Example 2 - Find limits (i) lim x->1 [x2 + 1 / x + 100] - Limits - 0/0 form

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  1. Class 11
  2. Important Question for exams Class 11
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Example , 2 Find the limits: (i) lim﷮x→1﷯ x2 + 1﷮x + 100﷯﷯ lim﷮x→1﷯ x2 + 1﷮x + 100﷯﷯ Putting x = 1 = 1﷯2 + 1﷮1 + 100﷯ = 1 + 1 ﷮101﷯ = 𝟐﷮𝟏𝟎𝟏﷯ Example, 2 Find the limits: (ii) lim﷮x→2﷯ x3 −4x2+4x﷮x2−4﷯﷯ lim﷮x→2﷯ x3 − 4x2 + 4x﷮x2 − 4﷯﷯ = lim﷮x→2﷯ 𝑥 𝑥﷮2﷯ − 4𝑥 + 4﷯﷮ 𝑥﷮2﷯ − (2)﷮2﷯﷯ = lim﷮x→2﷯ 𝑥 𝑥﷮2﷯ − 2𝑥 − 2𝑥 + 4﷯﷮ 𝑥 − 2﷯ (𝑥 + 2)﷯ = lim﷮x→2﷯ 𝑥 𝑥 𝑥 − 2﷯ − 2 (𝑥 − 2)﷯﷮ 𝑥 − 2﷯ (𝑥 + 2)﷯ = lim﷮x→2﷯ 𝑥 𝑥 − 2﷯ (𝑥 − 2)﷮ 𝑥 − 2﷯ (𝑥 + 2)﷯ = 𝐥𝐢𝐦﷮𝐱→𝟐﷯ 𝒙 𝒙 − 𝟐﷯﷮𝒙 + 𝟐﷯ Putting x = 2 = 2 (2 − 2)﷮2 + 2﷯ = 2 (0)﷮4﷯ = 0﷮4﷯ = 0 Example, 2 Find the limits: (iii) lim﷮x→2﷯ x2 −4﷮x3 − 4x2 + 4x﷯﷯ lim﷮x→2﷯ x2 − 4﷮x3− 4x2 + 4x﷯﷯ = lim﷮x→2﷯ x2 − 2﷯2﷮x(x2 − 4x + 4) ﷯﷯ = lim﷮x→2﷯ 𝑥 − 2﷯ (𝑥 + 2)﷮x(x x − 2﷯ − 2 x − 2﷯) ﷯﷯ = lim﷮x→2﷯ 𝑥 − 2﷯ (𝑥 + 2)﷮𝑥 x − 2﷯ x − 2﷯ ﷯﷯ = 𝐥𝐢𝐦﷮𝐱→𝟐﷯ 𝒙 + 𝟐﷮𝐱 𝐱 − 𝟐﷯ ﷯﷯ Putting x = 2 = 2 + 2﷮2 2 − 2﷯ ﷯ = 2 + 2﷮2 2 − 2﷯ ﷯ = 4﷮2(0)﷯ = 𝟒﷮𝟎﷯ = 唴 Which is not defined Example 2 Find the limits: (iv) lim﷮x→2﷯ x3 −2𝑥2﷮x2−5x+6﷯﷯ lim﷮x→2﷯ x3 − 2𝑥2﷮x2 − 5x + 6﷯﷯ = lim﷮x→2﷯ x2 x − 2﷯﷮𝑥2 − 3𝑥 − 2𝑥 + 6﷯﷯ = lim﷮x→2﷯ x2 x − 2﷯﷮x x − 3﷯ − 2 x − 3﷯﷯﷯ = lim﷮x→2﷯ x2 x − 2﷯﷮ x − 2﷯ x − 3﷯﷯﷯ = 𝐥𝐢𝐦﷮𝐱→𝟐﷯ 𝐱𝟐 ﷮ 𝐱 − 𝟑﷯﷯﷯ Putting x = 2 = 2﷯2﷮2 − 3﷯ = 2﷯2﷮2 − 3﷯ = 4﷮−1﷯ = – 4 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﷯ x − 2﷯2 − 1﷮𝑥 𝑥 − 1﷯ (𝑥 − 2)﷯﷯ = lim﷮x→1﷯ 𝑥2 + 2﷯2 − 2 𝑥﷯ 2﷯ − 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 − 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)﷯ = 1 − 3﷮1( − 1)﷯ = − 2﷮− 1﷯ = 2

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