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Im fairly sure it's $x$. Here's what I've got so far. $\lim_{n\to\infty}ne^{\frac{x}{n}}-n = \lim_{n\to\infty}n(e^{\frac{x}{n}}-1) = \lim_{n\to\infty}n(-1+\sum_{i=0}^\infty\frac{x^i}{i!n^i}) = \lim_{n\to\infty}x+\sum_{i=0}^\infty\frac{x^i}{i!n^{i-1}}$

mathemagiker
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  • It is possible to develop an entire theory of exponential and logarithmic functions starting from the definition $\log x=\lim_{n\to \infty} n(x^{1/n}-1)$. The result in your question is an immediate application of this well known limit. – Paramanand Singh Feb 13 '21 at 06:04

1 Answers1

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You have :

$$ne^{\frac{x}{n}}-n=n\left(1+\frac{x}{n}+o\left(\frac{1}{n}\right)\right)-n=x+o(1) $$

So you get :

$$\lim_{n\to\infty}ne^{\frac{x}{n}}-n = x $$

NHL
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  • Seems a bit off... That $o(1)$ term has nothing forcing it towards a specific limit, so how does it vanish? – abiessu Feb 12 '21 at 15:24
  • $o(1)$ means that the term is negligible compared to 1, so it's a function that goes to $0$ when $n$ goes to infinity. For example, $f=o(g) $, then $\lim f/g =0$, if you replace $g$ by $1$, you get that the small $o$ goes to 0 – NHL Feb 12 '21 at 15:25
  • That's weird, I guess my understanding that $o(1)$ is a term upper bounded by $1$ is incorrect. If this is proper terminology then no worries. – abiessu Feb 12 '21 at 15:28
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    By the conventions I know of, the upper bound by 1 is denoted by a big $o$ , and it would be $O(1)$ – NHL Feb 12 '21 at 15:35
  • Thank you for the clarification. – abiessu Feb 12 '21 at 16:47