Last updated at May 29, 2018 by Teachoo

Transcript

Ex 13.1, 24 (Method 1) Find limx→1 f(x), where f(x) = x2 −1,−x2 −1, x ≤1x>1 Finding limit at x = 1 lim x→1− f(x) = lim x→1+ f(x) = limx→1 f(x) Thus, lim x→1+f(x) = –2 & lim x→1−f(x) = 0 ⇒ 𝒍𝒊𝒎 𝐱→𝟏+f(x) ≠ 𝒍𝒊𝒎 𝐱→𝟏−f(x) So, left hand limit & right hand limit are not equal Hence 𝐥𝒊𝒎𝐱→𝟏 f(x) does not exit Ex13.1, 24 (Method 2) Find limx→1 f(x), where f(x) = x2 −1,−x2 −1, x ≤1x>1 We know that limx→𝑎 f(x) exists only if Left Hand limit = Right hand limit i.e. lim x→𝑎+f(x) = lim x→𝑎− f(x) Similarly in question , we have to find limits First we have to prove limx→ 1+ f(x) = limx→ 1− f(x) For 𝒍𝒊𝒎 𝒙→𝟏+f(x) , f(x) = – x2 – 1 Hence if we move value of x towards 1, value of f(x) tends towards – 2 Hence, limx→ 1+ f(x) = – 2 For 𝒍𝒊𝒎 𝒙→𝟏−f(x) , f(x) = x2 – 1 Hence if we move value of x towards 1, value tends of f(x) towards 0 Hence f(x)x→1−= 0 Since – 2 ≠ 0 ∴ 𝒍𝒊𝒎𝐱→ 𝟏+ f(x) ≠ 𝒍𝒊𝒎𝐱→ 𝟏− f(x) So, left hand limit & right hand limit are not equal Hence, limx→1 f(x) dues not exist

Ex 13.1

Ex 13.1, 1

Ex 13.1, 2

Ex 13.1, 3

Ex 13.1, 4

Ex 13.1, 5

Ex 13.1, 6 Important

Ex 13.1, 7

Ex 13.1, 8

Ex 13.1, 9

Ex 13.1,10 Important

Ex 13.1, 11

Ex 13.1, 12

Ex 13.1, 13 Important

Ex 13.1, 14

Ex 13.1, 15

Ex 13.1, 16 Important

Ex 13.1, 17

Ex 13.1, 18

Ex 13.1, 19

Ex 13.1, 20

Ex 13.1, 21

Ex 13.1, 22 Important

Ex 13.1, 23

Ex 13.1, 24 You are here

Ex 13.1, 25 Important

Ex 13.1, 26

Ex 13.1, 27

Ex 13.1, 28 Important

Ex 13.1, 29

Ex 13.1, 30 Important

Ex 13.1, 31

Ex 13.1, 32 Important

About the Author

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.