Limits 3.0

Calculus Level 4

The value of lim ⁡ x → 0 ∣ x ∣ 2 ( 1 + 2 + 3 + . . . . . . [ 1 ∣ x ∣ ] ) = A \displaystyle \lim_{x \to 0} |x|^{2}(1+2+3 +......[\frac{1}{|x|}])=A . where |x| means absolute value of x and [.] means Greatest Integer function .Find 1 A \frac{1}{A}


The answer is 2.

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Put ∣ x ∣ = 1 / n |x|=1/n where n → ∞ n\to\infty

⇒ A = ( n ) ( n + 1 ) / ( 2 n 2 ) = 1 / 2 \Rightarrow A=(n)(n+1)/(2n^2)=1/2

⇒ 1 / A = 2 \Rightarrow 1/A=2

3 Helpful 1 Interesting 0 Brilliant 0 Confused

though your answer is correct , the method is not elaborated. you have to mention how you removed the unbounded case of the floor function. you first have to write [1/x] = 1/x - {1/x}. where {▪}. is the fractional part function. now as fractional part is bounded and its range is [0,1]. so by your method {n}/n tends to 0 as n tends to infinity. and the ratio of the coefficient of n^2 is 0.5 so answer is 2.

Antara Chatterjee - 3 years ago

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