Evaluate : $\lim_{n\to \infty}e^{-2n}\left(1+{2\over n}\right)^{n^2}$ My attempt:
$$\lim_{n\to \infty}e^{-2n}\left(1+{2\over n}\right)^{n^2}\\ =\lim_{n\to \infty}e^{-2n}. \lim_{n\to \infty}\left(1+{2\over n}\right)^{n^2}\\=\lim_{n\to \infty}e^{-2n}.e^{2n}=1$$ Where did I go wrong? Please do help me to solve the problem. Thanks in advance.
 A: You can't write:
$$\lim f(n)g(n)=\lim f(n)\lim g(n)$$ when $f(n)\to 0$ and $g(n)\to \infty.$
You can't recombine terms $$\lim f(n)\lim h(n)=\lim f(n)h(n),$$ where again, $f(n)\to 0$ and $h(n)\to\infty.$
And you can't substitute $f(n)^n$ with $L^n$ just because $f(n)\to L.$
Basically, every step in this argument is wrong.
If we let $f(n)=e^{-2}(1+2/n)^n,$ then we want to compute $\lim_{n\to\infty} f(n)^n.$
It is true that $f(n)\to 1,$ but that doesn't mean $f(n)^n\to 1,$ which is the heart of your argument, minus the odd splitting of the limit into two limits.
If we could deduce that from $f(n)\to 1,$ we wouldn't be able to get that $(1+2/n)^n\to e^{2},$ since $1+2/n\to 1,$ also.
The problem is that $1^{\infty}$ is an indeterminate form.
What you need to know is how fast $f(n)$ converges to $1.$
In particular, if $n(f(n)-1)\to L,$ then $f(n)^n\to e^{L}.$ See this answer for a proof of that.
A: \begin{align}
&\lim_{n\to \infty}{e}^{-2n}\cdot\left(1+{2\over n}\right)^{n^2}\\
&=\lim_{n\to \infty}\LARGE{e}^{\large{-2n+n^2\ln\left(1+{2\over n}\right)}}\\
&=\lim_{n\to \infty}\LARGE{e}^{\large{-2n+2n}\Large{\frac{\ln\left(1+{2\over n}\right)}{2\over n}}}\\
&=\lim_{n\to \infty}\LARGE{e}^{\large{4}\Large{\frac{\frac{\ln\left(1+{2\over n}\right)}{2\over n}-1}{\frac2n}}{}}\\
&=\lim_{x\to 0}\LARGE{e}^{\large{4}\Large{\frac{\ln\left(1+x\right)-x}{x^2}}{}}\\
\text{(l'Hopital)}&=\lim_{x\to 0}\LARGE{e}^{\large{4}\Large{\frac{\frac1{1+x}-1}{2x}}{}}\\
\text{(l'Hopital)}&=\lim_{x\to 0}\LARGE{e}^{\large{4}\Large{\frac{-\frac1{(1+x)^2}}{2}}{}}\\
&=\LARGE{e}^{-2}
\end{align}

To answer to the OP's question,
$$\left(1+{2\over n}\right)^{n^2}\approx e^{2n-2}\quad,\quad n>>1$$
rather than $e^{2n}$.
In other words,
$$\lim_{n\to \infty} \left(-2n+n^2\ln\left(1+{2\over n}\right)\right)=-2$$
, not $0$.
A: Write $(1+2/n)^{n^2}=e^{n^2\log(1+2/n)}$. Now, use the fact, from a taylor expansion, that $$\log(1+2/n)=\frac{2}{n}-\frac{2}{n^2}+O(1/n^3)$$.
Thus, your expression is
$$e^{-2}+O(1/n).$$
Thus the limit is $e^{-2}$.
A: You cannot approximate
$$
\left(1+\frac{2}{n}\right)^{n^2} \sim e^{2n}
$$
because that doesn't take into account the higher order corrections properly. You can do that using Taylor's series.
Define
$$
f(n) = e^{-2n}\left(1+\frac{2}{n}\right)^{n^2}
$$
and consider
$$
g(n) = \log\left[
f(n)
\right]
=
n^2\log\left(1+\frac{2}{n}\right)-2n\,.
$$
Using the Taylor expansion of $\log(1+t)=t-\frac{t^2}{2}+\mathcal{O}(t^3)$ for small $t$ we have
$$
g(n)
=
2n-2-2n+\mathcal{O}(1/n) \to -2
$$
as $n\to\infty$.
Then
$$
f(n)=e^{g(n)}\to e^{-2}
$$
as $n\to\infty$.
In other words, the correct version of the approximation written at the beginning of the post is
$$
\left(1+\frac{2}{n}\right)^{n^2} 
\sim
e^{2n-2}\,.
$$
A: Let $$L= \lim_{n\to \infty}e^{-2n}\bigg(1+{2\over n}\bigg)^{n^2}$$
The way OP has done it, we have
$$L=\lim_{n\to\infty}e^{-2n}\lim_{n\to \infty} \left(1+\frac{2}{n}\right)^{n^2} \to 0 \lim_{n \to \infty}e^{-2n} \times \lim_{n\to \infty} e^{2n} \to 0 \times \infty \text{(indeterminate)}$$
Let us change it to the form $\infty/\infty$ as
$$L=\lim_{n\to \infty} \frac{\bigg(1+{2\over n}\bigg)^{n^2}
}{e^{2n}}=\lim_{n\to \infty}\frac{\exp\left[n^2\ln\bigg(1+{2\over n}\bigg)\right]}{e^{2n}}$$
Use $\log(1+z)=z-z^2/2+z^3/3..$ to get
$$L=\lim_{n\to \infty}\frac{\exp\left[n^2\bigg({2\over n}-{2\over n^2}+{4\over 3 n^3}+...\bigg)\right]}{e^{2n}}=\lim_{n\to \infty}\frac{\exp\left[2n-2+4/(3n)+...\bigg)\right]}{e^{2n}}=e^{-2}$$
Edit: $\lim_{x \to a} f(x)g(x)= \lim_{x \to a} f(x) \times \lim_{x \to a} g(x)$ holds only if both limits are finite.
A: As a complement to the very fine answer provided by Thomas Andrews, in order to better understand the issue, let also consider the following basic example

*

*$a_n =n\to \infty$

*$b_n =\frac1n \to 0$
then
$$\lim_{n\to \infty} a_n b_n = \lim_{n\to \infty} n\cdot \frac1n=\lim_{n\to \infty} 1=1$$
but
$$\left(\lim_{n\to \infty} a_n\right) \cdot \left(\lim_{n\to \infty} b_n\right) = \:"\infty \cdot 0"$$
is undefined.

You are also using that
$$\lim_{n\to \infty}\left(1+{2\over n}\right)^{n^2}=e^{2n}$$
which is meaningles (since limit value exists finite or infinite or doesn't exist but can't be a function itself), what is true is that
$$\lim_{n\to \infty}\left(1+{2\over n}\right)^{n^2}=\lim_{n\to \infty}e^{2n} =\infty$$

Also refer to the related:

*

*Analyzing limits problem Calculus (tell me where I'm wrong).
