Find $\lim_{n \to \infty} \left( \frac{n!}{n^n} \right)^{1/n}$.
Using Stirling's approximation or Cauchy's second limit theorem: $\lim (a_n)^{1/n} = \lim (a_{n+1}/a_n)$. Here a_n = n!/$n^n$. $\lim a_{n+1}/a_n = \lim (1 + 1/n)^{-n} = 1/e$.
Evaluate $\lim_{x \to 0^+} x^x$.
Let $L = x^x$. ln L = x ln x. As x -> 0+, $\lim x ln x = \lim (ln x)/(1/x) = \lim (-1/x)/(1/x^2) = 0$. Thus $L = e^0$ = 1.
Let $f(x) = x^3 - 3x + 2$. According to Rolle's Theorem on [-2, 1], there exists c $\in (-2, 1) such that f'(c) = 0$. What is the value of c?
f(-2) = -8 + 6 + 2 = 0 and f(1) = 1 - 3 + 2 = 0. f'(x) = $3x^2 - 3$ = 0 => $x^2$ = 1 => c = -1 $\in (-2, 1)$.
Applying Lagrange's Mean Value Theorem to f(x) = $\sqrt{x} on [1, 4], find$ the value of c in (1, 4) such that f'(c) = $\frac{f(4)-f(1)}{4-1}$.
f(4) = 2, f(1) = 1. Average rate = (2-1)/3 = 1/3. f'(c) = 1/(2$\sqrt{c}) = 1/3 => 2\sqrt{c} = 3 => \sqrt{c} = 1.5 => c = 2.25$.
The coefficient of (x - $1)^2$ in the Taylor series expansion of f(x) = x $\ln x about x = 1$ is:
f(x) = x ln x. f'(x) = 1 + ln x, f''(x) = 1/x. At x=1, f''(1) = 1. The coefficient of (x-$1)^2$ in Taylor series is f''(1)/2! = 1/2.
