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I came across the limit $$ \lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{n^2} $$ and immediately speculated that it is $$ \exp\left(x^2\right) $$ what e.g. Maple spits out perfectly. My idea was that the elementary formula $$ \left(\lim_{n\to\infty}\left(1+\frac{x^2}{n^2}\right)^{n}\right)^n =\exp\left(\frac{x^2}{n}\cdot n\right)=e^{x^2} $$ could be taken for a "proof" but i admit it seems more like a make-belief-argument when one knows the solution already.

How would one argue precisely to get a 100%-proof ? My analysis trickery is not so rich any more.

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  • $\begingroup$ Thanks to all for the clever and illuminating answers. I completely missed the argument with the subsequence; also I like the trick with the rule of L'Hopital as i am a fan of this rule... $\endgroup$ – Wolfgang Tintemann Jun 11 '15 at 20:42
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You only have to notice that $(n^2)$ is a subsequence of $(m)$, the sequence of nonnegative integers. Hence $$ \left( (1+x^2/n^2)^{n^2} \right) $$ is a subsequence of $$ \left( (1+x^2/m)^{m} \right), $$ and since the latter converges to $e^{x^2}$, so must the former.

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Take $n^2 = N$ and observe that

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

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A basic limit is $\,\lim\limits_{n\to\infty}\Bigl(1+\dfrac xn\Bigr)^n=\mathrm e^x$. So $\,\lim\limits_{n\to\infty}\Bigl(1+\dfrac{x^2}n\Bigr)^n=\mathrm e^{x^2}$.

Now $\Bigl(1+\dfrac{x^2}{n^2}\Bigr)^{\!n^2}$ is just a subsequence of the above sequence, hence it converges to the same limit.

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The limit is $e^{x^2}$. Observing that the exponential and logarithms are inverses of each other, we can write: $$ \lim_{n\to\infty}e^{\ln(1+\frac{x^2}{n^2})^{n^2}} =\lim_{n\to\infty}e^{n^2\ln(1+\frac{x^2}{n^2})} =e^{\lim_{n\to\infty}\frac{\ln(1+\frac{x^2}{n^2})}{1/n^2}} =e^{\lim_{n\to\infty}\frac{n^2x^2}{n^2+x^2}} $$ using L'Hopital's rule. Divide through by $n^2$ (the leading term) and take the limit to get the result.

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