We need to find the limit of the given functions in 2.1 (a), (b), (c), (d), and (e).

1. Problem Description

We need to find the limit of the given functions in 2.1 (a), (b), (c), (d), and (e).

2. Solution Steps

(a) $\lim_{x\to\infty} \sqrt{x^2 + ax} - \sqrt{x^2 + \beta x}$

$\lim_{x\to\infty} (\sqrt{x^2 + ax} - \sqrt{x^2 + \beta x}) \cdot \frac{\sqrt{x^2 + ax} + \sqrt{x^2 + \beta x}}{\sqrt{x^2 + ax} + \sqrt{x^2 + \beta x}}$

$= \lim_{x\to\infty} \frac{(x^2 + ax) - (x^2 + \beta x)}{\sqrt{x^2 + ax} + \sqrt{x^2 + \beta x}}$

$= \lim_{x\to\infty} \frac{(a - \beta)x}{\sqrt{x^2(1 + \frac{a}{x})} + \sqrt{x^2(1 + \frac{\beta}{x})}}$

$= \lim_{x\to\infty} \frac{(a - \beta)x}{x\sqrt{1 + \frac{a}{x}} + x\sqrt{1 + \frac{\beta}{x}}}$

$= \lim_{x\to\infty} \frac{(a - \beta)}{\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{\beta}{x}}}$

$= \frac{a - \beta}{\sqrt{1 + 0} + \sqrt{1 + 0}}$

$= \frac{a - \beta}{1 + 1}$

$= \frac{a - \beta}{2}$

(b) $\lim_{x\to 0} \begin{cases} \arctan(\frac{1}{x}), & x > 0 \\ \frac{\pi}{4}[2\cos x], & x \le 0 \end{cases}$

For $x > 0$, $\lim_{x\to 0^+} \arctan(\frac{1}{x}) = \frac{\pi}{2}$.

For $x \le 0$, $\lim_{x\to 0^-} \frac{\pi}{4} [2\cos x] = \frac{\pi}{4} [2\cos(0)] = \frac{\pi}{4}[2] = \frac{\pi}{4} \cdot 2 = \frac{\pi}{2}$

Since both limits agree, the limit exists and equals $\frac{\pi}{2}$.

(c) $\lim_{x\to\infty} x^2 \sin(\frac{1}{x})$

Let $y = \frac{1}{x}$. Then as $x \to \infty$, $y \to 0$.

$\lim_{x\to\infty} x^2 \sin(\frac{1}{x}) = \lim_{y\to 0} \frac{1}{y^2} \sin(y) = \lim_{y\to 0} \frac{\sin(y)}{y^2} = \lim_{y\to 0} \frac{\sin(y)}{y} \cdot \frac{1}{y}$

Since $\lim_{y\to 0} \frac{\sin(y)}{y} = 1$, the limit becomes $\lim_{y\to 0} \frac{1}{y}$, which does not exist.

To be precise:

$\lim_{x\to\infty} x^2 \sin(\frac{1}{x})$

Let $t = 1/x$. Then, as $x \to \infty$, $t \to 0^+$.

$\lim_{x\to\infty} x^2 \sin(1/x) = \lim_{t\to 0^+} (1/t^2) \sin(t) = \lim_{t\to 0^+} \frac{\sin(t)}{t^2}$

Since $\lim_{t\to 0^+} \sin(t) / t = 1$, we have $\lim_{t\to 0^+} \frac{1}{t} = +\infty$.

So the limit does not exist.

(d) $\lim_{x\to a} \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}}$

$\lim_{x\to a} \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}} \cdot \frac{\sqrt{a+2x} + \sqrt{3x}}{\sqrt{a+2x} + \sqrt{3x}} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{3a+x} + 2\sqrt{x}}$

$= \lim_{x\to a} \frac{(a+2x - 3x)(\sqrt{3a+x} + 2\sqrt{x})}{(3a+x - 4x)(\sqrt{a+2x} + \sqrt{3x})}$

$= \lim_{x\to a} \frac{(a-x)(\sqrt{3a+x} + 2\sqrt{x})}{(3a-3x)(\sqrt{a+2x} + \sqrt{3x})}$

$= \lim_{x\to a} \frac{(a-x)(\sqrt{3a+x} + 2\sqrt{x})}{3(a-x)(\sqrt{a+2x} + \sqrt{3x})}$

$= \lim_{x\to a} \frac{(\sqrt{3a+x} + 2\sqrt{x})}{3(\sqrt{a+2x} + \sqrt{3x})}$

$= \frac{\sqrt{3a+a} + 2\sqrt{a}}{3(\sqrt{a+2a} + \sqrt{3a})}$

$= \frac{\sqrt{4a} + 2\sqrt{a}}{3(\sqrt{3a} + \sqrt{3a})}$

$= \frac{2\sqrt{a} + 2\sqrt{a}}{3(2\sqrt{3a})}$

$= \frac{4\sqrt{a}}{6\sqrt{3a}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$

(e) $\lim_{x\to \frac{1}{3}} \frac{1}{[3x-4]}$

As $x\to \frac{1}{3}$, we have $3x-4\to 3(\frac{1}{3})-4 = 1-4=-3$.

Therefore, $[3x-4]\to [-3]=-3$.

$\lim_{x\to \frac{1}{3}} \frac{1}{[3x-4]}=\frac{1}{-3}=-\frac{1}{3}$

3. Final Answer

(a) $\frac{a-\beta}{2}$

(b) $\frac{\pi}{2}$

(c) Does not exist

(d) $\frac{2\sqrt{3}}{9}$

(e) $-\frac{1}{3}$