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Ex 12.1, 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 12.1, 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo