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Consider this limit: $$\lim_{n\rightarrow\infty} \left( 1+\frac{1}{n} \right) ^{n^2} = x$$

I thought the way to solve this for $x$ was to reduce it using the fact that as $n \rightarrow \infty$, $\frac{1}{n} \rightarrow 0$: $$\therefore \lim_{n\rightarrow\infty}(1+0)^{n^2} = x$$

Apparently this is wrong! Why is it wrong?

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By the same logic, $1=\lim_{n\rightarrow\infty}n\frac{1}{n}=\lim_{n\rightarrow\infty}n(0)=0$ which is clearly wrong. – Dan Rust May 13 '13 at 22:29
@DanielRust Clearly. Thanks! – xisk May 13 '13 at 22:31
If it was $n\longrightarrow -\infty$ there would be a finite answer. – fermesomme May 14 '13 at 10:03
up vote 3 down vote accepted

But as $n\rightarrow \infty $ that exponent gets ever larger. A smaller argument, yes, but raised to an ever larger power. Can't assume that the whole thing is going to go to one. Try a few examples on your calculator to see what happens.

For a little more detail, if $n$ is some big number, when you multiply everything out, the first two terms (look up the binomial expansion) will be

$1+ n^2 * \displaystyle\frac{1}{n} = 1+n \rightarrow \infty$

None of the remaining terms will be negative, so you know this is a bottom limit on what you are going to get.

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If your method isn't wrong so we have also

$$\lim_{n\to\infty}\frac{1}{n}\times n=\lim_{n\to\infty}0\times n=0???$$

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You have an indeterminate form: $1^\infty$, so need some finer analysis than just plugging in. We know $\displaystyle \lim_{n\rightarrow\infty} \left( 1+\frac{1}{n} \right) ^{n} = e$ so we might expect yours to diverge as $e^n$

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