𝑜𝑡〗⁡𝑥)/((𝑐𝑜𝑠𝑒𝑐 𝑥 −〖 𝑐𝑜𝑡〗⁡〖𝑥) (𝑐𝑜𝑠𝑒𝑐 𝑥 +〖 𝑐𝑜𝑡〗⁡〖𝑥)〗 〗 ) = (𝑙𝑖𝑚)┬(𝑥→0) 1/(𝑐𝑜𝑠𝑒𝑐 𝑥 +〖 𝑐𝑜𝑡〗⁡𝑥 ) (𝑈𝑠𝑖𝑛𝑔 𝑐𝑜𝑠𝑒𝑐2 𝜃−𝑐𝑜𝑡2 𝜃=1) Using cosec θ = (1 )/(sin θ) cot θ = (cos θ)/(sin θ) = (𝑙𝑖𝑚)┬(𝑥→0) 1/(1/𝑠𝑖𝑛⁡𝑥 + 𝑐𝑜𝑠⁡𝑥/𝑠𝑖𝑛⁡𝑥 ) = (𝑙𝑖𝑚)┬(𝑥→0) 1/((1 + 〖 𝑐𝑜𝑠〗⁡𝑥)/𝑠𝑖𝑛⁡𝑥 ) = (𝑙𝑖𝑚)┬(𝑥→0) 𝑠𝑖𝑛⁡𝑥/(1 + 𝑐𝑜𝑠⁡𝑥 ) Putting x = 0 = sin⁡0/(1 +〖 cos〗⁡0 ) = 0/(1 + 1) = 0/2 = 0 Ex 13.1, 21 (Method 3) Evaluate the Given limit: lim┬(x→0) (cosec x – cot x) (𝑙𝑖𝑚)┬(𝑥→0) (cosec 𝑥 − cot 𝑥) = (𝑙𝑖𝑚)┬(𝑥→0) 1/sin⁡𝑥 − cos⁡𝑥/sin⁡𝑥 = (𝑙𝑖𝑚)┬(𝑥→0) ((1 − cos⁡𝑥)/sin⁡𝑥 ) Using 1 − cos 2𝜃 = 2 sin2 𝜃 and sin 2𝜃 = 2 sin 𝜃 cos 𝜃 = (𝑙𝑖𝑚)┬(𝑥→0) (2 sin^2⁡〖 𝑥/2〗)/(2 sin⁡〖 𝑥/2〗 cos⁡〖 𝑥/2〗 ) = (𝑙𝑖𝑚)┬(𝑥→0) tan⁡〖𝑥/2〗 Putting x = 0 = tan 0 = 0

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.