Evaluating lim n → ∞ ( 1 − x n 1 + a ) n

That step you are uncomfortable with is indeed quite dodgy. Instead, consider the logarithm of your limit. Since log ⁡ ( 1 − t ) = − t + t 2 / 2 + t 3 / 3 + ⋯ , n log ⁡ ( 1 − x n 1 + a ) = − ( x n a + x 2 2 n 1 + 2 a + x 3 3 n 2 + 3 a + ⋯ ) so as n → ∞ , if a = 0 this goes to − x, if a &gt; 0 to goes to 0 and if a &lt; 0 this goes to − ∞ .. Exponentiating gives back the result.

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Garrett Sheppard

Open question

2022-08-14

Evaluating

lim_{n \to \infty} {(1 - \frac{x}{n^{1 + a}})}^{n}

Answer & Explanation

Brennan Parks

Beginner2022-08-15Added 14 answers

That step you are uncomfortable with is indeed quite dodgy. Instead, consider the logarithm of your limit. Since

\log (1 - t) = - t + t^{2} / 2 + t^{3} / 3 + \dots,

n \log (1 - \frac{x}{n^{1 + a}}) = - (\frac{x}{n^{a}} + \frac{x^{2}}{2 n^{1 + 2 a}} + \frac{x^{3}}{3 n^{2 + 3 a}} + \dots)

so as

n \to \infty,

a = 0

this goes to

- x

, if

a > 0

to goes to 0 and if

a < 0

this goes to

- \infty .

. Exponentiating gives back the result.

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