Question:

The value of $\displaystyle\lim_{n\to\infty}\dfrac{1+2-3+4+5-6+\cdots+(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$ is:

\dfrac{\sqrt{2}+1}{2}

3(\sqrt{2}+1)

\dfrac{3}{2}(\sqrt{2}+1)

\dfrac{3}{2\sqrt{2}}

Step-by-Step Solution

Key Concept: Group terms in triples: $(3k-2)+(3k-1)-3k=3(k-1)$ for $k=1,\ldots,n$. Numerator $=\sum_{k=1}^n3(k-1)=3n(n-1)/2\approx3n^2/2$.

$\dfrac{3}{2}(\sqrt{2}+1)$.

Correct Answer: 3