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…?
$$\left(1+\left(\frac1\infty\right)^2\right)^\infty = (1+0)^\infty = 1$$

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.
