What is the limit of this sequence? How does it relate to the exponential function?

Question

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

I particular, I am hoping to find the above to be $1+x^kg(x/n)$ where $g(\cdot)$ is a function with uniformly bounded derivatives.

Edit: I'm trying to prove a result that would be true if $(1+x^2/n^2)^n -1 = x^kO(x/n).$

Answer 1

The limit can be evaluated without appeal to the exponential function. To proceed we first note that

$$\left(1+\frac{x^2}{n^2}\right)^n\left(1-\frac{x^2}{n^2}\right)^n\le 1\tag1$$

Then, rearranging $(1)$, we obtain

$$ \left(1+\frac{x^2}{n^2}\right)^n\le \frac{1}{\left(1-\frac{x^2}{n^2}\right)^n}\tag 2$$

Using Bernoulli's Inequality on the right-hand side of $(2)$ reveals for $n>|x|$

$$1\le \left(1+\frac{x^2}{n^2}\right)^n\le\frac{1}{1-\frac{x^2}{n}}\tag3$$

whereupon applying the squeeze theorem to $(3)$ yields the coveted limit

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

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Now, we can relate this to the exponential function as follows. First, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality, that the exponential function satisfies the inequality

$$ \left(1+\frac xn\right)^n\le e^x \le \frac{1}{1-x} \tag 4$$

for $-n<x<1$.

From $(4)$ we see that

$$\begin{align} \left(1+\frac{x^2}{n^2}\right)^n&=\left(\left(1+\frac{x^2}{n^2}\right)^{n^2}\right)^{1/n}\\\\ &\le e^{x^2/n}\\\\ &\le \frac{1}{1-\frac {x^2}n}\tag 5 \end{align}$$

Note that $$\frac{1}{1-\frac {x^2}n}=1+\frac{x^2}{n}+O\left(\frac{x^4}{n^2}\right)$$

Answer 2

If $x\in\mathbb{R}$ then $$ 0\leq \log\Big(\Big(1+\frac{x^2}{n^2}\Big)^n\Big)=n\log\Big(1+\frac{x^2}{n^2}\Big)\leq n\frac{x^2}{n^2}=\frac{x^2}{n}$$ since $\log(1+t)\leq t$ for $t\geq 0$. Hence $$ \lim_{n\to\infty}\log\Big(\Big(1+\frac{x^2}{n^2}\Big)^n\Big)=0$$ for all $x\in\mathbb{R}$, or $$ \lim_{n\to\infty}\Big(1+\frac{x^2}{n^2}\Big)^n=1$$

Answer 3

Hint: $$1 \leq \left(1 + \frac{x^2}{n^2}\right)^n \leq \left(e^{x^2}\right)^{1/n}$$

Answer 4

$$A_n=\left(1+\frac{x^2}{n^2}\right)^n\implies \log(A_n)=n \log\left(1+\frac{x^2}{n^2}\right)$$ Now, using Taylor series for large values of $n$ $$\log\left(1+\frac{x^2}{n^2}\right)=\frac{x^2}{n^2}-\frac{x^4}{2 n^4}+O\left(\frac{1}{n^6}\right)$$ $$ \log(A_n)=\frac{x^2}{n}-\frac{x^4}{2 n^3}+O\left(\frac{1}{n^6}\right)$$ Taylor again using $$A_n=e^{\log(A_n)}=1+\frac{x^2}{n}+\frac{x^4}{2 n^2}+\frac{x^4 \left(x^2-3\right)}{6 n^3}+\frac{x^6 \left(x^2-12\right)}{24 n^4}+O\left(\frac{1}{n^5}\right)$$ For illustration purposes, let us use $x=10$ and $n=100$ $$A_{100}=\left(\frac{101}{100}\right)^{100}\approx 2.70481$$ while the above expansion would give $$\frac{1619}{600}\approx 2.69833$$

Answer 5

Hint:

It is a well-known fact that

$$\lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n = e$$

See e.g. http://aleph0.clarku.edu/~djoyce/ma122/elimit.pdf

Answer 6

We know that $$e^{x^2} = \lim_{n \rightarrow \infty}\left(1+\frac{x^2}{n}\right)^n = \lim_{n \rightarrow \infty}\left(1+\frac{x^2}{n^2}\right)^{n^2}$$ So $$\lim_{n \rightarrow \infty}\left(1+\frac{x^2}{n^2}\right)^{n} = \lim_{n \rightarrow \infty}\left( \left(1+\frac{x^2}{n^2}\right)^{n^2}\right)^{\frac{1}{n}} = \lim_{n \rightarrow \infty} (e^{x^2})^{\frac{1}{n}} = 1$$

Answer 7

This is a simple application of the following lemma (courtesy Thomas Andrews)

Lemma: If $a_{n} $ is a sequence of real or complex terms such that $n(a_{n} - 1)\to 0$ then $a_{n}^{n} \to 1$.

Now we just need to set $a_{n} =1+(x/n)^{2}$ and our job is done.