Question:

Step 1: Rewrite the given sum in summation notation. The given expression is a sum of $n$ terms: $$ L = \lim_{n\to\infty}\left(\dfrac{1}{1+n}+\dfrac{1}{2+n}+\dfrac{1}{3+n}+\cdots+\dfrac{1}{2n}\right) $$ The general term in the sum is $\dfrac{1}{k+n}$. The first term corresponds to $k=1$ (i.e., $\dfrac{1}{1+n}$), and the last term corresponds to $k=n$ (i.e., $\dfrac{1}{n+n} = \dfrac{1}{2n}$). So, the sum can be written using summation notation as: $$ L = \lim_{n\to\infty}\sum_{k=1}^{n} \dfrac{1}{k+n} $$ Step 2: Convert the sum into the form of a Riemann sum. To convert the sum into a definite integral using the definition of a Riemann sum, we need to express the general term in the form $\dfrac{1}{n} f\left(\dfrac{k}{n}\right)$. We can factor out $n$ from the denominator of each term: $$ \dfrac{1}{k+n} = \dfrac{1}{n\left(\dfrac{k}{n}+1\right)} = \dfrac{1}{n} \left(\dfrac{1}{1+\dfrac{k}{n}}\right) $$ Now, substitute this back into the limit expression: $$ L = \lim_{n\to\infty}\sum_{k=1}^{n} \dfrac{1}{n} \left(\dfrac{1}{1+\dfrac{k}{n}}\right) $$ Step 3: Identify the function and the limits of integration for the definite integral. The general form of a definite integral as a limit of a sum (Riemann sum) is: $$ \int_a^b f(x) dx = \lim_{n\to\infty}\sum_{k=1}^{n} \dfrac{1}{n} f\left(a + k\dfrac{b-a}{n}\right) $$ A common special case is for the interval $[0,1]$: $$ \int_0^1 f(x) dx = \lim_{n\to\infty}\sum_{k=1}^{n} \dfrac{1}{n} f\left(\dfrac{k}{n}\right) $$ By comparing our expression $L = \lim_{n\to\infty}\sum_{k=1}^{n} \dfrac{1}{n} \left(\dfrac{1}{1+\dfrac{k}{n}}\right)$ with this standard form, we can identify: The function $f(x) = \dfrac{1}{1+x}$. The limits of integration are from $x=0$ (corresponding to $k/n \to 0$ as $n \to \infty$) to $x=1$ (corresponding to $k/n \to 1$ when $k=n$). Thus, the limit of the sum can be expressed as the definite integral: $$ L = \int_0^1 \dfrac{1}{1+x} dx $$ Step 4: Evaluate the definite integral. Now, we evaluate the integral: $$ L = \int_0^1 \dfrac{1}{1+x} dx $$ The antiderivative of $\dfrac{1}{1+x}$ is $\ln|1+x|$. $$ L = \left[\ln|1+x|\right]_0^1 $$ Substitute the upper and lower limits: $$ L = \ln(1+1) - \ln(1+0) $$ $$ L = \ln(2) - \ln(1) $$ Since $\ln(1) = 0$: $$ L = \ln(2) - 0 $$ $$ L = \ln(2) $$ This can also be written as $\log_e 2$. Step 5: Conclude the final answer. The limit of the given sum is $\log_e 2$. The final answer is $\boxed{\log_e 2}$.