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

Examples

Example 1 (i)

Example 1 (ii)

Example 1 (iii)

Example 2 (i)

Example 2 (ii) Important

Example 2 (iii) Important

Example 2 (iv)

Example 2 (v) You are here

Example 3 (i) Important

Example 3 (ii) Important

Example 4 (i)

Example 4 (ii) Important

Example 5

Example 6

Example 7 Important

Example 8

Example 9

Example 10 Important

Example 11

Example 12

Example 13 Important

Example 14

Example 15 Important

Example 16

Example 17 Important

Example 18

Example 19 (i) Important

Example 19 (ii)

Example 20 (i)

Example 20 (ii) Important

Example 21 (i)

Example 21 (ii) Important

Example 22 (i)

Example 22 (ii) Important

Last updated at May 29, 2023 by Teachoo

Example 2 Find the limits: (v) lim┬(x→1) [(x −2)/(x2−x)− 1/(𝑥3 −3𝑥2+2𝑥)] lim┬(x→1) [(x − 2)/(x2 − x)− 1/(𝑥3 − 3𝑥2 + 2𝑥)] = lim┬(x→1) [(x − 2)/(x (x −1))− 1/(𝑥 (𝑥2 − 3𝑥 + 2))] = lim┬(x→1) [(x − 2)/(x (x −1))− 1/(𝑥 (𝑥2 − 2𝑥 − 𝑥 + 2))] = lim┬(x→1) [(x − 2)/(x (x −1))− 1/(𝑥 (𝑥 (𝑥 − 2) − 1 (𝑥 − 2)))] = lim┬(x→1) [(x − 2)/(x (x −1))− 1/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [((x − 2)(𝑥 − 2) − 1)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [((x − 2)2 − 1)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [(𝑥2 + 2^2 − 4𝑥 − 1)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [(𝑥2 + 4 − 4𝑥 − 1)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = (𝐥𝐢𝐦)┬(𝒙→𝟏) [(𝒙𝟐 − 𝟒𝒙 − 𝟑)/(𝒙 (𝒙 − 𝟏) (𝒙 − 𝟐))] Putting x = 1 = (1^2 − 4 (1) + 3)/(1(1 − 1)(1 − 2)) = 𝟎/𝟎 Since it is 0/0 form we can simplify = lim┬(x→1) [(𝑥2 − 4𝑥 − 3)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [(𝑥2 − 3𝑥 − 𝑥 + 3)/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [(𝑥 (𝑥 − 3) − 1(𝑥 − 3))/(𝑥 (𝑥 − 1) (𝑥 − 2))] = lim┬(x→1) [((𝑥 − 1) (𝑥 − 3))/(𝑥 (𝑥 − 1) (𝑥 − 2))] = (𝐥𝐢𝐦)┬(𝒙→𝟏) [( 𝒙 − 𝟑)/(𝒙 (𝒙 − 𝟐))] Putting x = 1 = (1 − 3)/(1 (1 − 2)) = (−2)/(1 × −1) = 2