1. Given limit

We need to evaluate

$$L=\lim_{x\to a}\frac{(a+2x)^{1/3}-(3x)^{1/3}}{(3a+x)^{1/3}-(4x)^{1/3}},\qquad a\neq 0.$$

As $x\to a$,

• numerator $\to (3a)^{1/3}-(3a)^{1/3}=0$
• denominator $\to (4a)^{1/3}-(4a)^{1/3}=0$

So this is of the form $\frac{0}{0}$, hence we can use L'Hospital's Rule.

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2. Differentiate numerator and denominator with respect to $x$

Let

$$N(x)=(a+2x)^{1/3}-(3x)^{1/3},$$ $$D(x)=(3a+x)^{1/3}-(4x)^{1/3}.$$

Then

$$N'(x)=\frac{1}{3}(a+2x)^{-2/3}\cdot 2-\frac{1}{3}(3x)^{-2/3}\cdot 3$$ $$=\frac{2}{3}(a+2x)^{-2/3}-(3x)^{-2/3}.$$

Also,

$$D'(x)=\frac{1}{3}(3a+x)^{-2/3}\cdot 1-\frac{1}{3}(4x)^{-2/3}\cdot 4$$ $$=\frac{1}{3}(3a+x)^{-2/3}-\frac{4}{3}(4x)^{-2/3}.$$

Therefore,

$$L=\lim_{x\to a}\frac{N'(x)}{D'(x)}.$$

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3. Substitute $x=a$

At $x=a$,

$$a+2x\to 3a,\qquad 3x\to 3a,\qquad 3a+x\to 4a,\qquad 4x\to 4a.$$

So

$$N'(a)=\frac{2}{3}(3a)^{-2/3}-(3a)^{-2/3} =\left(\frac{2}{3}-1\right)(3a)^{-2/3} =-\frac{1}{3}(3a)^{-2/3}.$$

And

$$D'(a)=\frac{1}{3}(4a)^{-2/3}-\frac{4}{3}(4a)^{-2/3} =\left(\frac{1}{3}-\frac{4}{3}\right)(4a)^{-2/3} =-(4a)^{-2/3}.$$

Thus,

$$L=\frac{-\frac{1}{3}(3a)^{-2/3}}{-(4a)^{-2/3}} =\frac{1}{3}\cdot \frac{(4a)^{-2/3}}{(3a)^{-2/3}}^{-1} =\frac{1}{3}\cdot \frac{(3a)^{-2/3}}{(4a)^{-2/3}}.$$

Using

$$\frac{(3a)^{-2/3}}{(4a)^{-2/3}}=\left(\frac{4a}{3a}\right)^{2/3}=\left(\frac{4}{3}\right)^{2/3},$$

we get

$$L=\frac{1}{3}\left(\frac{4}{3}\right)^{2/3}.$$

Now simplify:

$$\left(\frac{4}{3}\right)^{2/3}=\left(\frac{2^2}{3}\right)^{2/3}=\frac{2^{4/3}}{3^{2/3}}.$$

Hence

$$L=\frac{2^{4/3}}{3^{5/3}}.$$

Write this as

$$L=\frac{2}{3}\left(\frac{2}{9}\right)^{1/3}.$$

So the correct option is

$$\boxed{\text{B}}.$$

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4. Option check

Option B is

$$\left(\frac{2}{3}\right)\left(\frac{2}{9}\right)^{1/3},$$

which matches our result exactly.