Question

\[ \mathop {\lim }\limits_{n \to \infty }\left[\frac{1}{1-n^4}+\frac{8}{1-n^4}+\ldots \ldots \ldots \ldots .+\frac{n^3}{1-n^4}\right]= \]

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{2}\)
• C. \(-\frac{1}{2}\)
• D. \(-\frac{1}{4}\)

Hint:

The numerator is the sum of the first n cubes; use the sum formula.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
All terms share the denominator \(1-n^{4}\) and the numerators are \(1^{3},2^{3},\ldots,n^{3}\).

Step 2: Sum of cubes
\(1^{3}+2^{3}+\cdots+n^{3}=\dfrac{n^{2}(n+1)^{2}}{4}\).

Step 3: Form the limit
\(\lim_{n\to\infty}\dfrac{n^{2}(n+1)^{2}}{4(1-n^{4})}\). Divide the numerator and denominator by \(n^{4}\): \(\dfrac{(1+1/n)^{2}}{4(1/n^{4}-1)}\).

Step 4: Evaluate
As \(n\to\infty\) this tends to \(\dfrac{1}{4(0-1)}=-\dfrac{1}{4}\).

Final Answer:
The limit is -1/4. \[ \boxed{\text{(D) }-\dfrac{1}{4}} \]