Let’s solve the two limits from the image step by step.

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

Solution:

First, simplify the numerator.

$\frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2x}$

Thus, the expression becomes:

$\lim_{x \to 2} \frac{\frac{2 + x}{2x}}{x^3 + 8}$

Notice that $x^3 + 8$ can be factored as a sum of cubes:

$x^3 + 8 = (x + 2)(x^2 - 2x + 4)$

Now, substitute $x = 2$ directly in the limit since the function is defined at $x = 2$:

$\frac{\frac{2 + 2}{2 \cdot 2}}{2^3 + 8} = \frac{\frac{4}{4}}{8 + 8} = \frac{1}{16}$

So, the limit is:

$\boxed{\frac{1}{16}}$

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2. $\lim_{x \to 0} \frac{\tan(5x)}{\sin(2x)}$

Solution:

For small $x$, we can use the approximations:

$\tan(5x) \approx 5x \quad \text{and} \quad \sin(2x) \approx 2x$

Thus, the expression becomes:

$\lim_{x \to 0} \frac{5x}{2x}$

Canceling $x$ (as long as $x \neq 0$):

$\lim_{x \to 0} \frac{5}{2} = \frac{5}{2}$

So, the limit is:

$\boxed{\frac{5}{2}}$

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Do you need further clarifications or details?

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5 Related Questions:

1. How do you apply L'Hôpital's rule to evaluate limits?
2. What is the approximation for trigonometric functions near 0?
3. How do you factor sums and differences of cubes?
4. When can direct substitution be used to evaluate limits?
5. How are limits involving indeterminate forms handled?

Tip: For complicated limits, always check if factoring or approximations simplify the problem before applying advanced methods like L’Hôpital’s rule.