We are asked to solve three limit problems: a) $\lim_{x \to 0} (2e^{3x} - xe^x + 2e^x + 4)$ b) Find $a$ such that $\lim_{x \to \infty} \frac{8x^2 + x + 1}{ax^2 + 1} = 1$ c) $\lim_{x \to 0} \frac{\sin x - \sin x \cos x}{x^3}$

1. Problem Description

We are asked to solve three limit problems:

a) $\lim_{x \to 0} (2e^{3x} - xe^x + 2e^x + 4)$

b) Find $a$ such that $\lim_{x \to \infty} \frac{8x^2 + x + 1}{ax^2 + 1} = 1$

c) $\lim_{x \to 0} \frac{\sin x - \sin x \cos x}{x^3}$

2. Solution Steps

a) $\lim_{x \to 0} (2e^{3x} - xe^x + 2e^x + 4)$

Since the function is continuous, we can substitute $x=0$ directly:

$2e^{3(0)} - (0)e^0 + 2e^0 + 4 = 2e^0 - 0 + 2e^0 + 4 = 2(1) + 2(1) + 4 = 2 + 2 + 4 = 8$

b) $\lim_{x \to \infty} \frac{8x^2 + x + 1}{ax^2 + 1} = 1$

Divide the numerator and the denominator by $x^2$:

$\lim_{x \to \infty} \frac{8 + \frac{1}{x} + \frac{1}{x^2}}{a + \frac{1}{x^2}} = 1$

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

So, $\frac{8 + 0 + 0}{a + 0} = 1$, which means $\frac{8}{a} = 1$.

Therefore, $a = 8$.

c) $\lim_{x \to 0} \frac{\sin x - \sin x \cos x}{x^3} = \lim_{x \to 0} \frac{\sin x (1 - \cos x)}{x^3}$

We know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $1 - \cos x = 2 \sin^2(\frac{x}{2})$.

So, $\lim_{x \to 0} \frac{\sin x (1 - \cos x)}{x^3} = \lim_{x \to 0} \frac{\sin x (2 \sin^2(\frac{x}{2}))}{x^3} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{2 \sin^2(\frac{x}{2})}{x^2} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{2 \sin^2(\frac{x}{2})}{4(\frac{x}{2})^2} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{2}{4} \cdot (\frac{\sin(\frac{x}{2})}{\frac{x}{2}})^2$

Since $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{\sin(\frac{x}{2})}{\frac{x}{2}} = 1$,

$\lim_{x \to 0} \frac{\sin x - \sin x \cos x}{x^3} = 1 \cdot \frac{2}{4} \cdot (1)^2 = \frac{1}{2}$.

3. Final Answer

a) 8

b) 8

c) 1/2