Last updated at Nov. 30, 2019 by Teachoo

Transcript

Ex 13.1, 23 (Method 1) Find lim┬(x→0) f(x) and lim┬(x→1) f(x), where f(x) = {█(2x+3.@3(x+1),)┤ ■8(x ≤0@x>0) Finding limit at x = 0 lim┬(x→0) f(x) = lim┬(〖x→0〗^− ) f(x) =lim┬(〖x→0〗^+ ) f(x) ∴ (𝒍𝒊𝒎)┬(𝐱→𝟎) f(x) = 3 (𝒍𝒊𝒎)┬(〖𝐱→𝟎〗^− ) f(x) = lim┬(x→0) (2x + 3) = 2(0) + 3 = 0 + 3 = 3 (𝒍𝒊𝒎)┬(〖𝐱→𝟎〗^+ ) f(x) = lim┬(x→0) 3(x + 1) = 3(0 + 1) = 3(1) = 3 Finding limit at x = 1 "f(x) = " {█(2x+3.@3(x+1),)┤" " ■8(x ≤0@x>0) For x > 0, the function is same Therefore, (𝒍𝒊𝒎)┬(𝒙 → 𝟏) " f(x) " " = " lim┬(x→1) " 3(x + 1) " " = 3(1 + 1)" " = 3 (2)" " = 6" (𝐥𝐢𝐦)┬(𝐱→𝟏) f(x) = 6 Ex 13.1, 23 (Method 2) Find lim┬(x→0) f(x) and lim┬(x→1) f(x), where f(x) = {█(2x+3.@3(x+1),)┤ ■8(x ≤0@x>0) Finding limit at x = 0 lim┬(x→0) f(x) = lim┬(〖x→0〗^− ) f(x) =lim┬(〖x→0〗^+ ) f(x) LHL at x → 0 lim┬(x→0^− ) f(x) = lim┬(h→0) f(0 − h) = lim┬(h→0) f(−h) = lim┬(h→0) 2(–h) + 3 = 2(0) + 3 = 3 RHL at x → 0 lim┬(x→0^+ ) f(x) = lim┬(h→0) f(0 + h) = lim┬(h→0) f(h) = lim┬(h→0) 3(h + 1) = 3(0 + 1) = 3 ∴ (𝒍𝒊𝒎)┬(𝐱→𝟎) f(x) = 3 Finding limit at x = 1 "f(x) = " {█(2x+3.@3(x+1),)┤" " ■8(x ≤0@x>0) For x > 0, the function is same Therefore, (𝒍𝒊𝒎)┬(𝒙 → 𝟏) " f(x) " " = " lim┬(x→1) " 3(x + 1) " " = 3(1 + 1)" " = 3 (2)" " = 6" (𝐥𝐢𝐦)┬(𝐱→𝟏) f(x) = 6

Ex 13.1 (Term 1)

Ex 13.1, 1

Ex 13.1, 2

Ex 13.1, 3

Ex 13.1, 4 Important

Ex 13.1, 5

Ex 13.1, 6 Important

Ex 13.1, 7

Ex 13.1, 8 Important

Ex 13.1, 9

Ex 13.1,10 Important

Ex 13.1, 11

Ex 13.1, 12

Ex 13.1, 13

Ex 13.1, 14 Important

Ex 13.1, 15 Important

Ex 13.1, 16

Ex 13.1, 17 Important

Ex 13.1, 18

Ex 13.1, 19 Important

Ex 13.1, 20

Ex 13.1, 21 Important

Ex 13.1, 22 Important

Ex 13.1, 23 You are here

Ex 13.1, 24

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

Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.