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The sum limit law states that:

$$ \lim_{x \to a} f(x) + g(x) = \lim_{x \to a} f(x)+ \lim_{x \to a} g(x) $$ where $\lim_{x \to a} f(x)$ and $ \lim_{x \to a} g(x) $ exists. Now, using stiriling's approximation of the factorial, we can show that: $$ \lim_{x \to \infty} \frac{\log(x!)}{x} = \infty \quad (1) $$

However, \begin{align} \lim_{x \to \infty} \frac{\log(x!)}{x} =& \lim_{x \to \infty} \frac{\log(x) + \log(x-1) + \dots + \log(1)}{x} \\ = &\lim_{x \to \infty} \frac{\log(x)}{x} + \lim_{x \to \infty} \frac{\log(x-1)}{x} + \dots + \lim_{x \to \infty} \frac{\log(1)}{x} \quad (2) \end{align}

Now $x$ asymptotically dominates a logarithm of polynomials, hence any limit of the form:

$$ \lim_{x \to \infty} \frac{\log(x-i)}{x} = 0, \quad i \in N $$ Therefore, each limit in (2) computes to $0$. And as per sum law, $(2)$ is hence $0$. However this contradicts the actual result of $\infty$ in $(1)$

What am I missing?

EDIT: My question is different than the one mentioned in the close banner. That questions asks how to compute $\lim_{x \to \infty} \frac{\log(x!)}{x}$, my question asks why applying the sum limit law gives an incorrect limit for $\lim_{x \to \infty} \frac{\log(x!)}{x}$.

Karan Mehta
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1 Answers1

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The sum limit law applies to a fixed, nonvarying, number of summands. For example, you cannot invoke the sum limit law to deduce $$ \lim_{n\to\infty}\frac nn = \lim \frac{1+\cdots+1}n=\lim\frac 1n +\cdots+\lim\frac1n= 0.$$

grand_chat
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