We need to solve four problems related to limits and derivatives. 2.1. Prove that $\lim_{x\to -2} x^2 - 1 = 3$ using the precise definition of the limit. 2.2. Prove that $\lim_{x\to 4^+} \frac{2}{\sqrt{x-4}} = +\infty$ using the precise definition of the limit. 2.3. Find the limit $\lim_{x\to \frac{\pi}{4}} \frac{1-\tan x}{\sin x - \cos x}$. 2.4. Find the derivative $f'(x)$ of $f(x) = \sqrt{9-x}$ using the first principle definition of the derivative.

1. Problem Description

We need to solve four problems related to limits and derivatives.

2.

1. Prove that $\lim_{x\to -2} x^2 - 1 = 3$ using the precise definition of the limit.

2.

2. Prove that $\lim_{x\to 4^+} \frac{2}{\sqrt{x-4}} = +\infty$ using the precise definition of the limit.

2.

3. Find the limit $\lim_{x\to \frac{\pi}{4}} \frac{1-\tan x}{\sin x - \cos x}$.

2.

4. Find the derivative $f'(x)$ of $f(x) = \sqrt{9-x}$ using the first principle definition of the derivative.

2. Solution Steps

2.

1. Using the precise definition of the limit to prove $\lim_{x\to -2} x^2 - 1 = 3$.

We need to show that for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - (-2)| < \delta$, then $|x^2 - 1 - 3| < \epsilon$.

$|x^2 - 1 - 3| = |x^2 - 4| = |(x - 2)(x + 2)| = |x - 2| |x + 2|$.

We want $|x - 2| |x + 2| < \epsilon$.

Let's assume that $|x + 2| < 1$, then $-1 < x + 2 < 1$, which implies $-3 < x < -1$.

Therefore, $-5 < x - 2 < -3$, so $|x - 2| < 5$.

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

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

Let $\delta = \min(1, \frac{\epsilon}{5})$.

Then, if $0 < |x + 2| < \delta$, we have $|x^2 - 4| = |x - 2| |x + 2| < 5 |x + 2| < 5 \cdot \frac{\epsilon}{5} = \epsilon$.

Therefore, $\lim_{x\to -2} x^2 - 1 = 3$.

2.

2. Using the precise definition of the limit to prove $\lim_{x\to 4^+} \frac{2}{\sqrt{x-4}} = +\infty$.

We need to show that for every $M > 0$, there exists a $\delta > 0$ such that if $0 < x - 4 < \delta$, then $\frac{2}{\sqrt{x-4}} > M$.

We want $\frac{2}{\sqrt{x-4}} > M$, so $\sqrt{x-4} < \frac{2}{M}$, which means $x - 4 < \frac{4}{M^2}$.

Let $\delta = \frac{4}{M^2}$.

Then, if $0 < x - 4 < \delta$, we have $\frac{2}{\sqrt{x-4}} > \frac{2}{\sqrt{\frac{4}{M^2}}} = \frac{2}{\frac{2}{M}} = M$.

Therefore, $\lim_{x\to 4^+} \frac{2}{\sqrt{x-4}} = +\infty$.

2.

3. Find the limit $\lim_{x\to \frac{\pi}{4}} \frac{1-\tan x}{\sin x - \cos x}$.

$\lim_{x\to \frac{\pi}{4}} \frac{1-\tan x}{\sin x - \cos x} = \lim_{x\to \frac{\pi}{4}} \frac{1-\frac{\sin x}{\cos x}}{\sin x - \cos x} = \lim_{x\to \frac{\pi}{4}} \frac{\frac{\cos x - \sin x}{\cos x}}{\sin x - \cos x} = \lim_{x\to \frac{\pi}{4}} \frac{\cos x - \sin x}{\cos x (\sin x - \cos x)} = \lim_{x\to \frac{\pi}{4}} \frac{-(\sin x - \cos x)}{\cos x (\sin x - \cos x)} = \lim_{x\to \frac{\pi}{4}} \frac{-1}{\cos x} = \frac{-1}{\cos(\frac{\pi}{4})} = \frac{-1}{\frac{\sqrt{2}}{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.

2.

4. Find the derivative $f'(x)$ of $f(x) = \sqrt{9-x}$ using the first principle definition of the derivative.

$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{\sqrt{9-(x+h)} - \sqrt{9-x}}{h}$.

Multiply by the conjugate:

$f'(x) = \lim_{h\to 0} \frac{(\sqrt{9-(x+h)} - \sqrt{9-x})(\sqrt{9-(x+h)} + \sqrt{9-x})}{h(\sqrt{9-(x+h)} + \sqrt{9-x})} = \lim_{h\to 0} \frac{9-(x+h) - (9-x)}{h(\sqrt{9-(x+h)} + \sqrt{9-x})} = \lim_{h\to 0} \frac{9 - x - h - 9 + x}{h(\sqrt{9-(x+h)} + \sqrt{9-x})} = \lim_{h\to 0} \frac{-h}{h(\sqrt{9-(x+h)} + \sqrt{9-x})} = \lim_{h\to 0} \frac{-1}{\sqrt{9-(x+h)} + \sqrt{9-x}} = \frac{-1}{\sqrt{9-x} + \sqrt{9-x}} = \frac{-1}{2\sqrt{9-x}}$.

3. Final Answer

2.

1. Proof complete.

2.

2. Proof complete.

2.

3. $-\sqrt{2}$

2.

4. $\frac{-1}{2\sqrt{9-x}}$