The image contains four problems related to calculus:

1. Problem Description

The image contains four problems related to calculus:

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

3. 2 Prove that $\lim_{x\to 4^+} \frac{2}{\sqrt{x-4}} = +\infty$ using the precise definition of the limit.

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

5. 4 Find the derivative $f'(x)$ of the function $f(x) = \sqrt{9-x}$ using the first principle definition of the derivative.

6. Solution Steps

7. 1: Proving $\lim_{x\to -2} (x^2-1) = 3$ using the $\epsilon$-$\delta$ definition.

We need to show that for any $\epsilon > 0$, 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 to bound $|x-2|$. Let's assume that $|x+2| < 1$. Then $-1 < x+2 < 1$, which means $-3 < x < -1$. Therefore, $-5 < x-2 < -3$, so $|x-2| < 5$.

Now, $|(x^2 - 1) - 3| = |x-2| |x+2| < 5 |x+2|$.

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

Choose $\delta = \min\{1, \frac{\epsilon}{5}\}$. Then if $0 < |x+2| < \delta$, we have:

$|(x^2 - 1) - 3| = |x^2 - 4| = |x-2||x+2| < 5 |x+2| < 5 \cdot \frac{\epsilon}{5} = \epsilon$.

Thus, $\lim_{x\to -2} (x^2-1) = 3$.

8. 2: Proving $\lim_{x\to 4^+} \frac{2}{\sqrt{x-4}} = +\infty$.

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

We want $\frac{2}{\sqrt{x-4}} > M$. This is equivalent to $\sqrt{x-4} < \frac{2}{M}$, which implies $x-4 < \frac{4}{M^2}$.

So, $x < 4 + \frac{4}{M^2}$. Let $\delta = \frac{4}{M^2}$.

If $4 < x < 4 + \delta$, then $0 < x-4 < \delta = \frac{4}{M^2}$.

Then $\sqrt{x-4} < \sqrt{\frac{4}{M^2}} = \frac{2}{M}$.

Therefore, $\frac{2}{\sqrt{x-4}} > \frac{2}{\frac{2}{M}} = M$.

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

9. 3: Finding the limit $\lim_{x\to \frac{\pi}{4}} \frac{1-\tan x}{\sin x - \cos x}$.

We can rewrite $\tan x$ as $\frac{\sin x}{\cos x}$.

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

Since $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$, the limit is $\frac{-1}{\frac{\sqrt{2}}{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.

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0. 4: Finding the derivative of $f(x) = \sqrt{9-x}$ using the first principle definition.

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

Multiply by the conjugate:

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

As $h \to 0$, we have:

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

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1. Final Answer

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2. 1: The precise definition of the limit is proved.

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3. 2: The precise definition of the limit is proved.

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4. 3: $-\sqrt{2}$

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5. 4: $f'(x) = \frac{-1}{2\sqrt{9-x}}$