1. Class 11
2. Important Question for exams Class 11

Transcript

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┬(z→1) 𝑥^15 – 〖(1)〗^15 ÷ lim┬(z→1) x10 – (1)10 Multiplying and dividing by x – 1 = lim┬(z→1) (𝑥^15 − 1^15)/(𝑥 − 1) ÷ lim┬(z→1) (𝑥^10 − 〖(10)〗^10)/(𝑥 − 1) Comparing lim┬(x→1) (x^15 − 〖(1)〗^15)/(x − 1) with lim┬(x→a) (xn − an)/(x − a) = nan – 1 Here a = 1 , n = 15 , x = x lim┬(x→1) (𝑥^15 − 〖(10)〗^15)/(𝑥 − 1) = 15(1)15 – 1 = 15 (1)14 = 15 Also comparing lim┬(x→1) (x^10 − 〖(1)〗^10)/(x − 1) with lim┬(x→a) (x^n − a^n)/(x − a) Here x = x , n = 10 , a = 1 So, lim┬(z→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┬(x→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) = 𝟏/𝟐

Class 11
Important Question for exams Class 11