Limits and exponentials Asked to find $\lim_{n\to\infty}a_n$ where
$$a_n = \left(1+\dfrac1{n^2}\right)^n$$
I know that the limit = 1, and can get to this by saying $\ln a_n=n\ln\left(1+\dfrac1{n^2}\right)$ and going from there.
My question is: would it also be enough to simply direct substitute and say that…?
$$\left(1+\left(\frac1\infty\right)^2\right)^\infty = (1+0)^\infty = 1$$
 A: $$\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^n=\left[\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^{n^2}\right]^{\left(\lim_{n\to\infty}1/n\right)}$$
Remember $\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$
A: To see this, do as what you did, but use the fact:
$$\ln(1 + \frac{1}{n^2}) \sim_{n \to \infty} \frac{1}{n^2}$$
Then, 
$$n\ln(1 + \frac{1}{n^2}) \sim_{n \to \infty} \frac{1}{n} \longrightarrow 0 \ \ (n \to \infty)$$
Hence the limit is $e^0 = 1$.
AS for the second question, it's not possible to do that. This is because $1^{\infty}$ is an indeterminant form, like $\frac{\infty}{\infty}$ and $\frac00$
A: Your simplification is not correct since, for instance, $\frac{1}{\sqrt{n}}$ tends to $0$ but $(1+\frac{1}{\sqrt{n}})^n \geq 1 + \frac{n}{\sqrt{n}}$ which tends to $+\infty$. Therefore, you need to be more precise.
Example: as $n$ tends to $+\infty$,\begin{align}
\left(1+\frac{1}{n^2}\right)^n &= \exp\left[n\log\left(1+\frac{1}{n^2}\right)\right]\\
& = \exp\left[n\left(\frac{1}{n^2} + o(\frac{1}{n^2})\right)\right]\\
& = \exp\left[\frac{1}{n} + o(\frac{1}{n})\right]\\
& = \exp\left[o(1)\right]\\
& = 1 + o(1).
\end{align}
A: In short no. If you look at the limit that yields $e$ And did the same trick what would you get?
$$
\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e\neq \left(1+\frac{1}{\infty}\right)^\infty = 1
$$
The last equality is more to match with your statement question than being a proper definition of 1. 
A: Here are the steps 
$$\lim\limits_{n\to\infty} \left(1+\frac{1}{n^2}\right)^n $$
$$= \lim\limits_{n\to\infty} \exp\left(\ln\left(1+\frac{1}{n^2}\right)^n\right) $$
$$= \lim\limits_{n\to\infty} \exp\left(n\ln\left(1+\frac{1}{n^2}\right)\right) $$
$$= \exp\left(\lim\limits_{n\to\infty} \left[n\ln\left(1+\frac{1}{n^2}\right)\right]\right) $$
$$= \exp\left(\lim\limits_{n\to\infty} \left[\frac{\ln\left(1+\frac{1}{n^2}\right)}{\frac{1}{n}}\right]\right) $$
Let $h=\frac{1}{n}$. Note that $h$ approaches $0$ as $n$ approaches $\infty$. So now we have
$$= \exp\left(\lim\limits_{h\to 0} \left[\frac{\ln\left(1+h^2\right)}{h}\right]\right) $$
$$= \exp\left(\lim\limits_{h\to 0} \left[\frac{\frac{d}{dh}\left[\ln\left(1+h^2\right)\right]}{\frac{d}{dh}[h]}\right]\right) $$
$$= \exp\left(\lim\limits_{h\to 0} \left[\frac{\frac{1}{1+h^2}\frac{d}{dh}\left[1+h^2\right]}{1}\right]\right) $$
$$= \exp\left(\lim\limits_{h\to 0} \left[\frac{2h}{1+h^2}\right]\right) $$
$$= \exp\left(\frac{0}{1+0}\right) = \exp\left(0\right)=1 $$
Note that you cannot treat infinity as a quantity, for infinity is not quantifiable.
A: This is precisely because $1^\infty$ is an indeterminate form.
See this very relevant question.
User's 'Blue' comment:

"Consider the purpose of the list of indeterminate forms. (Barring pathologies...) The first thing to try in an $x\to a$ limit is to "plug in" $a$ for $x$; if you get an expression that evaluates to $3$ or $\sqrt{\pi}$ or even $-\infty$, you're done.
The "indeterminate forms" are labels (and/or warnings) for cases where there's more work to do. They capture the essence of the problem and guide you to appropriate follow-up strategies ."

Also see the most upvoted and accepted answer there.
