Let's analyze the limit from the provided image:

$\lim_{{x \to \infty}} \frac{21x^{12} + 12}{7x^{12} + x^6}$

Step 1: Identify dominant terms

As $x \to \infty$, the highest power of $x$ dominates the behavior of the function. In both the numerator and the denominator, the term with $x^{12}$ will dominate.

• Numerator: $21x^{12} + 12$ → dominated by $21x^{12}$
• Denominator: $7x^{12} + x^6$ → dominated by $7x^{12}$

Step 2: Simplify the expression

We can factor out $x^{12}$ from both the numerator and denominator:

$\frac{21x^{12} + 12}{7x^{12} + x^6} = \frac{x^{12}(21 + \frac{12}{x^{12}})}{x^{12}(7 + \frac{1}{x^6})}$

Step 3: Evaluate the limit as $x \to \infty$

As $x \to \infty$, the terms $\frac{12}{x^{12}}$ and $\frac{1}{x^6}$ both approach zero, so the expression simplifies to:

$\lim_{{x \to \infty}} \frac{21 + \frac{12}{x^{12}}}{7 + \frac{1}{x^6}} = \frac{21}{7} = 3$

Thus, the limit is 3.

Let me know if you need more details or have any questions!

Here are 5 related questions for further practice:

1. How do you find limits involving square roots at infinity?
2. What is the behavior of rational functions at infinity when the degree of the numerator is higher than the denominator?
3. How do you apply L'Hopital's Rule to limits at infinity?
4. What happens to a limit when the highest power in the denominator is greater than the numerator?
5. How do you solve limits involving exponential functions as $x \to \infty$?

Tip: Always compare the highest powers of $x$ in both the numerator and denominator when dealing with limits at infinity.