$$\lim_{n \to \infty}\left(\frac{(n+1)^{1/3}}{n^{4/3}} + \frac{(n+2)^{1/3}}{n^{4/3}} + \ldots + \frac{(2n)^{1/3}}{n^{4/3}}\right)$$ is equal to

We have to compute the limit

$$\lim_{n\to\infty}\left(\frac{(n+1)^{1/3}}{n^{4/3}}+\frac{(n+2)^{1/3}}{n^{4/3}}+\ldots+\frac{(2n)^{1/3}}{n^{4/3}}\right).$$

Observe that the summation index $$k$$ can run from $$1$$ to $$n$$ if we write the general term in the compact form

$$\sum_{k=1}^{n}\frac{(n+k)^{1/3}}{n^{4/3}}.$$

Now we separate the factor $$n$$ inside the power of the numerator. We write

$$ (n+k)^{1/3}=n^{1/3}\left(1+\frac{k}{n}\right)^{1/3}.$$

Substituting this expression into every fraction we get

$$\frac{(n+k)^{1/3}}{n^{4/3}} \;=\; \frac{n^{1/3}\left(1+\dfrac{k}{n}\right)^{1/3}}{n^{4/3}}.$$

Because powers of the same base add their exponents, $$n^{1/3}/n^{4/3}=n^{(1/3-4/3)}=n^{-1}=1/n$$. Therefore

$$\frac{(n+k)^{1/3}}{n^{4/3}}=\frac{1}{n}\left(1+\frac{k}{n}\right)^{1/3}.$$

Hence the entire sum becomes

$$\sum_{k=1}^{n}\frac{(n+k)^{1/3}}{n^{4/3}} \;=\;\sum_{k=1}^{n}\frac{1}{n}\left(1+\frac{k}{n}\right)^{1/3}.$$

The factor $$1/n$$ suggests a Riemann sum. Set

$$x_k=\frac{k}{n},\quad \Delta x=\frac{1}{n}.$$

As $$n\to\infty$$, the points $$x_k$$ fill the interval $$[0,1]$$, and we recognise

$$\sum_{k=1}^{n}\frac{1}{n}\left(1+\frac{k}{n}\right)^{1/3} \;\longrightarrow\; \int_{0}^{1}(1+x)^{1/3}\,dx.$$

Now we evaluate this definite integral. We first state the power rule for integration:

$$\int u^{m}\,du=\frac{u^{m+1}}{m+1}+C,\qquad\text{provided }m\neq-1.$$

In our case $$m=\dfrac{1}{3}$$ and $$u=1+x$$. Making the substitution $$u=1+x\Longrightarrow du=dx$$, we have

$$\int_{0}^{1}(1+x)^{1/3}\,dx=\int_{u=1}^{2}u^{1/3}\,du.$$

Applying the power rule,

$$\int u^{1/3}\,du=\frac{u^{1/3+1}}{1/3+1}=\frac{u^{4/3}}{4/3}=\frac{3}{4}u^{4/3}.$$

Therefore

$$\int_{1}^{2}u^{1/3}\,du=\left.\frac{3}{4}u^{4/3}\right|_{1}^{2} =\frac{3}{4}\left(2^{4/3}-1^{4/3}\right) =\frac{3}{4}\left(2^{4/3}-1\right).$$

Thus the limit that we were asked to find equals

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

This exact value matches Option A, written as $$\frac{3}{4}(2)^{4/3}-\frac{3}{4}.$$

Hence, the correct answer is Option A.