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  • Evaluating     <img alt="\mathrm{\lim _{n \rightarrow \infty}(\sqrt{(n+a)(n+b)}-n)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bn%20%5Crightarrow%2

Evaluating     \mathrm{\lim _{n \rightarrow \infty}(\sqrt{(n+a)(n+b)}-n)}

Option: 1

\mathrm{\frac{a}{b}}


Option: 2

\mathrm{\frac{ab}{2}}


Option: 3

\mathrm{\frac{a+b}{2}}


Option: 4

0


Answers (1)

Let   \mathrm{a+b=p, a b=q}  , where   \mathrm{p^2-4 q \geqslant 0} , then we have:
\mathrm{\lim _{n \rightarrow \infty} \sqrt{n^2+p n+q}-n }
First, we observe for all   \mathrm{n>|p|+\sqrt{|q|}}   the following inequality holds:
\mathrm{0 \leqslant-n+\sqrt{n^2+p n+q}
\mathrm{ \leqslant-n+\sqrt{\left(n+\frac{p}{2}\right)^2-\frac{\Delta_n}{4}} }\mathrm{ \leqslant-n+\sqrt{\left(n+\frac{p}{2}\right)^2-\frac{\Delta_n}{4}} }
\leqslant \frac{|p|}{2}

Then, define the function    \mathrm{\omega: \mathbb{R}_{>|p|+\sqrt{|q|}} \longrightarrow \mathbb{R}_{\geqslant 0},} where,  \mathrm{\omega(n):=\sqrt{n^2+p n+q}-n}


\mathrm{ \sqrt{n^2+p n+q}=\omega(n)+n }


\mathrm{\Longrightarrow p n+q=\omega^2(n)+2 n \omega(n) }\mathrm{ \sqrt{n^2+p n+q}=\omega(n)+n }


\mathrm{\Longrightarrow \frac{\omega^2(n)-q}{n}=p-2 \omega(n) }\mathrm{\Longrightarrow \frac{\omega^2(n)-q}{n}=p-2 \omega(n) }

Since

 \mathrm{lim _{n \rightarrow \infty} \frac{\omega^9(n)-q}{n}=0} ,

this leads to:
\mathrm{ \lim _{n \rightarrow \infty}(p-2 \omega(n))=0 }
\mathrm{\Longrightarrow \lim _{n \rightarrow \infty} \omega(n)=\frac{p}{2}=\frac{a+b}{2} }

Posted by

Sumit Saini

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