To solve the limit using the Squeeze Theorem, let's first express the limit in a more manageable form. We are given:

$\lim_{x \to 0^+} \sqrt{3x} \cdot e^{\cos\left(\frac{8\pi}{x}\right)}$

Step 1: Analyzing the components of the expression

1. $\sqrt{3x}$:

   • As $x$ approaches 0 from the right, $\sqrt{3x}$ approaches 0.

2. $e^{\cos\left(\frac{8\pi}{x}\right)}$:

   • The term $\cos\left(\frac{8\pi}{x}\right)$ oscillates between $-1$ and $1$ as $x$ approaches 0 because $\cos(\theta)$ always lies between $-1$ and $1$ for any $\theta$.
   • Thus, $e^{\cos\left(\frac{8\pi}{x}\right)}$ oscillates between $e^{-1}$ and $e^1$.

Step 2: Applying the Squeeze Theorem

Since $e^{-1} \leq e^{\cos\left(\frac{8\pi}{x}\right)} \leq e$, we can write:

$\sqrt{3x} \cdot e^{-1} \leq \sqrt{3x} \cdot e^{\cos\left(\frac{8\pi}{x}\right)} \leq \sqrt{3x} \cdot e$

Now, consider the limits of the bounding functions as $x \to 0^+$:

$\lim_{x \to 0^+} \sqrt{3x} \cdot e^{-1} = e^{-1} \cdot \lim_{x \to 0^+} \sqrt{3x} = 0$

$\lim_{x \to 0^+} \sqrt{3x} \cdot e = e \cdot \lim_{x \to 0^+} \sqrt{3x} = 0$

Conclusion:

Since both bounds approach 0 as $x \to 0^+$, by the Squeeze Theorem, the original limit also approaches 0.

$\lim_{x \to 0^+} \sqrt{3x} \cdot e^{\cos\left(\frac{8\pi}{x}\right)} = 0$

Would you like further details or have any other questions?

Here are some related questions for further exploration:

1. What is the Squeeze Theorem and how is it applied in different types of limits?
2. How does the behavior of oscillating functions affect limits?
3. Can the Squeeze Theorem be applied to non-oscillating functions?
4. What are some examples of limits that require the Squeeze Theorem for evaluation?
5. How does the exponential function behave when its exponent is oscillating?

Tip: When dealing with limits involving oscillating functions, consider the Squeeze Theorem to evaluate the behavior of the function at the limit point.