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I want to solve the following problem:

Find $B \in \mathbb{R}$ such that

$$\lim_{n \rightarrow \infty} \left( \frac{2}{\sqrt{n^2 + 4n} - n} \right) ^{B(n+2)} \in \left] \frac{1}{2}, 2 \right[ \quad.$$

My attempt is as follows:

$$ \left( \frac{2}{\sqrt{x^2 + 4x} - x} \right) ^{B(x+2)} = \frac{2^{B(x+2)}}{2^{B(x+2)\log_{2} \left( \sqrt{x² + 4x} - x \right)}} = 2^{B(x+2) \left(1- \log_{2} \left( \sqrt{x² + 4x} - x \right) \right)}.$$

Thus,

$$\lim_{n \rightarrow \infty} \left( \frac{2}{\sqrt{n^2 + 4n} - n} \right) ^{B(n+2)} = 2^{B \lim_{x \rightarrow \infty}(x+2) \left(1- \log_{2} \left( \sqrt{x² + 4x} - x \right) \right)} $$

and I'm left with

$$\lim_{x \rightarrow \infty}(x+2) \left(1- \log_{2} \left( \sqrt{x² + 4x} - x \right) \right) = \lim_{x \rightarrow \infty} \frac{1- \log_{2} \left( \sqrt{x² + 4x} - x \right)}{\frac{1}{x+2}}$$

that I can compute using L'Hôpital's rule and really boring computations, arriving at

$$\tag{*} \lim_{x \rightarrow \infty} \frac{1- \log_{2} \left( \sqrt{x² + 4x} - x \right)}{\frac{1}{x+2}} = \frac{1}{\log(2)}.$$

From here I'd compute a valid interval for $B$.

Question: Computing (*) is a horrible hassle. Is there an easier way to solve this problem?

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  • $\begingroup$ Is B a real number ? $\endgroup$ – Rene Schipperus Nov 30 '17 at 0:04
  • $\begingroup$ @ReneSchipperus, yes. I forgot to mention it, sorry. $\endgroup$ – Maya Nov 30 '17 at 0:09
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$$\frac{2}{\sqrt{n^2+4n}-n}=1+\frac{\sqrt{n^2+4n}-n}{2n}$$ $$=1+\frac{a_n}{n}$$ where $a_n\to 1$

It follows that $$(1+\frac{a_n}{n})^{n+2}\to e.$$ Does this help ?

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  • $\begingroup$ Elegant and simple. Thank you very much. $\endgroup$ – Maya Nov 30 '17 at 0:43

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