We need to solve the following problems: 1.1 (a) Find the limit $\lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$. 1.1 (b) Find the limit $\lim_{x \to 3} (2x + |x-3|)$. 1.2 1 (a) Find the domain of $f(x) = \frac{x^2}{\lfloor x+1 \rfloor - 2}$. 1.2 1 (b) Discuss the continuity of $f(x)$ at $x=1$. 1.2 2 (a) Find the domain of $g(x) = \sqrt{\sec(x+\pi)}$. 1.2 2 (b) Discuss the continuity of $g(x)$ at $a = \pi$. 1.3 (a) Prove that $\frac{d}{dx}(\cos x) = -\sin x$ using the definition $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$. 1.3 (b) Find the domain of $y = \sec^{-1} x$. Find the first derivative of the function. 1.4 Find the equation of the tangent line to the curve $y = g(x)$ at $x = 0$, if $y + x \cos y = x^2$.

1. Problem Description

We need to solve the following problems:

1.1 (a) Find the limit $\lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$.

1.1 (b) Find the limit $\lim_{x \to 3} (2x + |x-3|)$.

1.2 1 (a) Find the domain of $f(x) = \frac{x^2}{\lfloor x+1 \rfloor - 2}$.

1.2 1 (b) Discuss the continuity of $f(x)$ at $x=1$.

1.2 2 (a) Find the domain of $g(x) = \sqrt{\sec(x+\pi)}$.

1.2 2 (b) Discuss the continuity of $g(x)$ at $a = \pi$.

1.3 (a) Prove that $\frac{d}{dx}(\cos x) = -\sin x$ using the definition $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$.

1.3 (b) Find the domain of $y = \sec^{-1} x$. Find the first derivative of the function.

1.4 Find the equation of the tangent line to the curve $y = g(x)$ at $x = 0$, if $y + x \cos y = x^2$.

2. Solution Steps

1.1 (a)

$\lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \lim_{h \to 0} \frac{x^2 - (x+h)^2}{h(x+h)^2 x^2} = \lim_{h \to 0} \frac{x^2 - (x^2 + 2xh + h^2)}{h(x+h)^2 x^2} = \lim_{h \to 0} \frac{-2xh - h^2}{h(x+h)^2 x^2} = \lim_{h \to 0} \frac{-2x - h}{(x+h)^2 x^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}$.

The limit exists for $x \neq 0$.

1.1 (b)

$\lim_{x \to 3} (2x + |x-3|)$.

When $x > 3$, $|x-3| = x-3$, so $\lim_{x \to 3^+} (2x + x-3) = 2(3) + 3 - 3 = 6$.

When $x < 3$, $|x-3| = -(x-3) = 3-x$, so $\lim_{x \to 3^-} (2x + 3-x) = 2(3) + 3 - 3 = 6$.

Since the left-hand limit and the right-hand limit are both equal to 6, the limit exists and equals

6.

1.2 1 (a)

$f(x) = \frac{x^2}{\lfloor x+1 \rfloor - 2}$. The domain is all $x$ such that $\lfloor x+1 \rfloor - 2 \neq 0$, or $\lfloor x+1 \rfloor \neq 2$.

So $\lfloor x+1 \rfloor$ cannot be 2, which means $2 \leq x+1 < 3$ must be avoided. Thus $1 \leq x < 2$ must be avoided.

Therefore, the domain is $x \in (-\infty, 1) \cup [2, \infty)$.

1.2 1 (b)

We need to discuss the continuity of $f(x)$ at $x=1$.

$f(1)$ is not defined, since $1$ is not in the domain.

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^2}{\lfloor x+1 \rfloor - 2} = \frac{1^2}{1-2} = -1$.

$\lim_{x \to 1^+} f(x)$ does not exist, since $f(x)$ is not defined on $(1,2)$.

Since $f(1)$ is not defined, $f(x)$ is discontinuous at $x=1$.

1.2 2 (a)

$g(x) = \sqrt{\sec(x+\pi)}$.

For the square root to be defined, $\sec(x+\pi) \geq 0$.

Since $\sec(x+\pi) = \frac{1}{\cos(x+\pi)}$, we need $\cos(x+\pi) \geq 0$.

$\cos(x+\pi) \geq 0$ when $-\frac{\pi}{2} + 2n\pi \leq x + \pi \leq \frac{\pi}{2} + 2n\pi$ for some integer $n$.

$-\frac{3\pi}{2} + 2n\pi \leq x \leq -\frac{\pi}{2} + 2n\pi$.

Therefore, the domain is $x \in \bigcup_{n \in \mathbb{Z}} [-\frac{3\pi}{2} + 2n\pi, -\frac{\pi}{2} + 2n\pi]$.

1.2 2 (b)

We need to discuss the continuity of $g(x)$ at $a = \pi$.

$g(\pi) = \sqrt{\sec(2\pi)} = \sqrt{1} = 1$.

We need to check if $\lim_{x \to \pi} g(x) = g(\pi)$.

First, we need to check if $\pi$ is in the domain.

For $n=1$, we have $-\frac{3\pi}{2} + 2\pi = \frac{\pi}{2} \leq \pi \leq -\frac{\pi}{2} + 2\pi = \frac{3\pi}{2}$. So $\pi$ is in the domain.

$\cos(x+\pi)$ is continuous, $\sec(x+\pi)$ is continuous where $\cos(x+\pi) \neq 0$, and $\sqrt{x}$ is continuous for $x \geq 0$.

Thus, $g(x) = \sqrt{\sec(x+\pi)}$ is continuous at $x = \pi$.

1.3 (a)

$f(x) = \cos x$. $f(x+h) = \cos(x+h)$.

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

We know that $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$ and $\lim_{h \to 0} \frac{\sin h}{h} = 1$.

$f'(x) = \cos x (0) - \sin x (1) = -\sin x$.

1.3 (b)

$y = \sec^{-1} x$.

The domain of $\sec^{-1} x$ is $|x| \geq 1$, which means $x \in (-\infty, -1] \cup [1, \infty)$.

$y = \sec^{-1} x \implies x = \sec y$.

$\frac{dx}{dy} = \sec y \tan y$.

$\frac{dy}{dx} = \frac{1}{\sec y \tan y} = \frac{1}{\sec y \sqrt{\sec^2 y - 1}} = \frac{1}{x \sqrt{x^2 - 1}}$.

1.4

$y + x \cos y = x^2$.

We need to find the equation of the tangent line at $x=0$.

When $x=0$, $y + 0 \cos y = 0^2$, so $y = 0$.

The point is $(0,0)$.

Differentiate the equation with respect to $x$:

$\frac{dy}{dx} + \cos y - x \sin y \frac{dy}{dx} = 2x$.

$\frac{dy}{dx}(1 - x \sin y) = 2x - \cos y$.

$\frac{dy}{dx} = \frac{2x - \cos y}{1 - x \sin y}$.

At $(0,0)$, $\frac{dy}{dx} = \frac{2(0) - \cos 0}{1 - 0 \sin 0} = \frac{-1}{1} = -1$.

The equation of the tangent line is $y - 0 = -1(x - 0)$, which is $y = -x$.

3. Final Answer

1.1 (a) $-\frac{2}{x^3}$ for $x \neq 0$.

1.1 (b) $6$.

1.2 1 (a) $x \in (-\infty, 1) \cup [2, \infty)$.

1.2 1 (b) $f(x)$ is discontinuous at $x=1$.

1.2 2 (a) $x \in \bigcup_{n \in \mathbb{Z}} [-\frac{3\pi}{2} + 2n\pi, -\frac{\pi}{2} + 2n\pi]$.

1.2 2 (b) $g(x)$ is continuous at $x = \pi$.

1.3 (a) Proof given above.

1.3 (b) Domain: $x \in (-\infty, -1] \cup [1, \infty)$. Derivative: $\frac{1}{x \sqrt{x^2 - 1}}$.

1.4 $y = -x$.