I want to compute the limit $\left(1+\frac{1}{n^2}\right)^n$. One way to do this is to take logs. So $$x_n=\left(1+\frac{1}{n^2}\right)^n$$ Then $$\log x_n = n\log\left(1+\frac{1}{n^2}\right) = n\left(\frac{1}{n^2}-\frac{1}{2n^4}+O\left(\frac{1}{n^6}\right)\right)$$So $\log x_n$ converges to $0$, and $x_n$ converges to $1$.
Is there a simpler way to do this, maybe without Taylor log expansion?
On
Claim: Let $\,\{a_n\}\,$ be a non-negative convergent sequence s.t.
$$\lim_{n\to\infty}a_n=a>0\;,\;\;\text{then}\;\;\lim_{n\to\infty}\sqrt[n]{a_n}=1$$
Proof: there exists $\,N\in\Bbb N\,$ s.t.
$$n>N\implies |a_n-a|<\frac a2\iff 0<\frac a2<a_n<\frac{3a}2\implies$$
$$\sqrt[n]\frac a2\le\sqrt[n]{a_n}\le\sqrt[n]\frac{3a}2$$
and since $\,\displaystyle{\sqrt[n]x\xrightarrow[n\to\infty]{}1\;\;\;\forall\,x>0}\,$ , we're done by the squeeze theorem.$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\square\;$
Thus, applying the above lemma we get that
$$\left(1+\frac1{n^2}\right)^{n^2}\xrightarrow[n\to\infty]{}e>0\implies\left(1+\frac1{n^2}\right)^n=\left[\left(1+\frac1{n^2}\right)^{n^2}\right]^{\frac1n}\xrightarrow[n\to\infty]{}1$$
From the Taylor Series, $$1+\frac{1}{n^2}\leq \exp\left(\frac{1}{n^2}\right)$$
Thus, $$1\leq \left(1+\frac{1}{n^2}\right)^n \leq \exp\left(\frac{1}{n}\right)$$ so by the Squeeze Theorem, the result follows $\square$