Let's analyze the two limits presented in the image.

1. Limit 1:

$\lim_{x \to -1} \frac{x^{15} + 1}{x^9 + 1}$

To solve this, let's substitute $x = -1$ directly:

$\frac{(-1)^{15} + 1}{(-1)^9 + 1} = \frac{-1 + 1}{-1 + 1} = \frac{0}{0}$

Since this results in the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule, which involves differentiating the numerator and denominator.

• The derivative of the numerator $x^{15} + 1$ is $15x^{14}$.
• The derivative of the denominator $x^9 + 1$ is $9x^8$.

Now, applying L'Hopital's Rule:

$\lim_{x \to -1} \frac{15x^{14}}{9x^8} = \frac{15(-1)^{14}}{9(-1)^8} = \frac{15}{9} = \frac{5}{3}$

Thus, the value of the limit is:

$\boxed{\frac{5}{3}}$

2. Limit 2:

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

This limit is in a form suitable for the derivative definition, where:

$\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

The function $f(x) = x^{\frac{2}{7}}$. The limit can be evaluated as the derivative of $f(x)$ at $x = a$.

The derivative of $f(x) = x^{\frac{2}{7}}$ is:

$f'(x) = \frac{2}{7}x^{-\frac{5}{7}}$

Evaluating this derivative at $x = a$:

$f'(a) = \frac{2}{7}a^{-\frac{5}{7}}$

Thus, the value of the second limit is:

$\boxed{\frac{2}{7} a^{-\frac{5}{7}}}$

Would you like more details on either limit or the steps involved?

Here are 5 related questions for further exploration:

1. How is L'Hopital's Rule derived and when can it be applied?
2. What is the geometric interpretation of a limit?
3. How can we apply Taylor series to approximate limits near indeterminate forms?
4. Why is the derivative definition useful in evaluating limits?
5. What happens if a limit does not exist, and how can we identify such cases?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ or $\infty/\infty$ before applying L'Hopital's Rule!