The problem asks to evaluate the limits of the function $f(x) = 2x + \sqrt{6x^2 - 2}$ as $x$ approaches positive infinity ($\infty$) and negative infinity ($-\infty$).

Let's break it down step by step.

1. Limit as $x$ approaches $+\infty$:

$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \left( 2x + \sqrt{6x^2 - 2} \right)$

First, notice that as $x$ becomes very large, $6x^2 - 2$ behaves very similarly to $6x^2$. So, the square root $\sqrt{6x^2 - 2}$ behaves like $\sqrt{6x^2} = \sqrt{6} \cdot |x|$.

Since $x$ is approaching $+\infty$, $|x| = x$, so: $\lim_{x \to +\infty} \sqrt{6x^2 - 2} = \lim_{x \to +\infty} \sqrt{6} \cdot x = \sqrt{6} \cdot x$

Thus: $\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} \left( 2x + \sqrt{6} \cdot x \right) = \lim_{x \to +\infty} \left( (2 + \sqrt{6})x \right) = +\infty$

2. Limit as $x$ approaches $-\infty$:

$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \left( 2x + \sqrt{6x^2 - 2} \right)$

Again, using the approximation $\sqrt{6x^2 - 2} \approx \sqrt{6} \cdot |x|$, but this time as $x$ approaches $-\infty$, $|x| = -x$.

So: $\lim_{x \to -\infty} \sqrt{6x^2 - 2} = \lim_{x \to -\infty} \sqrt{6} \cdot (-x) = -\sqrt{6} \cdot x$

Thus: $\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \left( 2x - \sqrt{6} \cdot x \right) = \lim_{x \to -\infty} \left( (2 - \sqrt{6})x \right)$

Here, $2 - \sqrt{6}$ is a negative number (approximately -0.45), so: $\lim_{x \to -\infty} f(x) = -\infty$

Summary of Answers:

• $\lim_{x \to +\infty} f(x) = +\infty$
• $\lim_{x \to -\infty} f(x) = -\infty$

Would you like further details on the steps, or do you have any additional questions?

Here are five related questions you might consider exploring:

1. How would the function $f(x) = 2x + \sqrt{6x^2 - 2}$ behave if the square root term was instead $\sqrt{6x^2 + 2}$?
2. What is the significance of the coefficient $\sqrt{6}$ in this context?
3. How would you find the horizontal asymptotes of $f(x)$ for other similar functions?
4. Can you graph $f(x)$ to visually confirm the limits?
5. What happens to the function $f(x) = 2x + \sqrt{6x^2 - 2}$ as $x$ approaches zero?

Tip: When dealing with limits involving square roots, factor out the highest power of $x$ to simplify the expression.