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$$\lim_{n\to \infty}\left\{\left(1+\frac{2}{n}\right)^n,\;\sqrt{\frac{n+1}{4n-1}}\right\}$$

$$\lim_{n\to \infty}\left(1+\frac{2}{n}\right)^n=\lim_{n\to \infty}\left(1+\frac{2}{n}\right)^{\frac{n}{2}\cdot 2}$$

$m=\frac{n}{2}$

$$\lim_{n\to \infty}(1+\frac{2}{n})^{\frac{n}{2}\cdot 2}=\lim_{n\to \infty}(1+\frac{1}{m})^{m\cdot 2}=\lim_{n\to \infty}(1+\frac{1}{m})^{{m}^{2}}=e^2$$

$$\lim_{n\to \infty}\sqrt{\frac{n+1}{4n-1}}=\lim_{n\to \infty}\sqrt{\frac{\frac{n}{n}+\frac{1}{n}}{\frac{4n}{n}-\frac{1}{n}}}$$

Because square root is a continuos function

$$\sqrt {lim_{n\to \infty}{\frac{\frac{n}{n}+\frac{1}{n}}{\frac{4n}{n}-\frac{1}{n}}}}=\sqrt {{\frac{1+0}{4-0}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$$

So

$$\lim_{n\to \infty}\{(1+\frac{2}{n})^n,(\sqrt{\frac{n+1}{4n-1}})\}=(e^{2},\frac{1}{2})$$

Are all the moves correct?

DonAntonio
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gbox
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1 Answers1

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Almost everything is correct, just a slight slip in the $e^2$.

$$\lim_{n\to\infty} \left(1+ \frac2n\right)^{\frac n2 \cdot 2} = \left(\lim_{n\to\infty} \left(1+ \frac2n\right)^{\frac n2}\right)^2 \\ =\left(\lim_{m\to\infty} \left(1+ \frac1m\right)^m\right)^2 \\ = e^2$$

B. Mehta
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