The image contains the following limit expression:

$\left( \lim_{{x \to 1}} \frac{{x^2 - 64}}{{x - 8}} \right)^3 + \left( \lim_{{x \to \infty}} \frac{{8x^2 - 5x}}{{x^2}} \right)^2 + \lim_{{x \to 0}} \frac{{e^{13x} - 1}}{{x}}$

Let's break it down and calculate each limit:

1. First Limit: $\lim_{{x \to 1}} \frac{{x^2 - 64}}{{x - 8}}$

We can factor $x^2 - 64$ as $(x - 8)(x + 8)$: $\lim_{{x \to 1}} \frac{{(x - 8)(x + 8)}}{{x - 8}} = \lim_{{x \to 1}} (x + 8)$ Substituting $x = 1$: $1 + 8 = 9$ So, this limit evaluates to 9.

2. Second Limit: $\lim_{{x \to \infty}} \frac{{8x^2 - 5x}}{{x^2}}$

We can divide each term by $x^2$: $\lim_{{x \to \infty}} \left( 8 - \frac{5}{x} \right)$ As $x \to \infty$, $\frac{5}{x} \to 0$, so the limit is: $8$

3. Third Limit: $\lim_{{x \to 0}} \frac{{e^{13x} - 1}}{{x}}$

This is a standard limit, which is recognized as the derivative of $e^{13x}$ at $x = 0$. Using the known result: $\lim_{{x \to 0}} \frac{{e^{13x} - 1}}{{x}} = 13$

Now, combining all three parts: $\left( 9 \right)^3 + \left( 8 \right)^2 + 13 = 729 + 64 + 13 = 806$

Thus, the value of the entire expression is: $806$

Would you like further details or explanations for any of the steps? Here are some related questions you might find helpful:

1. What is the formal definition of a limit?
2. How do you evaluate indeterminate forms like $\frac{0}{0}$?
3. How can you use L'Hopital's rule to simplify limits?
4. What is the derivative of $e^{13x}$, and how does it relate to the third limit?
5. How does factoring help when evaluating limits involving polynomials?

Tip: When dealing with limits of polynomials, always look for opportunities to factor expressions, especially when direct substitution results in indeterminate forms.