We are asked to solve the following problems from the image: a. $\lim_{x \to 0} (2e^x - xe^x + 2e^0 + 4)$ b. Find $a$ if $\lim_{x \to \infty} \frac{8x^2+x+1}{ax^2+1} = 1$ c. $\lim_{x \to -\infty} (x + \sqrt{x^2+9})$ d. $\lim_{x \to 0} \frac{\sin x - \sin x \cos x}{x^3}$

1. Problem Description

We are asked to solve the following problems from the image:

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

b. Find $a$ if $\lim_{x \to \infty} \frac{8x^2+x+1}{ax^2+1} = 1$

c. $\lim_{x \to -\infty} (x + \sqrt{x^2+9})$

d. $\lim_{x \to 0} \frac{\sin x - \sin x \cos x}{x^3}$

2. Solution Steps

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

Since $e^0 = 1$, substitute $x=0$:

$2e^0 - 0 \cdot e^0 + 2e^0 + 4 = 2(1) - 0 + 2(1) + 4 = 2 + 2 + 4 = 8$

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

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

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

Since the limit is equal to 1, $\frac{8}{a} = 1$, which means $a=8$.

c. $\lim_{x \to -\infty} (x + \sqrt{x^2+9})$

Multiply by the conjugate:

$\lim_{x \to -\infty} (x + \sqrt{x^2+9}) \cdot \frac{x - \sqrt{x^2+9}}{x - \sqrt{x^2+9}} = \lim_{x \to -\infty} \frac{x^2 - (x^2+9)}{x - \sqrt{x^2+9}} = \lim_{x \to -\infty} \frac{-9}{x - \sqrt{x^2+9}}$

Since $x \to -\infty$, we can write $\sqrt{x^2} = |x| = -x$. Then

$\lim_{x \to -\infty} \frac{-9}{x - \sqrt{x^2+9}} = \lim_{x \to -\infty} \frac{-9}{x - \sqrt{x^2(1+\frac{9}{x^2})}} = \lim_{x \to -\infty} \frac{-9}{x - |x|\sqrt{1+\frac{9}{x^2}}} = \lim_{x \to -\infty} \frac{-9}{x - (-x)\sqrt{1+\frac{9}{x^2}}} = \lim_{x \to -\infty} \frac{-9}{x + x\sqrt{1+\frac{9}{x^2}}} = \lim_{x \to -\infty} \frac{-9}{x(1 + \sqrt{1+\frac{9}{x^2}})}$

As $x \to -\infty$, $\frac{9}{x^2} \to 0$, so $\sqrt{1+\frac{9}{x^2}} \to 1$.

Therefore, $\lim_{x \to -\infty} \frac{-9}{x(1+1)} = \lim_{x \to -\infty} \frac{-9}{2x} = 0$

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

$\lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \frac{1 - \cos x}{x^2}$

We know $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

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

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

3. Final Answer

a. 8

b. 8

c. 0

d. 1/2