The problem consists of multiple limit calculations. (a) $\lim_{h\to 0} \frac{(2+h)^3 - 8}{h}$ (b) $\lim_{x\to -\infty} \frac{\sqrt{1+4x^6}}{2-x^3}$ (c) $\lim_{x\to 1} \arcsin(\frac{1-\sqrt{x}}{1-x})$ (d) $\lim_{x\to 4^+} \frac{4-x}{|4-x|}$ (e) $\lim_{x\to -\frac{5}{2}} [x]$

1. Problem Description

The problem consists of multiple limit calculations.

(a) $\lim_{h\to 0} \frac{(2+h)^3 - 8}{h}$

(b) $\lim_{x\to -\infty} \frac{\sqrt{1+4x^6}}{2-x^3}$

(c) $\lim_{x\to 1} \arcsin(\frac{1-\sqrt{x}}{1-x})$

(d) $\lim_{x\to 4^+} \frac{4-x}{|4-x|}$

(e) $\lim_{x\to -\frac{5}{2}} [x]$

2. Solution Steps

(a) $\lim_{h\to 0} \frac{(2+h)^3 - 8}{h}$

Expand $(2+h)^3$:

$(2+h)^3 = 2^3 + 3(2^2)h + 3(2)h^2 + h^3 = 8 + 12h + 6h^2 + h^3$

So,

$(2+h)^3 - 8 = 12h + 6h^2 + h^3 = h(12 + 6h + h^2)$

Then,

$\lim_{h\to 0} \frac{h(12 + 6h + h^2)}{h} = \lim_{h\to 0} (12 + 6h + h^2) = 12 + 6(0) + (0)^2 = 12$

(b) $\lim_{x\to -\infty} \frac{\sqrt{1+4x^6}}{2-x^3}$

Divide both the numerator and the denominator by $x^3$. Since $x \to -\infty$, we have $\sqrt{x^6} = |x^3| = -x^3$.

$\lim_{x\to -\infty} \frac{\sqrt{1+4x^6}}{2-x^3} = \lim_{x\to -\infty} \frac{\frac{\sqrt{1+4x^6}}{-x^3}}{\frac{2-x^3}{-x^3}} = \lim_{x\to -\infty} \frac{-\sqrt{\frac{1}{x^6}+4}}{\frac{2}{-x^3}+1}$

As $x \to -\infty$, $\frac{1}{x^6} \to 0$ and $\frac{2}{-x^3} \to 0$.

So, $\lim_{x\to -\infty} \frac{-\sqrt{\frac{1}{x^6}+4}}{\frac{2}{-x^3}+1} = \frac{-\sqrt{0+4}}{0+1} = \frac{-2}{1} = -2$

(c) $\lim_{x\to 1} \arcsin(\frac{1-\sqrt{x}}{1-x})$

We can rewrite $\frac{1-\sqrt{x}}{1-x}$ as follows:

$\frac{1-\sqrt{x}}{1-x} = \frac{1-\sqrt{x}}{(1-\sqrt{x})(1+\sqrt{x})} = \frac{1}{1+\sqrt{x}}$

Then,

$\lim_{x\to 1} \arcsin(\frac{1-\sqrt{x}}{1-x}) = \lim_{x\to 1} \arcsin(\frac{1}{1+\sqrt{x}})$

As $x\to 1$, $\frac{1}{1+\sqrt{x}} \to \frac{1}{1+\sqrt{1}} = \frac{1}{2}$.

$\lim_{x\to 1} \arcsin(\frac{1}{1+\sqrt{x}}) = \arcsin(\frac{1}{2}) = \frac{\pi}{6}$

(d) $\lim_{x\to 4^+} \frac{4-x}{|4-x|}$

As $x\to 4^+$, $x > 4$, so $4-x < 0$. Then, $|4-x| = -(4-x) = x-4$.

So, $\frac{4-x}{|4-x|} = \frac{4-x}{x-4} = -1$.

$\lim_{x\to 4^+} \frac{4-x}{|4-x|} = \lim_{x\to 4^+} (-1) = -1$

(e) $\lim_{x\to -\frac{5}{2}} [x]$

Here, $[x]$ represents the greatest integer less than or equal to $x$.

$-\frac{5}{2} = -2.5$

Since we are taking the limit as $x$ approaches $-\frac{5}{2}$, we consider values of $x$ close to $-2.5$.

As $x \to -2.5$, $[x] = -3$.

$\lim_{x\to -\frac{5}{2}} [x] = -3$

3. Final Answer

(a) 12

(b) -2

(c) $\frac{\pi}{6}$

(d) -1

(e) -3