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Ex 13.1, 10 - Evaluate: lim z->1 z1/3 - 1 / z1/6 - 1 - Limits - x^n formula

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  1. Class 11
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Ex 13.1, 10 Evaluate the Given limit: lim┬(zβ†’1) (𝑧^(1/3) βˆ’ 1)/(𝑧^(1/6) βˆ’ 1) lim┬(zβ†’1) (𝑧^(1/3) βˆ’ 1)/(𝑧^(1/6) βˆ’ 1) = (γ€–(1)γ€—^(1/3) βˆ’ 1)/(γ€–(1)γ€—^(1/6) βˆ’ 1) = (1 βˆ’ 1)/(1 βˆ’ 1) = 0/0 Since it is form 0/0, we can solve by using theorem lim┬(xβ†’a) (π‘₯^𝑛 βˆ’ π‘Ž^𝑛)/(π‘₯ βˆ’ π‘Ž) = na n – 1 Hence, lim┬(zβ†’1) (𝑧^(1/3) βˆ’ 1)/(𝑧^(1/6) βˆ’ 1) = lim┬(zβ†’1) 𝑧^(1/3) – 1 Γ· lim┬(zβ†’1) 𝑧^(1/6) βˆ’ 1 = lim┬(zβ†’1) 𝑧^(1/3) – γ€–(1)γ€—^(1/3) Γ· lim┬(zβ†’1) 𝑧^(1/6) – γ€–(1)γ€—^(1/6) Multiplying and dividing by z – 1 = lim┬(zβ†’1) (𝑧^(1/3) βˆ’ γ€–(1)γ€—^(1/3))/(𝑧 βˆ’ 1) Γ· lim┬(zβ†’1) (𝑧^(1/6) βˆ’γ€– (1)γ€—^(1/6))/(𝑧 βˆ’ 1) It is of form lim┬(x β†’ a) (π‘₯^𝑛 βˆ’ π‘Ž^𝑛)/(π‘₯ βˆ’ π‘Ž ) = nan – 1 Replacing x with z, a with 1, n with 1/3 lim┬(zβ†’1) (𝑧^(1/3) βˆ’ γ€–(1)γ€—^(1/3))/(𝑧 βˆ’1) = 1/3 γ€–(1)γ€—^(1/3 βˆ’ 1) = 1/3 Γ— 1 = 1/3 Similarly replacing x with z, a with 1 and n = 1/6 lim┬(zβ†’1) (𝑧^(1/6) βˆ’ 1^(1/6))/(𝑧 βˆ’ 1) = 1/6 γ€–(1)γ€—^(1/6 βˆ’ 1) = 1/6 Γ— 1 = 1/6 Hence our equation = lim┬(zβ†’1) (𝑧^(1/3) βˆ’ γ€–(1)γ€—^(1/3))/(𝑧 βˆ’ 1) Γ· lim┬(zβ†’1) (𝑧^(1/6) βˆ’ 6)/(𝑧 βˆ’ 1) = 1/3 Γ·1/6 = 1/3 Γ— 6/1 = 2

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    Anuj Chowdhary
    Sept. 17, 2017, 2 p.m.

    when f(x) is continuos for all real values of x ; and f(x) = acosx ; x<0 ; F(x)= ax^1000000 ; x≥0 what is f(1000000000000000000000000000000) ?

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