# Limits of sequence $(1 + \frac{3}{n^2})^{n^2}$ as n tends to infinity

I need to find $\lim_{n \to \infty}$ $(1 + \frac{3}{n^2})^{n^2}$ and I've been given the following:

$\lim_{n \to \infty}$ $n^{1/n}$ = 1, $\lim_{n \to \infty}$ $a^{1/n}$ = 1 and $\lim_{n \to \infty}$ $(1 + \frac{1}{n})^{n}$ = e.

My first thoughts were to use the 3rd limit so $(1 + \frac{3}{n^2})^{n^2}$ <= 3e$^{n}$ and then using the squeeze theorem to show as n tends to infinity the sequence is null, but I think I'm missing something out.

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Hint:Let $N=\frac{n^2}{3}$,then try to use $\lim(1 + \frac{1}{n})^{n}=e$ – 89085731 Apr 13 '12 at 12:46
@Gingerjin so ${e^(n^{2})/3}$ and as n tends to infinity e = 1? – franky Apr 13 '12 at 13:01

Hint: $$\Bigl(1+\textstyle{3\over n^2}\Bigr)^{n^2}=\Bigl(\Bigl(1+{1\over n^3/3}\Bigr)^{n^2/3}\Bigr)^3$$ Note that $n^2/3\rightarrow\infty$ as $n\rightarrow\infty$.
You might also need to show that, for $x$ a real variable $$\tag{1} \lim_{x\rightarrow\infty} (1+\textstyle{1\over x})^x =e.$$ One way to show this is the following: for $x>1$, $$\textstyle \bigl( 1+{1\over x}\bigr)^x \le \bigl(1+{1\over \lfloor x\rfloor} \bigr)^{\lceil x\rceil}= \bigl(1+{1\over \lfloor x\rfloor} \bigr)^{\lfloor x\rfloor +1}= \bigl(1+{1\over \lfloor x\rfloor} \bigr)^{\lfloor x\rfloor } \bigl(1+{1\over \lfloor x\rfloor} \bigr)^{1}$$ and $$\textstyle \bigl( 1+{1\over x}\bigr)^x \ge \bigl(1+{1\over \lceil x\rceil} \bigr)^{\lfloor x\rfloor} =\bigl(1+{1\over \lceil x\rceil} \bigr)^{\lceil x\rceil -1} =\bigl(1+{1\over \lceil x\rceil} \bigr)^{\lceil x\rceil } \bigl(1+{1\over \lceil x\rceil} \bigr)^{-1}.$$ Apply the Squeeze Theorem to show that $(1)$ holds.