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$\lim _{n \rightarrow \infty} n\left(\sqrt{n^2+9}-n\right)=$
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Verified Answer
The correct answer is:
$\frac{9}{2}$
$\lim _{n \rightarrow \infty} n\left(\sqrt{n^2+9}-n\right)$
$\lim _{n \rightarrow \infty} \frac{n\left(n^2+9-n^2\right)}{\sqrt{n^2+9}+n}$
$\lim _{n \rightarrow \infty} \frac{9}{\sqrt{1+\frac{9}{n^2}}+1}=\frac{9}{2}$
$\lim _{n \rightarrow \infty} \frac{n\left(n^2+9-n^2\right)}{\sqrt{n^2+9}+n}$
$\lim _{n \rightarrow \infty} \frac{9}{\sqrt{1+\frac{9}{n^2}}+1}=\frac{9}{2}$
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