Question

The value of $\lim_{n \to \infty} (\frac{n!}{n^n})^{\frac{3n^3+4}{4n^4-1}}$, $n \in N$ is equal to :

• A. $(\frac{1}{e})^{3/4}$
• B. $e^{3/4}$
• C. $e^{-1}$
• D. 0

Answer 1

Answer: (A)

$(\frac{1}{e})^{3/4}$

Answer 2

Let the given limit be $L$. The expression is $L = \lim_{n \to \infty} (\frac{n!}{n^n})^{\frac{3n^3+4}{4n^4-1}}$.

This is of the form $\lim_{n \to \infty} (a_n)^{b_n}$, where $a_n = \frac{n!}{n^n}$ and $b_n = \frac{3n^3+4}{4n^4-1}$.

First, let's find the limit of the base $a_n$: $a_n = \frac{n!}{n^n} = \frac{1 \cdot 2 \cdot 3 \cdots n}{n \cdot n \cdot n \cdots n} = \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n}{n}$.

For $n > 1$, $\frac{k}{n} \le 1$ for all $k=1, 2, \dots, n$. So, $0 < a_n = \frac{1}{n} \cdot \frac{2}{n} \cdots \frac{n}{n} \le \frac{1}{n} \cdot 1 \cdots 1 = \frac{1}{n}$. As $n \to \infty$, $\frac{1}{n} \to 0$. By the Squeeze Theorem, $\lim_{n \to \infty} a_n = 0$.

Next, let's find the limit of the exponent $b_n$: $b_n = \frac{3n^3+4}{4n^4-1}$. To find the limit as $n \to \infty$, we divide the numerator and the denominator by the highest power of $n$ in the denominator, which is $n^4$:

$\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{\frac{3n^3}{n^4}+\frac{4}{n^4}}{\frac{4n^4}{n^4}-\frac{1}{n^4}} = \lim_{n \to \infty} \frac{\frac{3}{n}+\frac{4}{n^4}}{4-\frac{1}{n^4}} = \frac{0+0}{4-0} = 0$.

The limit is of the indeterminate form $0^0$. To evaluate this limit, we consider the limit of the logarithm of the expression:

$\ln L = \lim_{n \to \infty} \ln \left[ (\frac{n!}{n^n})^{\frac{3n^3+4}{4n^4-1}} \right] = \lim_{n \to \infty} \left( \frac{3n^3+4}{4n^4-1} \right) \ln \left( \frac{n!}{n^n} \right)$.

Let's evaluate the limit of the term $\ln(\frac{n!}{n^n})$. $\ln(\frac{n!}{n^n}) = \ln(n!) - \ln(n^n) = \ln(n!) - n \ln n$.

We use the approximation for $\ln(n!)$ for large $n$ from Stirling's formula: $\ln(n!) = n \ln n - n + \frac{1}{2} \ln(2\pi n) + O(\frac{1}{n})$.

So, $\ln(\frac{n!}{n^n}) = (n \ln n - n + \frac{1}{2} \ln(2\pi n) + O(\frac{1}{n})) - n \ln n = -n + \frac{1}{2} \ln(2\pi n) + O(\frac{1}{n})$.

Now, substitute this back into the limit of $\ln L$: $\ln L = \lim_{n \to \infty} \left( \frac{3n^3+4}{4n^4-1} \right) \left( -n + \frac{1}{2} \ln(2\pi n) + O(\frac{1}{n}) \right)$.

Let's multiply the leading terms: $\lim_{n \to \infty} \left( \frac{3n^3}{4n^4} \right) (-n) = \lim_{n \to \infty} \left( \frac{3}{4n} \right) (-n) = \lim_{n \to \infty} -\frac{3}{4} = -\frac{3}{4}$.

Let's verify that the other terms go to zero. The expression is $\frac{3n^3+4}{4n^4-1} = \frac{n^3(3+4/n^3)}{n^4(4-1/n^4)} = \frac{1}{n} \frac{3+4/n^3}{4-1/n^4}$. The term $\ln(\frac{n!}{n^n}) = -n + \frac{1}{2} \ln(2\pi n) + O(\frac{1}{n})$.

The product is $\left( \frac{1}{n} \frac{3+4/n^3}{4-1/n^4} \right) \left( -n + \frac{1}{2} \ln(2\pi n) + O(\frac{1}{n}) \right)$ $= \frac{3+4/n^3}{4-1/n^4} \cdot \frac{1}{n} \left( -n + \frac{1}{2} \ln(2\pi n) + O(\frac{1}{n}) \right)$ $= \frac{3+4/n^3}{4-1/n^4} \cdot \left( -1 + \frac{\ln(2\pi n)}{2n} + O(\frac{1}{n^2}) \right)$.

As $n \to \infty$: $\frac{3+4/n^3}{4-1/n^4} \to \frac{3+0}{4-0} = \frac{3}{4}$. $\frac{\ln(2\pi n)}{2n} \to 0$ (since $\ln x$ grows slower than $x$). $O(\frac{1}{n^2}) \to 0$.

So, $\lim_{n \to \infty} \ln L = \frac{3}{4} \cdot (-1 + 0 + 0) = -\frac{3}{4}$. Therefore, $\ln L = -\frac{3}{4}$. $L = e^{-3/4}$. This can also be written as $L = (e^{-1})^{3/4} = (\frac{1}{e})^{3/4}$.

Our result matches option (a).