Key Concepts and Formulas

• Limit of a Sum as a Definite Integral: The fundamental theorem connecting sums and integrals states that for a function $f(x)$ continuous on $[0, 1]$: $\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {f\left( {{r \over n}} \right)} = \int\limits_0^1 {f\left( x \right)} \,\,dx$ This formula is crucial for converting limits involving sums into definite integrals.
• Summation Formulas: Knowledge of standard summation formulas is helpful, though not strictly required if using the integral approach. For example, the sum of the first $n$ fourth powers is $\sum_{r=1}^n r^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$, and the sum of the first $n$ cubes is $\sum_{r=1}^n r^3 = \left(\frac{n(n+1)}{2}\right)^2$. However, the limit of these sums divided by an appropriate power of $n$ can be evaluated more efficiently using the definite integral method.
• Limits involving powers of n: When evaluating limits of the form $\mathop {\lim }\limits_{n \to \infty } \frac{P(n)}{Q(n)}$ where $P(n)$ and $Q(n)$ are polynomials in $n$, the limit is zero if the degree of $Q(n)$ is greater than the degree of $P(n)$.

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Step-by-Step Solution

The problem asks for the evaluation of the difference between two limits: $L = \mathop {\lim }\limits_{n \to \infty } {{1 + {2^4} + {3^4} + .... + {n^4}} \over {{n^5}}} - \mathop {\lim }\limits_{n \to \infty } {{1 + {2^3} + {3^3} + .... + {n^3}} \over {{n^5}}}$ We will evaluate each limit separately and then subtract the results.

1. Evaluating the First Limit ($L_1$)

The first limit is: $L_1 = \mathop {\lim }\limits_{n \to \infty } {{1 + {2^4} + {3^4} + .... + {n^4}} \over {{n^5}}}$

• Step 1.1: Rewrite the numerator in sigma notation. The sum of the fourth powers of the first $n$ natural numbers is given by $\sum_{r=1}^n r^4$. $L_1 = \mathop {\lim }\limits_{n \to \infty } {{\sum_{r=1}^n r^4} \over {{n^5}}}$

• Step 1.2: Manipulate the expression to fit the limit of a sum formula. To apply the formula $\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {f\left( {{r \over n}} \right)} = \int\limits_0^1 {f\left( x \right)} \,\,dx$, we need to extract a factor of $\frac{1}{n}$ and express the terms inside the sum as a function of $\frac{r}{n}$. We rewrite $\frac{1}{n^5}$ as $\frac{1}{n} \cdot \frac{1}{n^4}$. $L_1 = \mathop {\lim }\limits_{n \to \infty } {1 \over n} \left( {{\sum_{r=1}^n r^4} \over {{n^4}}} \right)$ Now, we move the $n^4$ from the denominator of the fraction into the summation term, making it $\frac{r^4}{n^4} = \left(\frac{r}{n}\right)^4$. $L_1 = \mathop {\lim }\limits_{n \to \infty } {1 \over n} \sum_{r=1}^n \left( {{r \over n}} \right)^4$ This manipulation is done to transform the expression into the standard form $\frac{1}{n}\sum f(\frac{r}{n})$.

• Step 1.3: Convert the limit of a sum to a definite integral. By comparing $L_1 = \mathop {\lim }\limits_{n \to \infty } {1 \over n} \sum_{r=1}^n \left( {{r \over n}} \right)^4$ with the general formula, we identify $f\left( {{r \over n}} \right) = \left( {{r \over n}} \right)^4$. Thus, the function is $f(x) = x^4$. The limit of the sum is then equivalent to the definite integral: $L_1 = \int\limits_0^1 {{x^4}} \,\,dx$

• Step 1.4: Evaluate the definite integral. $L_1 = \left[ {{{{x^5}} \over 5}} \right]_0^1 = \left( {{{{1^5}} \over 5}} \right) - \left( {{{{0^5}} \over 5}} \right) = {1 \over 5} - 0 = {1 \over 5}$

2. Evaluating the Second Limit ($L_2$)

The second limit is: $L_2 = \mathop {\lim }\limits_{n \to \infty } {{1 + {2^3} + {3^3} + .... + {n^3}} \over {{n^5}}}$

• Step 2.1: Rewrite the numerator in sigma notation. The sum of the cubes of the first $n$ natural numbers is given by $\sum_{r=1}^n r^3$. $L_2 = \mathop {\lim }\limits_{n \to \infty } {{\sum_{r=1}^n r^3} \over {{n^5}}}$

• Step 2.2: Manipulate the expression to fit the limit of a sum formula. We need to extract a factor of $\frac{1}{n}$ and express the terms inside the sum as a function of $\frac{r}{n}$. The sum $\sum_{r=1}^n r^3$ has a leading term of order $n^4$ (since $\sum r^3 \approx \frac{n^4}{4}$). We have $n^5$ in the denominator. To form the integral $\int_0^1 x^3 dx$, we need $\frac{1}{n} \sum (\frac{r}{n})^3$, which implies a total denominator of $n \cdot n^3 = n^4$. Since we have $n^5$, there will be an extra factor of $\frac{1}{n}$. We rewrite $\frac{1}{n^5}$ as $\frac{1}{n} \cdot \frac{1}{n^4}$. Then, we move $\frac{1}{n^4}$ inside the summation as $\frac{r^3}{n^3} \cdot \frac{1}{n}$, which is not directly $(\frac{r}{n})^3$. Instead, we rewrite it as: $L_2 = \mathop {\lim }\limits_{n \to \infty } {1 \over n} \left( {{\sum_{r=1}^n r^3} \over {{n^4}}} \right)$ Now, move $n^4$ into the summation: $L_2 = \mathop {\lim }\limits_{n \to \infty } {1 \over n} \sum_{r=1}^n \left( {{r^3} \over {{n^4}}} \right)$ This is not yet in the form $\sum (\frac{r}{n})^3$. Let's re-evaluate the manipulation. We need $\sum (\frac{r}{n})^3$. This requires $\frac{r^3}{n^3}$ inside the sum. $L_2 = \mathop {\lim }\limits_{n \to \infty } {1 \over n} \sum_{r=1}^n \left( {{r^3} \over {{n^3}}} \cdot {1 \over n} \right)$ This still doesn't look right. The correct way is to isolate the term that becomes the integral. The sum $\sum_{r=1}^n r^3$ is approximately $\frac{n^4}{4}$. So the expression is approximately $\frac{n^4/4}{n^5} = \frac{1}{4n}$, which tends to 0. Let's use the integral form directly by identifying $f(x) = x^3$. To do this, we need $\frac{1}{n} \sum (\frac{r}{n})^3$. $L_2 = \mathop {\lim }\limits_{n \to \infty } {1 \over n} \sum_{r=1}^n \left( {{r^3} \over {{n^3}}} \right) \cdot {1 \over n^2}$ The term $\mathop {\lim }\limits_{n \to \infty } {1 \over n} \sum_{r=1}^n \left( {{r \over n}} \right)^3$ converges to $\int_0^1 x^3 dx$. So, $L_2 = \mathop {\lim }\limits_{n \to \infty } \left( {1 \over n} \sum_{r=1}^n \left( {{r \over n}} \right)^3 \right) \cdot \mathop {\lim }\limits_{n \to \infty } \left( {1 \over n^2} \right)$ The first part is $\int_0^1 x^3 dx = \frac{1}{4}$. The second part is $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n^2} = 0$. Therefore, $L_2 = \frac{1}{4} \cdot 0 = 0$.

  Alternatively, we can express the sum of cubes as $\sum_{r=1}^n r^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4} = \frac{n^4 + 2n^3 + n^2}{4}$. Then, $L_2 = \mathop {\lim }\limits_{n \to \infty } {{n^4 + 2n^3 + n^2} \over {4n^5}}$ To evaluate this limit, we divide the numerator and denominator by the highest power of $n$ in the denominator, which is $n^5$: $L_2 = \mathop {\lim }\limits_{n \to \infty } {{{n^4 \over n^5} + {2n^3 \over n^5} + {n^2 \over n^5}} \over {{4n^5 \over n^5}}} = \mathop {\lim }\limits_{n \to \infty } {{{1 \over n} + {2 \over n^2} + {1 \over n^3}} \over 4}$ As $n \to \infty$, the terms $\frac{1}{n}$, $\frac{2}{n^2}$, and $\frac{1}{n^3}$ all approach 0. $L_2 = {{0 + 0 + 0} \over 4} = {0 \over 4} = 0$

3. Combine the Results

The total limit $L$ is the difference between $L_1$ and $L_2$. $L = L_1 - L_2 = {1 \over 5} - 0 = {1 \over 5}$

Common Mistakes & Tips

• Incorrectly matching the limit of a sum formula: Ensure the expression is precisely in the form $\frac{1}{n}\sum f(\frac{r}{n})$. Misidentifying $f(x)$ or the limits of integration is a common error.
• Algebraic errors in manipulation: Carefully manipulate the terms to isolate the $\frac{1}{n}$ factor and the $\frac{r}{n}$ argument for the function.
• Mistaking the order of growth: For the second limit, recognizing that the sum of cubes grows as $n^4$ while the denominator grows as $n^5$ is a quick way to see the limit is zero. The integral method confirms this rigorously.

Summary

The problem involves evaluating the difference of two limits. Each limit is transformed into a definite integral using the limit of a sum as a definite integral concept. The first limit, involving the sum of fourth powers divided by $n^5$, is converted to $\int_0^1 x^4 dx$, which evaluates to $\frac{1}{5}$. The second limit, involving the sum of cubes divided by $n^5$, evaluates to zero because the denominator's growth rate ($n^5$) is higher than the effective growth rate of the numerator's summation part when structured for integration ($\approx n^4$). Subtracting the two results gives the final answer.

The final answer is \boxed{\frac{1}{5}}. This corresponds to option (A).