$$\lim_{x \to a}\frac{(a+2x)^{\frac{1}{3}} - (3x)^{\frac{1}{3}}}{(3a+x)^{\frac{1}{3}} - (4x)^{\frac{1}{3}}}$$ $$(a \neq 0)$$ is equal to:

We have to evaluate the limit

$$L=\lim_{x \to a}\dfrac{(a+2x)^{\frac13}-(3x)^{\frac13}}{(3a+x)^{\frac13}-(4x)^{\frac13}}, \qquad (a\neq 0).$$

First we check what happens when we directly put $$x=a$$:

$$\text{Numerator at }x=a=(a+2a)^{\frac13}-(3a)^{\frac13}=(3a)^{\frac13}-(3a)^{\frac13}=0,$$

$$\text{Denominator at }x=a=(3a+a)^{\frac13}-(4a)^{\frac13}=(4a)^{\frac13}-(4a)^{\frac13}=0.$$

Thus the limit is of the indeterminate form $$0/0$$. Consequently we can apply L’Hospital’s Rule, which states:

$$\text{If }\displaystyle\lim_{x\to c}\dfrac{f(x)}{g(x)}=\dfrac{0}{0}\text{ or }\dfrac{\infty}{\infty},\ \text{then }\displaystyle\lim_{x\to c}\dfrac{f(x)}{g(x)}=\lim_{x\to c}\dfrac{f'(x)}{g'(x)},$$

provided the latter limit exists.

So we differentiate the numerator and the denominator with respect to $$x$$. We recall the derivative formula

$$\dfrac{d}{dx}\bigl(u^{1/3}\bigr)=\dfrac{1}{3}\,u^{-2/3}\,\dfrac{du}{dx}.$$

Numerator:

$$f(x)=(a+2x)^{\frac13}-(3x)^{\frac13}.$$

Using the above formula,

$$f'(x)=\dfrac{1}{3}(a+2x)^{-\frac23}\cdot 2-\dfrac{1}{3}(3x)^{-\frac23}\cdot 3.$$

Simplifying each term gives

$$f'(x)=\dfrac{2}{3\,(a+2x)^{\frac23}}-(3x)^{-\frac23}.$$

Now we evaluate at $$x=a$$:

$$f'(a)=\dfrac{2}{3\,(a+2a)^{\frac23}}-(3a)^{-\frac23} =\dfrac{2}{3\,(3a)^{\frac23}}-(3a)^{-\frac23}.$$

Writing the second term with the same base factor,

$$f'(a)=\dfrac{2}{3}(3a)^{-\frac23}-(3a)^{-\frac23} =\left(\dfrac{2}{3}-1\right)(3a)^{-\frac23} =-\dfrac{1}{3}(3a)^{-\frac23}.$$

So

$$f'(a)=-\dfrac{1}{3(3a)^{\frac23}}.$$

Denominator:

$$g(x)=(3a+x)^{\frac13}-(4x)^{\frac13}.$$

Differentiating,

$$g'(x)=\dfrac{1}{3}(3a+x)^{-\frac23}\cdot 1-\dfrac{1}{3}(4x)^{-\frac23}\cdot 4.$$

Hence

$$g'(x)=\dfrac{1}{3(3a+x)^{\frac23}}-\dfrac{4}{3(4x)^{\frac23}}.$$

Putting $$x=a$$ gives

$$g'(a)=\dfrac{1}{3(3a+a)^{\frac23}}-\dfrac{4}{3(4a)^{\frac23}} =\dfrac{1}{3(4a)^{\frac23}}-\dfrac{4}{3(4a)^{\frac23}} =\dfrac{1-4}{3(4a)^{\frac23}} =-\dfrac{1}{(4a)^{\frac23}}.$$

Applying L’Hospital’s Rule:

$$L=\dfrac{f'(a)}{g'(a)} =\dfrac{-\dfrac{1}{3(3a)^{\frac23}}}{-\dfrac{1}{(4a)^{\frac23}}} =\dfrac{(4a)^{\frac23}}{3(3a)^{\frac23}}.$$

We separate the powers of $$a$$ and simplify the numerical ratio:

$$L=\dfrac{1}{3}\,\dfrac{(4a)^{\frac23}}{(3a)^{\frac23}} =\dfrac{1}{3}\left(\dfrac{4a}{3a}\right)^{\frac23} =\dfrac{1}{3}\left(\dfrac{4}{3}\right)^{\frac23}.$$

Since $$4=2^{2}$$, we write

$$\left(\dfrac{4}{3}\right)^{\frac23} =\left(\dfrac{2^{2}}{3}\right)^{\frac23} =\dfrac{2^{\frac43}}{3^{\frac23}}.$$

Therefore

$$L=\dfrac{1}{3}\cdot\dfrac{2^{\frac43}}{3^{\frac23}} =\dfrac{2^{\frac43}}{3^{1+\frac23}} =\dfrac{2^{\frac43}}{3^{\frac53}}.$$

We now express this in the form appearing in the options. Notice that

$$\dfrac{2^{\frac43}}{3^{\frac53}} =\dfrac{2}{3}\cdot\dfrac{2^{\frac13}}{3^{\frac23}} =\left(\dfrac{2}{3}\right)\left(\dfrac{2}{9}\right)^{\frac13}.$$

This matches Option D.

Hence, the correct answer is Option D.