1. Chapter 13 Class 11 Limits and Derivatives
  2. Serial order wise
  3. Ex 13.1

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

Last updated at Nov. 30, 2019 by

Slide6.JPG

Slide7.JPG
Slide8.JPG

 

  1. Chapter 13 Class 11 Limits and Derivatives
  2. Serial order wise

    3. Ex 13.1

    Ex 13.2 →

Transcript

Ex 13.1, 6 Evaluate the Given limit: lim┬(x→0) ((x +1)5 −1)/x lim┬(x→0) ((x + 1)5 − 1)/x = ((0 + 1)5 −1)/0 = (15 − 1)/0 = (1 − 1)/0 = 0/0 Since it is of from 0/0 Hence, we simplify lim┬(x→0) ((x +1)5 −1)/x Putting y = x + 1 ⇒ x = y – 1 As x → 0 y → 0 + 1 y → 1 Our equation becomes lim┬(x→0) ((x +1)5 −1)/x = lim┬(y→1) (𝑦5 − 1)/(y − 1) = (𝐥𝐢𝐦)┬(𝐲→𝟏) (𝒚𝟓 − 𝟏^𝟓)/(𝐲 − 𝟏) = lim┬(y→1) (y5−15)/(y−1) = 5 × 15-1 = 5 × 14 = 5 ∴ lim┬(x→0) ((x + 1)5 − 1)/x = 5 We know that (𝑙𝑖𝑚)┬(𝑥→𝑎) ( 𝑥^𝑛 − 𝑎^𝑛)/(𝑥 − 𝑎) = nan – 1 Comparing (𝑙𝑖𝑚)┬(𝑦→1) ( 𝑦^5 − 1^5)/(𝑦 − 1) Here x = y , n = 5 , a = 1

Chapter 13 Class 11 Limits and Derivatives
Serial order wise

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.