The image presents a set of math problems involving limits, probability, complex numbers, integrals, and vectors. I will solve the limit problems first. a. Evaluate $\lim_{x\to3} \sqrt{3x^2-11}$ b. Evaluate $\lim_{x\to\frac{\pi}{3}} \frac{\sqrt{3}\sin(x-\frac{\pi}{3})}{\frac{\pi}{3} - x}$ c. Evaluate $\lim_{x\to\infty} (2x-7-11\ln x)$

1. Problem Description

The image presents a set of math problems involving limits, probability, complex numbers, integrals, and vectors. I will solve the limit problems first.

a. Evaluate $\lim_{x\to3} \sqrt{3x^2-11}$

b. Evaluate $\lim_{x\to\frac{\pi}{3}} \frac{\sqrt{3}\sin(x-\frac{\pi}{3})}{\frac{\pi}{3} - x}$

c. Evaluate $\lim_{x\to\infty} (2x-7-11\ln x)$

2. Solution Steps

a. For the first limit, we can directly substitute $x=3$ into the expression since the function is continuous at $x=3$.

$\lim_{x\to3} \sqrt{3x^2-11} = \sqrt{3(3)^2-11} = \sqrt{3(9)-11} = \sqrt{27-11} = \sqrt{16} = 4$.

b. For the second limit, let $u = x - \frac{\pi}{3}$. Then as $x\to \frac{\pi}{3}$, $u\to 0$. Also, $\frac{\pi}{3}-x = -u$.

The limit becomes:

$\lim_{u\to0} \frac{\sqrt{3}\sin(u)}{-u} = -\sqrt{3}\lim_{u\to0} \frac{\sin(u)}{u}$.

We know that $\lim_{u\to0} \frac{\sin(u)}{u} = 1$. Therefore, the limit is $-\sqrt{3}$.

c. For the third limit, $\lim_{x\to\infty} (2x-7-11\ln x)$, we can consider the growth rates of $x$ and $\ln x$. As $x \to \infty$, $x$ grows faster than $\ln x$.

Rewrite the expression as: $x(2 - \frac{7}{x} - \frac{11\ln x}{x})$.

As $x \to \infty$, $\frac{7}{x} \to 0$ and $\frac{\ln x}{x} \to 0$ (since $x$ grows faster than $\ln x$). Therefore, the expression inside the parentheses approaches $2$.

Since $x$ goes to infinity, the entire limit goes to infinity. However, since $x$ is positive and $\ln(x)$ is positive when x > 1, and the coefficient of $ln(x)$ is negative, we have:

Let $f(x) = 2x - 7 - 11ln(x)$

$f'(x) = 2 - 11/x$

$f'(x) = 0 \Rightarrow x = 11/2$

$f''(x) = 11/x^2$ which is always positive for positive x, thus $x = 11/2$ is a minimum.

As $x$ goes to infinity, since $x$ grows much faster than $ln(x)$, the $ln(x)$ term is less important than the $x$ term, therefore, the expression grows to infinity. However, since $2x - 11lnx = x(2 - 11lnx/x)$, by L'Hopital's rule, $lim_{x->infinity} ln(x)/x = lim_{x->infinity} 1/x = 0$. Thus, the limit should be $+\infty$.

We have $\lim_{x\to\infty} \frac{\ln x}{x} = 0$. Thus, $\lim_{x\to\infty} (2x - 7 - 11\ln x) = \lim_{x\to\infty} x(2-\frac{7}{x} - 11\frac{\ln x}{x}) = \infty(2 - 0 - 0) = \infty$.

3. Final Answer

a. 4

b. $-\sqrt{3}$

c. $\infty$