The limit expression provided in the image is:

$\lim_{x \to +\infty} \left( 1 + \frac{x}{2} \right)^{\frac{1}{x}}$

Step-by-Step Solution:

We are tasked with finding:

$\lim_{x \to +\infty} \left( 1 + \frac{x}{2} \right)^{\frac{1}{x}}$

Step 1: Rewrite the expression

Let us rewrite the base of the exponent in a more convenient form:

$\left( 1 + \frac{x}{2} \right)^{\frac{1}{x}}$

As $x \to +\infty$, the term $\frac{x}{2}$ grows large, making $1 + \frac{x}{2}$ also large. However, the exponent $\frac{1}{x}$ tends to zero, so this is an indeterminate form of type $1^{0}$.

Step 2: Apply logarithms

To resolve the indeterminate form, we can apply logarithms. Let $L = \lim_{x \to +\infty} \left( 1 + \frac{x}{2} \right)^{\frac{1}{x}}$. Taking the natural logarithm of both sides:

$\ln(L) = \lim_{x \to +\infty} \frac{1}{x} \ln\left( 1 + \frac{x}{2} \right)$

Step 3: Simplify the logarithm

For large $x$, we can approximate $1 + \frac{x}{2}$ as approximately $\frac{x}{2}$. Hence:

$\ln\left( 1 + \frac{x}{2} \right) \approx \ln\left( \frac{x}{2} \right) = \ln(x) + \ln\left( \frac{1}{2} \right)$

Thus, the expression becomes:

$\ln(L) = \lim_{x \to +\infty} \frac{1}{x} \left( \ln(x) + \ln\left( \frac{1}{2} \right) \right)$

Since $\ln\left( \frac{1}{2} \right)$ is a constant, the limit will depend on the behavior of $\frac{\ln(x)}{x}$.

Step 4: Evaluate the limit

We now need to evaluate:

$\lim_{x \to +\infty} \frac{\ln(x)}{x}$

This is a well-known limit, and it evaluates to 0, since $\ln(x)$ grows slower than $x$:

$\lim_{x \to +\infty} \frac{\ln(x)}{x} = 0$

Thus, we have:

$\ln(L) = 0$

Step 5: Final result

Exponentiating both sides gives:

$L = e^0 = 1$

Therefore, the original limit evaluates to:

$\lim_{x \to +\infty} \left( 1 + \frac{x}{2} \right)^{\frac{1}{x}} = 1$

Conclusion:

The value of the limit is $\boxed{1}$.

Would you like further details on this solution, or do you have any specific questions? Here are some follow-up questions you might find useful:

1. What is the significance of the logarithmic approach in limit problems?
2. Why is the form $1^0$ indeterminate, and how do we resolve such cases?
3. How does L'Hôpital's Rule relate to evaluating limits like this?
4. What are other common techniques to handle limits at infinity?
5. How do logarithmic approximations help in simplifying limit problems?

Tip: Whenever you encounter an indeterminate form like $1^0$, $0^0$, or $\infty^0$, applying logarithms can often simplify the expression and make it easier to compute the limit.