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The given expression is:

\[
\lim_{{x \to 0}} \left( \frac{1}{x} - \frac{2 + x}{x^2} \right)
\]

### Solution:

1. First, split the limit into two parts:
   \[
   \lim_{{x \to 0}} \left( \frac{1}{x} - \frac{2 + x}{x^2} \right) = \lim_{{x \to 0}} \frac{1}{x} - \lim_{{x \to 0}} \frac{2 + x}{x^2}
   \]

2. Evaluate each limit separately.

   - For the first part:
     \[
     \lim_{{x \to 0}} \frac{1}{x}
     \]
     As \(x \to 0\), \(\frac{1}{x}\) approaches infinity or negative infinity depending on the direction of the limit. So, this part diverges.

   - For the second part:
     \[
     \lim_{{x \to 0}} \frac{2 + x}{x^2}
     \]
     We can split it as:
     \[
     \frac{2 + x}{x^2} = \frac{2}{x^2} + \frac{x}{x^2} = \frac{2}{x^2} + \frac{1}{x}
     \]
     Both terms here also tend toward infinity as \(x \to 0\).

### Conclusion:
Since both parts of the limit approach infinity (or negative infinity depending on the direction), the limit does not exist. Therefore, the given expression diverges as \(x \to 0\).

Would you like more details or have any questions?

Here are 5 related questions for further practice:

1. What is the limit of \(\frac{1}{x}\) as \(x \to 0^+\)?
2. How can you determine when a limit does not exist?
3. What techniques can be used to solve limits that involve indeterminate forms like \(\frac{0}{0}\)?
4. How do one-sided limits differ from two-sided limits?
5. How can L'Hôpital's Rule be applied to evaluate limits?

**Tip**: When solving limits that lead to infinity, always check if the function behaves differently as \(x\) approaches from the left versus the right!

Learn how to solve the limit of the expression (1/x - (2+x)/x^2) as x approaches 0. This problem covers limits, indeterminate forms, and algebraic manipulation, with key focus on the behavior of functions near zero.

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