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  4. displaystyle lim x arrow ∞ (1-2+3-4+5-6+⋅s-2 n/√n2+1+√4 n2-1) is equal to:

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Q. $\displaystyle\lim _{x \rightarrow \infty} \frac{1-2+3-4+5-6+\cdots-2 n}{\sqrt{n^{2}+1}+\sqrt{4 n^{2}-1}}$ is equal to:

Limits and Derivatives

Solution:

$\displaystyle\lim _{x \rightarrow \infty} \frac{1-2+3-4+5-6+\cdots-2 n}{\sqrt{n^{2}+1}+\sqrt{4 n^{2}-1}}$
$=\displaystyle\lim _{x \rightarrow \infty} \frac{-1-1-1 \cdots n \text { times }}{\sqrt{n^{2}+1}+\sqrt{4 n^{2}-1}}$
$=\displaystyle\lim _{x \rightarrow \infty} \frac{-1}{\sqrt{1+\frac{1}{n^{2}}}+\sqrt{4-\frac{1}{n^{2}}}}=-\frac{1}{3}$