$$\lim_{x \to 0^+} \frac{\tan\left(5(x)^{1/3}\right) \cdot \log_e(1+3x^2)}{\left(\tan^{-1}3\sqrt{x}\right)^2 \left(e^{5(x)^{4/3}} - 1\right)}$$ is equal to

The limit is

$$L=\lim_{x\to 0^{+}} \frac{\tan\!\bigl(5x^{1/3}\bigr)\;\ln(1+3x^{2})} {\bigl(\tan^{-1} 3\sqrt{x}\bigr)^{2}\; \bigl(e^{\,5x^{4/3}}-1\bigr)}$$

When a variable $$t$$ approaches $$0$$, the following first-order approximations are standard:
$$\tan t \approx t,$$ $$\ln(1+t) \approx t,$$ $$\tan^{-1} t \approx t,$$ $$e^{\,t}-1 \approx t.$$

All the arguments in our limit are powers of $$x$$, so we may replace each function by its first-order term.

Step 1: Expand the numerator

Argument of tangent: $$5x^{1/3}\to 0$$, hence
$$\tan\!\bigl(5x^{1/3}\bigr)\approx 5x^{1/3}.$$

Argument of logarithm: $$3x^{2}\to 0$$, hence
$$\ln(1+3x^{2})\approx 3x^{2}.$$

Therefore the numerator is approximately
$$5x^{1/3}\;\cdot\;3x^{2}=15\,x^{\,1/3+2}=15\,x^{7/3}.$$

Step 2: Expand the denominator

Argument of inverse tangent: $$3\sqrt{x}=3x^{1/2}\to 0$$, so
$$\tan^{-1}\!\bigl(3\sqrt{x}\bigr)\approx 3x^{1/2}.$$

Squaring it gives
$$(\tan^{-1}3\sqrt{x})^{2}\approx (3x^{1/2})^{2}=9x.$$

Argument of the exponential: $$5x^{4/3}\to 0$$, hence
$$e^{\,5x^{4/3}}-1\approx 5x^{4/3}.$$

Thus the denominator is approximately
$$9x\;\cdot\;5x^{4/3}=45\,x^{\,1+4/3}=45\,x^{7/3}.$$

Step 3: Form the ratio

$$ \frac{15\,x^{7/3}}{45\,x^{7/3}}=\frac{15}{45}=\frac13. $$

Since both numerator and denominator share the same power $$x^{7/3}$$, the factor cancels exactly as $$x\to 0^{+}$$. Hence the original limit equals $$\frac13$$.

Final answer: $$\displaystyle \frac13$$ (Option C).