The problem consists of three questions. Question A1 requires the definition of an accumulation point of a subset of real numbers and the definition of a limit. Question A2 asks to determine the truth value of two statements about continuity/differentiability and differentiability of products. Question A3 requires proving a limit using the epsilon-delta definition and using the first principle to find the derivative of $f(x) = \sin(2x)$.

1. Problem Description

The problem consists of three questions. Question A1 requires the definition of an accumulation point of a subset of real numbers and the definition of a limit. Question A2 asks to determine the truth value of two statements about continuity/differentiability and differentiability of products. Question A3 requires proving a limit using the epsilon-delta definition and using the first principle to find the derivative of $f(x) = \sin(2x)$.

2. Solution Steps

A

1. a) $c \in \mathbb{R}$ is an accumulation point of a subset $A$ of $\mathbb{R}$ if for every $\delta > 0$, there exists an $x \in A$ such that $0 < |x - c| < \delta$. In other words, every neighborhood of $c$ contains a point of $A$ other than $c$ itself.

b) $\lim_{x \to c} f(x) = L$ means that for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - c| < \delta$ and $x \in D_f$, then $|f(x) - L| < \epsilon$, where $D_f$ is the domain of $f$.

A

2. a) The statement "If $f$ is continuous at $c \in I$, then $f$ is differentiable at $c$" is FALSE.

Counterexample: Let $f(x) = |x|$ and $I = [-1, 1]$. Then $f(x)$ is continuous at $x = 0$, but $f(x)$ is not differentiable at $x = 0$.

The reason is because the limit $\lim_{h\to 0} \frac{f(0+h)-f(0)}{h} = \lim_{h \to 0} \frac{|h|}{h}$ does not exist. The limit from the right is 1, and the limit from the left is -

1.

b) The statement: "If $\frac{d}{dx}[f(x)]$ and $\frac{d}{dx}[g(x)]$ both exist, then $\frac{d}{dx}(f(x)g(x)) = (\frac{d}{dx}[f(x)])(\frac{d}{dx}[g(x)])$" is FALSE.

The correct formula is the product rule:

$\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$.

A counterexample is $f(x) = x$ and $g(x) = x$.

Then $f'(x) = 1$ and $g'(x) = 1$. Thus $f'(x)g'(x) = 1$.

However, $\frac{d}{dx}(f(x)g(x)) = \frac{d}{dx}(x^2) = 2x$. These are not equal unless $x=1/2$.

A

3. a) Show $\lim_{x \to 2} (x^2 - 5) = -1$.

Let $\epsilon > 0$. We want to find a $\delta > 0$ such that if $0 < |x - 2| < \delta$, then $|(x^2 - 5) - (-1)| < \epsilon$.

$|(x^2 - 5) + 1| = |x^2 - 4| = |(x - 2)(x + 2)| = |x - 2||x + 2|$.

We want to bound $|x + 2|$. Assume $|x - 2| < 1$. Then $-1 < x - 2 < 1$, so $1 < x < 3$. Thus $3 < x + 2 < 5$, so $|x + 2| < 5$.

Then $|x - 2||x + 2| < 5|x - 2|$.

We want $5|x - 2| < \epsilon$, so $|x - 2| < \frac{\epsilon}{5}$.

Choose $\delta = \min\{1, \frac{\epsilon}{5}\}$.

If $|x - 2| < \delta$, then $|x - 2| < 1$ and $|x - 2| < \frac{\epsilon}{5}$.

Thus $|(x^2 - 5) - (-1)| = |x - 2||x + 2| < |x - 2| \cdot 5 < \frac{\epsilon}{5} \cdot 5 = \epsilon$.

Therefore, $\lim_{x \to 2} (x^2 - 5) = -1$.

b) Let $f(x) = \sin(2x)$. We want to find $f'(x)$ using the first principle.

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{\sin(2(x + h)) - \sin(2x)}{h}$.

$\sin(2(x + h)) = \sin(2x + 2h) = \sin(2x)\cos(2h) + \cos(2x)\sin(2h)$.

$f'(x) = \lim_{h \to 0} \frac{\sin(2x)\cos(2h) + \cos(2x)\sin(2h) - \sin(2x)}{h} = \lim_{h \to 0} \frac{\sin(2x)(\cos(2h) - 1) + \cos(2x)\sin(2h)}{h} = \lim_{h \to 0} \sin(2x)\frac{(\cos(2h) - 1)}{h} + \lim_{h \to 0} \cos(2x)\frac{\sin(2h)}{h}$.

We know $\lim_{\theta \to 0} \frac{\cos(\theta) - 1}{\theta} = 0$ and $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$.

$\lim_{h \to 0} \frac{\cos(2h) - 1}{h} = \lim_{h \to 0} 2 \cdot \frac{\cos(2h) - 1}{2h} = 2 \cdot 0 = 0$.

$\lim_{h \to 0} \frac{\sin(2h)}{h} = \lim_{h \to 0} 2 \cdot \frac{\sin(2h)}{2h} = 2 \cdot 1 = 2$.

Therefore, $f'(x) = \sin(2x) \cdot 0 + \cos(2x) \cdot 2 = 2\cos(2x)$.

3. Final Answer

A

1. a) $c \in \mathbb{R}$ is an accumulation point of a subset $A$ of $\mathbb{R}$ if for every $\delta > 0$, there exists an $x \in A$ such that $0 < |x - c| < \delta$.

b) $\lim_{x \to c} f(x) = L$ means that for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - c| < \delta$ and $x \in D_f$, then $|f(x) - L| < \epsilon$, where $D_f$ is the domain of $f$.

A

2. a) False. Counterexample: $f(x) = |x|$ at $x=0$.

b) False. The correct formula is $\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$.

A

3. a) $\lim_{x \to 2} (x^2 - 5) = -1$ (Proof shown in solution steps).

b) $f'(x) = 2\cos(2x)$.