Ex 13.1, 6 - Evaluate: lim x->0  (x + 1)5 -1/x - Class 11

Ex 13.1, 6 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Ex 13.1, 6 - Chapter 13 Class 11 Limits and Derivatives - Part 3

 

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 12.1, 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

Ask a doubt
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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.