Test $f_n = (1+\frac{1}{n^2})^n$ for convergence and give its limit if it exists.

Bernoulli's inequality is a very powerful tool as seen below.

We have via Bernoulli's inequality for $n>1$ $$1-\frac{1}{n^3}\leq \left(1-\frac{1}{n^4}\right)^{n}\leq 1\tag{1}$$ and hence by Squeeze Theorem we have $$\lim_{n\to\infty} \left(1-\frac{1}{n^4}\right)^{n}=1\tag{2}$$ In exactly the same manner we can show that $$\lim_{n\to\infty} \left(1-\frac{1}{n^2}\right)^{n}=1\tag{3}$$ Dividing equation $(2)$ by equation $(3)$ we get $$\lim_{n\to\infty} \left(1+\frac{1}{n^{2}}\right)^{n}=1\tag{4}$$ The same technique can be used to prove the general result that $(1\pm n^{p}) ^{n} \to 1$ as $n\to\infty$ if $p<-1$ and one observes that the above makes no use of symbols like $e, \exp, \log$.

$$\left(1+\frac{1}{n^2}\right)^n\geq 1+\frac{1}{n}\quad\text{by Bernoulli's inequality or the binomial theorem}$$ $$ \left(1+\frac{1}{n^2}\right)^n\leq \exp\frac{n}{n^2}\leq \frac{1}{1-\frac{1}{n}}\quad\text{by }e^x\geq x+1\text{ and }e^x\leq\frac{1}{1-x}\text{ for }x\in(0,1) $$ so the limit is $1$ by squeezing.

$$\ln f_n=n\ln\left(1+\frac1{n^2}\right)=n\left(\frac1{n^2}+O(n^{-4})\right)=\frac1n+O(n^{-3})\to0$$ as $n\to\infty$. So $f_n\to1$.

You have $$ f_n=\exp\left[n\log\left (1+\frac1 {n^2}\right)\right]= \exp\left[\frac 1n+o \left(\frac1 {n^3}\right)\right]\to e^0=1 $$ as $n \to \infty$.

The expression equals

$$\left [\left(1+\frac{1}{n^2}\right)^{n^2}\right]^{1/n}$$

Inside the brackets the limit is $e.$ Thus for large $n,$

$$2 < \left(1+\frac{1}{n^2}\right)^{n^2}<3.$$

For such $n,$

$$2^{1/n} < \left[\left(1+\frac{1}{n^2}\right)^{n^2}\right]^{1/n}<3^{1/n}.$$

Since the left side and the right side both $\to 1,$ our limit is $1$ by the squeeze theorem.