Evaluate the limit: $\lim_{x \to \infty} \frac{\tan\left(5x^{1/3}\right) \log_e(1 + 3x^2)}{\left(\tan^{-1}(3\sqrt{x})\right)^2 \left(e^{5x^{4/3}} - 1\right)}$ is equal to:

Step 1: Simplify terms using standard limits

When evaluating limits as $x \to \infty$, we often use standard limit approximations. For very large $x$, we can simplify terms like $\tan^{-1}(3\sqrt{x})$ and $e^{5x^{4/3}} - 1$.

Step 2: Approximate terms for large x

As $x \to \infty$, $3\sqrt{x} \to \infty$, so $\tan^{-1}(3\sqrt{x})$ approaches $\frac{\pi}{2}$. Also, $e^{5x^{4/3}}$ becomes much larger than 1, so $e^{5x^{4/3}} - 1 \approx e^{5x^{4/3}}$. For $\log_e(1 + 3x^2)$, as $x \to \infty$, $3x^2$ dominates 1, so $\log_e(1 + 3x^2) \approx \log_e(3x^2)$.

Step 3: Substitute approximations into the limit expression

Now we substitute these approximations back into the original limit expression. The denominator simplifies to $(\frac{\pi}{2})^2 e^{5x^{4/3}}$.

Step 4: Analyze the behavior of the terms

As $x \to \infty$, $5x^{1/3} \to \infty$. The tangent function $\tan(y)$ oscillates between $-\infty$ and $\infty$ as $y \to \infty$. It does not approach a single finite value.

Step 5: Determine the limit

The numerator contains an oscillating term $\tan(5x^{1/3})$ and a term $(\log_e(3) + 2\log_e(x))$ that goes to $\infty$. The denominator $e^{5x^{4/3}}$ goes to $\infty$ much faster than the numerator. However, due to the oscillating nature of the tangent function, the limit does not exist.

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