Limit's theorems

Calculus Level 3

lim ⁡ x → 0 ( ( 1 + x ) 2018 − 1 x ⋅ ( 1 + x ) 2017 − 1 x ⋅ ( 1 + x ) 2016 − 1 x ⋅ ⋯ ( 1 + x ) 2 − 1 x ⋅ 1 ) \lim_{x\to 0 }\left(\dfrac{\,(1+x)^{2018}-1}{x}\cdot \dfrac{\,(1+x)^{2017}-1}{x}\cdot \dfrac{\,(1+x)^{2016}-1}{x}\cdot\cdots \dfrac{\,(1+x)^{2}-1}{x}\cdot 1 \right) If the value of the above limit can be expressed as a ! a! . Find the value of a a .


Notation: ! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 2018.

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

The given limit is:

L = lim ⁡ x → 0 ∏ n = 2 2018 ( 1 + x ) n − 1 x By binomial expansion = ∏ n = 2 2018 lim ⁡ x → 0 ( 1 + n x + n ( n − 1 ) 2 ! x 2 + ⋯ ) − 1 x Divide up and down by x = ∏ n = 2 2018 lim ⁡ x → 0 ( n + n ( n − 1 ) 2 ! x + n ( n − 1 ) ( n − 2 ) 3 ! x 2 ⋯ ) = ∏ n = 2 2018 n = 2018 ! \begin{aligned} L & = \lim_{x \to 0} \prod_{n=2}^{2018} \frac {{\color{#3D99F6}(1+x)^n}-1}x & \small \color{#3D99F6} \text{By binomial expansion} \\ & = \prod_{n=2}^{2018} \lim_{x \to 0} \frac {{\color{#3D99F6}\left(1+nx + \frac {n(n-1)}{2!}x^2 + \cdots \right)}-1}x & \small \color{#3D99F6} \text{Divide up and down by }x \\ & = \prod_{n=2}^{2018} \lim_{x \to 0} \left(n + \frac {n(n-1)}{2!}x + \frac {n(n-1)(n-2)}{3!}x^2 \cdots \right) \\ & = \prod_{n=2}^{2018} n = 2018! \end{aligned}

Therefore, a = 2018 a=\boxed{2018} .

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