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!