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$\lim_{x\rightarrow \infty }2[ \sqrt{n+6} \sqrt{n+10}-n]$ find value
in Calculus by (139 points) | 16 views

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Assuming it as $ \lim_{n→∞} instead\ of\ \lim_{x→∞}$

$\lim_{n→∞} 2(\sqrt{n+6}\sqrt{n+10}−n)$

= $2\lim_{n→∞} (\sqrt{n+6}\sqrt{n+10}−n)$

Multiplying & dividing with $(\sqrt{n+6}\sqrt{n+10}+n)$

= $2\lim_{n→∞} \frac{(\sqrt{n+6}\sqrt{n+10})−n)*(\sqrt{n+6}\sqrt{n+10})+n)}{(\sqrt{n+6}\sqrt{n+10}+n)}$

= $2\lim_{n→∞} \frac{(n+6)*(n+10)−n^2}{\sqrt{n+6}\sqrt{n+10}+n}$

= $2\lim_{n→∞} \frac{(n^2+16n+60)−n^2}{\sqrt{(n+6)(n+10)}+n}$ $(\because \sqrt{a} *\sqrt{b}=\sqrt{a*b})$

= $2\lim_{n→∞} \frac{16n+60}{\sqrt{(n^2+16n+60)}+n}$

Taking out $n$ common

= $2\lim_{n→∞} \frac{n(16+\frac{60}{n})}{\sqrt{n^2(1+\frac{16}{n}+\frac{60}{n^2})}+n}$

= $2\lim_{n→∞} \frac{n(16+\frac{60}{n})}{n(\sqrt{(1+\frac{16}{n}+\frac{60}{n^2})}+\frac{1}{n})}$

= $2\lim_{n→∞} \frac{16+\frac{60}{n}}{\sqrt{(1+\frac{16}{n}+\frac{60}{n^2})}+\frac{1}{n}}$

= $2*16$

= 32

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