MathsMaths QuestionsLimits And Derivatives Questions for CBSE Class 11th

Limits And Derivatives Questions for CBSE Class 11th

Let f”(x) be continuous at x = 0.If lim x 0 2 f ( x ) − 3 af ( 2 x ) + bf ( 8 x ) sin 2 ⁡ x exists and f ( 0 ) ≠ 0 , f ‘ ( 0 ) ≠ 0 , then the value of 3a/b is .

If L = lim x ∞ x 3 + 2 x 2 3 − x 2 + x then the value of 3/L is .

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    If lim x ∞ x α x 2 + x 4 + 1 – 2 x exists and has value non-zero finite real number L , then the value of 100 – α L 2 is

    Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit lim x 0 + ( 1 − x ) 1 x − e − 1 x a is equal to a nonzero real number, is

    If L = lim x ∞ x − x 2 log e ⁡ 1 + 1 x , then the value of 40L is .

    If L = lim x 2 60 + x 2 3 − 4 sin ⁡ ( x − 2 ) , then the value of 1/L is .

    The value of the limit lim x π 2 4 2 ( sin ⁡ 3 x + sin ⁡ x ) 2 sin ⁡ 2 x sin ⁡ 3 x 2 + cos ⁡ 5 x 2 − 2 + 2 cos ⁡ 2 x + cos ⁡ 3 x 2 is

    Let f p ( α ) = e i α p 2 ⋅ e 2 i α p 2 … … . e i α p , p ∈ N where i = − 1 , then find the value of lim n ∞ f n ( π )

    The value of the limit lim x 0 1 1 / sin 2 ⁡ x + 2 1 / sin 2 ⁡ x + … + n 1 / sin 2 ⁡ x sin 2 ⁡ x is

    The value of 78.6 e lim x 0 1 + x 1 x e 1 x i s

    Match the following lists: Column -I Column -II A. If L = lim x − 1 ( 7 − x ) 3 − 2 ( x + 1 ) , then 12L = P. -2 B. If L = lim x π / 4 tan 3 ⁡ x − tan ⁡ x cos ⁡ x + π 4 then -L/4= Q. 2 C. If L = lim x 1 ( 2 x − 3 ) ( x − 1 ) 2 x 2 + x − 3 , then 20L= R. 1 D. If L = lim x ∞ log ⁡ x n − [ x ] [ x ] , where n ∈ N ([x] denotes greatest integer less than or equal to x),then -2L= S. -1

    If y = x ( log x ) log ( log x ) ,   t h e n    d y d x i s

    If lim x 0 1 + x log ⁡ 1 + b 2 1 / x = 2 b sin 2 ⁡ θ , b > 0 and θ ∈ [ − π , π ] , then the value of θ is

    If m , n ∈ N , lim x 0 sin ⁡ x n ( sin ⁡ x ) m is

    f ( x ) = 1 − x ( 1 + | 1 − x | ) | 1 − x | cos ⁡ 1 1 − x x ≠ 1 . Then

    Let f(x) be a polynomial satisfying lim x ∞ x 2 f ( x ) 2 x 5 + 3 = 6 and f ( 1 ) = 3 , f ( 3 ) = 7 and f ( 5 ) = 11 Then the value of | f ( 0 ) | is .

    If lim x 0 ( 4 x − 1 ) 1 3 + a + bx x = 1 3 , then the value of | ab |

    L = lim x 0 sin ⁡ x + ae x + be − x + cln ⁡ ( 1 + x ) x 3 = ∞

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