Example 3 - Chapter 13 Class 11 Limits and Derivatives

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Example 3 Evaluate: (i) (𝑙𝑖𝑚)┬(𝑥→1) (𝑥 15 − 1)/(𝑥10 − 1) (𝑙𝑖𝑚)┬(𝑥→1) (𝑥 15 − 1)/(𝑥10 − 1) = (〖(1)〗^15 − 1)/(〖(1)〗^10 − 1) = (1 − 1)/(1 − 1) = 0/0 Since it is form 0/0, We can solve by using theorem (𝑙𝑖𝑚)┬(𝑥→𝑎) (𝑥^𝑛 − 𝑎^𝑛)/(𝑥 − 𝑎) = na n – 1 Hence, (𝑙𝑖𝑚)┬(𝑥→1) (𝑥^15 − 1)/(𝑥^10 − 1) = (𝑙𝑖𝑚)┬(𝑥→1) 𝑥^15 – 1 ÷lim┬(x→1) x10 – 1 = (𝑙𝑖𝑚)┬(𝑥→1) 𝑥^15 – 〖(1)〗^15 ÷ lim┬(x→1) x10 – (1)10 Multiplying and dividing by x – 1 = (𝑙𝑖𝑚)┬(𝑥→1) (𝑥^15 − 1^15)/(𝑥 − 1) ÷ (𝑙𝑖𝑚)┬(𝑧→1) (𝑥^10 − 〖(10)〗^10)/(𝑥 − 1) Using (𝑙𝑖𝑚)┬(𝑥→𝑎) ( 𝑥^𝑛 − 𝑎^𝑛)/(𝑥 − 𝑎) = nan – 1 Using (𝑙𝑖𝑚)┬(𝑥→𝑎) ( 𝑥^𝑛 − 𝑎^𝑛)/(𝑥 − 𝑎) = nan – 1 (𝑙𝑖𝑚)┬(𝑥→1) (𝑥^15 − 〖(1)〗^15)/(𝑥 − 1) = 15(1)15 – 1 = 15 (1)14 = 15 (𝑙𝑖𝑚)┬(𝑥→1) (𝑥^10 − 〖(1)〗^10)/(𝑥 − 1) = 10(1)10 – 1 = 10 (1)9 = 10 Hence , (𝑙𝑖𝑚)┬(𝑥→1) (𝑥^15 − 1^15)/(𝑥 − 1) ÷ (𝑙𝑖𝑚)┬(𝑥→1) (𝑥^10 −110)/(𝑥 − 1) = 15 ÷ 10 = 15/10 = 3/2 ∴ (𝒍𝒊𝒎)┬(𝒙→𝟏) (𝒙^𝟏𝟓 − 𝟏)/(𝒙^𝟏𝟎 − 𝟏) = 𝟑/𝟐 Example 3 Evaluate: (ii) (𝑙𝑖𝑚)┬(𝑥→0) (√(1 + 𝑥) − 1)/𝑥 (𝑙𝑖𝑚)┬(𝑥→0) (√(1 + x )− 1)/x Putting x = 0 = (√(1 + 0) − 1)/0 = (√(1 ) − 1)/0 = (1 − 1)/0 = 0/0 Since it is a 0/0 form We simplify the equation Putting y = 1 + x ⇒ y – 1 = x As x → 0 y → 1 + 0 y → 1 So, our equation becomes (𝑙𝑖𝑚)┬(𝑥→0) (√(1 + 𝑥 )− 1)/𝑥 = (𝑙𝑖𝑚)┬(𝑦→1) (√𝑦 − 1)/(𝑦 − 1) = (𝑙𝑖𝑚)┬(𝑥→1) ( 𝑦^((−1)/2) − 1)/(𝑦 − 1) = (𝑙𝑖𝑚)┬(𝑥→1) ( 𝑦^((−1)/2) − 1^((−1)/2))/(𝑦 − 1) = 1/2 × 1^((−1)/2 − 1) = 1/2 × 1 = 𝟏/𝟐