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I trying to calculate $$ \lim_{n\to\infty}\big(\frac{n^2+2}{2n^2+1}\big)^{n^2}.\tag{1}$$ Of course I cannot use theorem that if $a_n\to\infty$ then $\lim_{n\to\infty}(1+\frac{1}{a_n})^{a_n}=e,$ because in this particular situation we have $$\lim_{n\to\infty}\big(\frac{n^2+2}{2n^2+1}\big)^{n^2}=\lim_{n\to\infty}\big(1+\frac{1-n^2}{2n^2+1}\big)^{n^2}=\lim_{n\to\infty}\big(1+\frac{1}{\frac{(2n^2+1)}{(1-n^2)}}\big)^{n^2},$$ and $\lim_{n\to\infty}\frac{(2n^2+1)}{(1-n^2)}=-2.$

I would be grateful of any hints and comments.

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3 Answers 3

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i would write $$\left(\frac{1+\frac{2}{n^2}}{2+\frac{1}{n^2}}\right)^{n^2}$$ and tis tends to zero if $n$ tends to infinity

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$$ \Big( \frac{n^2+2}{2n^2+1} \Big)^{n^2} = \exp\Bigg( n^2 \ln\Big( \frac{n^2+2}{2n^2+1} \Big) \Bigg). $$

Also:

$$ \ln\Big( \frac{n^2+2}{2n^2 + 1} \Big) \; \mathop{\longrightarrow} \limits_{n \to +\infty} \; \ln\Big(\frac{1}{2} \Big) = -\ln(2). $$

Therefore:

$$ n^2 \ln\Big( \frac{n^2+2}{2n^2 +1} \Big) \; \mathop{\longrightarrow} \limits_{n \to +\infty} \; -\infty $$

and

$$ \Big( \frac{n^2+2}{2n^2 + 1} \Big)^{n^2} \; \mathop{\longrightarrow} \limits_{n \to +\infty} \; 0. $$

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Note that: $$\frac{n^2}{2n^2}<\frac{n^2+2}{2n^2+1}<\frac{2n^2}{3n^2},n>1 \Rightarrow $$ $$\lim_{n\to\infty}\big(\frac{n^2}{2n^2}\big)^{n^2}\le \lim_{n\to\infty}\big(\frac{n^2+2}{2n^2+1}\big)^{n^2} \le \lim_{n\to\infty}\big(\frac{2n^2}{3n^2}\big)^{n^2} \Rightarrow$$ $$\lim_{n\to\infty}\big(\frac{n^2+2}{2n^2+1}\big)^{n^2}=0.$$

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