Question

The value of $\displaystyle\lim _{n \rightarrow \infty} \frac{1+2-3+4+5-6+\ldots +(3 n-2)+(3 n-1)-3 n}{\sqrt{2 n^4+4 n+3-} \sqrt{n^4+5 n+4}}$ is :

• A. $3(\sqrt{2}+1)$
• B. $\frac{3}{2}(\sqrt{2}+1)$
• C. $\frac{\sqrt{2}+1}{2}$
• D. $\frac{3}{2 \sqrt{2}}$

Answer 1

Answer: (B)

$\frac{3}{2}(\sqrt{2}+1)$

Answer 2

n→∞Lim​2n4+4n+3​−n4+5n+4​0+3+6+9+….n terms ​
n→∞Lim​2(2n4+4n+3​−n4+5n+4​)3n(n−1)​
$=\frac3{2(\sqrt2-1)}=\frac3{2}(\sqrt2+1)$