`lim_(xtooo) f(x)," where "(2x-3)/(x)ltf(x)lt(2x^(2)+5x)/(x^(2))AAxgt0," is "`_____.

Explanation

To solve the limit problem \( \lim_{x \to \infty} f(x) \) given the inequalities \( \frac{2x - 3}{x} < f(x) < \frac{2x^2 + 5x}{x^2} \), we will follow these steps:

### Step 1: Simplify the lower bound
We start with the lower bound of the inequality:

\[ \frac{2x - 3}{x} \]

This can be simplified as follows:

\[ \frac{2x - 3}{x} = \frac{2x}{x} - \frac{3}{x} = 2 - \frac{3}{x} \]

### Step 2: Find the limit of the lower bound as \( x \to \infty \)
Now, we take the limit of the lower bound as \( x \) approaches infinity:

\[ \lim_{x \to \infty} \left( 2 - \frac{3}{x} \right) = 2 - 0 = 2 \]

### Step 3: Simplify the upper bound
Next, we simplify the upper bound of the inequality:

\[ \frac{2x^2 + 5x}{x^2} \]

This can be simplified as follows:

\[ \frac{2x^2 + 5x}{x^2} = \frac{2x^2}{x^2} + \frac{5x}{x^2} = 2 + \frac{5}{x} \]

### Step 4: Find the limit of the upper bound as \( x \to \infty \)
Now, we take the limit of the upper bound as \( x \) approaches infinity:

\[ \lim_{x \to \infty} \left( 2 + \frac{5}{x} \right) = 2 + 0 = 2 \]

### Step 5: Apply the Squeeze Theorem
Since we have established that:

\[ 2 - \frac{3}{x} < f(x) < 2 + \frac{5}{x} \]

and both the lower and upper bounds approach 2 as \( x \to \infty \), we can apply the Squeeze Theorem:

\[ \lim_{x \to \infty} f(x) = 2 \]

### Conclusion
Thus, the limit is:

\[ \lim_{x \to \infty} f(x) = 2 \]