The image contains a series of limit problems. Let's solve these problems one by one. I'll start with a detailed solution to a few of them. Please let me know if you'd like explanations for all of them or specific ones.

6. $\lim_{x \to 0} \frac{x}{x^4 - 4x^3}$

Factor the denominator: $x^4 - 4x^3 = x^3(x - 4)$ The limit becomes: $\lim_{x \to 0} \frac{x}{x^3(x - 4)}$ Simplify: $\lim_{x \to 0} \frac{1}{x^2(x - 4)}$ As $x \to 0$, the denominator $x^2(x - 4) \to 0$, leading to divergence. The limit does not exist.

7. $\lim_{x \to \infty} \frac{x - 2}{x^2 - x - 2}$

Divide numerator and denominator by $x^2$: $\lim_{x \to \infty} \frac{\frac{x - 2}{x^2}}{1 - \frac{x}{x^2} - \frac{2}{x^2}}$ Simplify: $\lim_{x \to \infty} \frac{\frac{1}{x} - \frac{2}{x^2}}{1 - \frac{1}{x} - \frac{2}{x^2}}$ As $x \to \infty$, all terms with $\frac{1}{x}$ or $\frac{2}{x^2}$ approach 0: $\text{Limit} = \frac{0}{1} = 0$

8. $\lim_{x \to \infty} \frac{2x^3}{x^3 - 4x + 1}$

Factor $x^3$ in the denominator: $\frac{2x^3}{x^3(1 - \frac{4}{x^2} + \frac{1}{x^3})}$ Simplify: $\lim_{x \to \infty} \frac{2}{1 - \frac{4}{x^2} + \frac{1}{x^3}}$ As $x \to \infty$, terms with $\frac{1}{x^2}$ and $\frac{1}{x^3}$ approach 0: $\text{Limit} = \frac{2}{1} = 2$

9. $\lim_{x \to \infty} \frac{x^3}{4x^2 + 12x + 16}$

Divide numerator and denominator by $x^3$: $\lim_{x \to \infty} \frac{1}{\frac{4}{x} + \frac{12}{x^2} + \frac{16}{x^3}}$ As $x \to \infty$, all terms with $\frac{1}{x}$, $\frac{1}{x^2}$, and $\frac{1}{x^3}$ approach 0: $\text{Limit} = \infty$

10. $\lim_{x \to \infty} \frac{\sqrt{x^2 + 16} - 5}{2x + 6}$

Rationalize the numerator: $\frac{\sqrt{x^2 + 16} - 5}{2x + 6} \cdot \frac{\sqrt{x^2 + 16} + 5}{\sqrt{x^2 + 16} + 5}$ This becomes: $\frac{x^2 + 16 - 25}{(2x + 6)(\sqrt{x^2 + 16} + 5)}$ Simplify: $\frac{x^2 - 9}{(2x + 6)(\sqrt{x^2 + 16} + 5)}$ Factor numerator: $\frac{(x - 3)(x + 3)}{(2x + 6)(\sqrt{x^2 + 16} + 5)}$ Divide by $x$ as $x \to \infty$: $\lim_{x \to \infty} \frac{1}{2\sqrt{1 + \frac{16}{x^2}} + 0} = \frac{1}{2}$

Would you like all of them solved? Let me know! 😊

Related Questions:

1. Can you identify conditions where limits do not exist?
2. How does rationalizing the numerator simplify limits involving square roots?
3. When dividing polynomials, how do you decide which terms dominate as $x \to \infty$?
4. What is the difference between indeterminate forms like $\frac{0}{0}$ and divergence to infinity?
5. How can L'Hôpital's Rule help simplify tricky limit problems?

Tip:

Always look for dominant terms in the numerator and denominator when $x \to \infty$. These terms often dictate the behavior of the limit!