Breaking my limits

Calculus Level 3

lim ⁡ x → ∞ n x n e x = 0 \large \lim_{x\to \infty} \frac{nx^n}{e^x} = 0

Let n n be a certain positive integer satisfying the equation above, which of the following options must be true?

Clarification : e e denotes the Euler's number , e ≈ 2.71828 e \approx 2.71828 .

Source : IIT - 1992.
n = 0 n= 0 only n = 2 n=2 only None of these choices n n is any positive integer There is no value of n n

Sabhrant Sachan by Sabhrant Sachan , 21, India

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2 solutions

Chew-Seong Cheong
May 22, 2016

lim ⁡ x → ∞ n x n e x = lim ⁡ x → ∞ n x n 1 + x 1 ! + x 2 2 ! + . . . + x n n ! + . . . Maclaurin series = lim ⁡ x → ∞ n x n x n 1 x n + 1 1 ! x n + x 2 2 ! x n + . . . + x n n ! x n + . . . = lim ⁡ x → ∞ n 1 x n + 1 x n − 1 + 1 2 x n − 2 + . . . + 1 n ! + 1 ( n + 1 ) ! x + 1 ( n + 2 ) ! x 2 + . . . Red terms → 0 and blue terms → ∞ = 0 \begin{aligned} \lim_{x \to \infty} \frac{nx^n}{\color{#3D99F6}{e^x}} & = \lim_{x \to \infty} \frac{nx^n}{\color{#3D99F6}{1+\frac{x}{1!} + \frac{x^2}{2!}+ ... + \frac{x^n}{n!} + ...}} \quad \quad \small \color{#3D99F6}{\text{Maclaurin series}} \\ & = \lim_{x \to \infty} \frac{\frac{nx^n}{x^n}}{\frac{1}{x^n}+\frac{1}{1!x^n} + \frac{x^2}{2!x^n}+ ... + \frac{x^n}{n!x^n} + ...} \\ & = \lim_{x \to \infty} \frac{n}{\color{#D61F06}{\frac{1}{x^n}+\frac{1}{x^{n-1}} + \frac{1}{2x^{n-2}}+ ... }+ \frac{1}{n!} + \color{#3D99F6}{\frac{1}{(n+1)!}x + \frac{1}{(n+2)!}x^2 + ...}} \quad \quad \small \color{#D61F06}{\text{Red terms}\to 0} \text{ and } \color{#3D99F6}{\text{blue terms}\to \infty} \\ & = \boxed{0} \end{aligned}

4 Helpful 0 Interesting 0 Brilliant 0 Confused
J D
May 22, 2016

An exponential will always grow faster than a polynomial or a constant, so no matter what n is, the limit will always be zero as X approaches infinity.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

exactly ... nice observation, you can even solve the problem using LH rule

lim ⁡ x → ∞ n x n e x lim ⁡ x → ∞ n ⋅ n ! e x ⟹ 0 \displaystyle \lim_{x\to \infty} \frac{nx^n}{e^x} \\ \displaystyle \lim_{x\to \infty} \frac{n\cdot n!}{e^x} \implies 0

Sabhrant Sachan - 5 years ago

Yeah , same approach!

Samanvay Vajpayee - 5 years ago

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