To solve the limit problem

$\lim_{{x \to 0}} \frac{x^2}{2} \cdot \cos \left( x + \frac{1}{x} \right),$

we’ll analyze the behavior of each component as $x \to 0$.

Step-by-Step Solution

1. Examine $\frac{x^2}{2}$: As $x \to 0$, $x^2$ also approaches $0$. Thus, $\frac{x^2}{2} \to 0$ as $x \to 0$.

2. Examine $\cos \left( x + \frac{1}{x} \right)$:

   • The term $x + \frac{1}{x}$ oscillates wildly as $x \to 0$ because $\frac{1}{x}$ grows infinitely large in magnitude, causing $x + \frac{1}{x}$ to vary between large positive and negative values.
   • The cosine of a wildly oscillating argument like this does not approach a specific value, but it remains bounded between -1 and 1.

3. Analyze the product:

   • Since $\cos \left( x + \frac{1}{x} \right)$ is bounded, we only need to focus on $\frac{x^2}{2}$, which approaches $0$.
   • The product of $\frac{x^2}{2}$ and a bounded function will approach $0$ as $x \to 0$.

Conclusion

$\lim_{{x \to 0}} \frac{x^2}{2} \cdot \cos \left( x + \frac{1}{x} \right) = 0.$

Would you like further details on any steps?

Here are 5 related questions:

1. What happens if we replace $\cos$ with $\sin$ in this limit?
2. How does the behavior of $\frac{1}{x}$ as $x \to 0$ affect other functions within limits?
3. Can we use the squeeze theorem for this type of limit problem?
4. What if the power of $x$ were different (e.g., $x^3$) in this limit?
5. How does the limit change if we include an absolute value around $x^2$?

Tip: When dealing with limits involving oscillatory functions, look for boundedness and approach of other factors to zero for simplification.