The value of operatorname{Lim}_{n rightarrow | vedclass

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(C) The numerator is a sum of $n$ groups of the form $(3k-2) + (3k-1) - 3k = 3k-3$ for $k=1$ to $n$.Sum $= \sum_{k=1}^{n} (3k-3) = 3 \sum_{k=1}^{n} (k

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The numerator is a sum of $n$ groups of the form $(3k-2) + (3k-1) - 3k = 3k-3$ for $k=1$ to $n$.Sum $= \sum_{k=1}^{n} (3k-3) = 3 \sum_{k=1}^{n} (k-1) = 3 \frac{(n-1)n}{2} = \frac{3n^2-3n}{2}$.Now,consider the limit: $\operatorname{Lim}_{n ightarrow \infty} \frac{\frac{3n^2-3n}{2}}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$.Divide numerator and denominator by $n^2$: $\operatorname{Lim}_{n}$ ${ightarrow \infty} \frac{\frac{3}{2} - \frac{3}{2n}}{\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}}$.As $n ightarrow \infty$,the expression approaches $\frac{3/2}{\sqrt{2}-1}$.Rationalizing the denominator: $\frac{3}{2(\sqrt{2}-1)} 	imes \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{3(\sqrt{2}+1)}{2(2-1)} = \frac{3}{2}(\sqrt{2}+1)$.

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

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