Check the limits using Taylor expansion. I want to check the below limit using Taylor expansion.
$\displaystyle{\lim_{n\rightarrow +\infty}\frac{1}{n}\sqrt[n]{n^n-1}}$
For that how many terms do we have to consider at the expansion?
Also do we expand each term separately? I mean do we expand the whole function $\frac{1}{n}\sqrt[n]{n^n-1}$ ? Or do we expand $\frac{1}{n}$ and $\sqrt[n]{n^n-1}$ separately?
$$a_n=\frac{1}{n}\sqrt[n]{n^n-1}\implies na_n=\sqrt[n]{n^n-1}\implies \log(n a_n)=\frac 1 n\log(n^n-1)$$
$$\log(n^n-1)=\log(n^n)+\log \left(1-n^{-n}\right)=n \log(n)+\log \left(1-n^{-n}\right)$$
$$\log \left(1-n^{-n}\right)\sim -\frac 1{n^n}$$ which is very small.
So, $$\log(na_n) \sim \frac 1 n \times n \log(n)=\log(n)\implies n a_n=n-\frac 1{n^n}$$ which makes $$a_n \sim 1-\frac 1 {n^{n+1}}$$
Using Taylor series, you have
$$\log \left(1-n^{-n}\right)=-n^{-n}-\frac{1}{2}n^{-2 n}-\frac{1}{3} n^{-3 n}+\cdots$$  Computing for small values of $n$
$$\left(
\begin{array}{ccc}
n &amp;  1-\frac 1 {n^{n+1}} &amp; a_n \\
 2 &amp; 0.875000000000000 &amp; 0.866025403784439 \\
 3 &amp; 0.987654320987654 &amp; 0.987498689469124 \\
 4 &amp; 0.999023437500000 &amp; 0.999022003720117 \\
 5 &amp; 0.999936000000000 &amp; 0.999935991806427 \\
 6 &amp; 0.999996427754915 &amp; 0.999996427723013 \\
 7 &amp; 0.999999826533474 &amp; 0.999999826533384 \\
 8 &amp; 0.999999992549419 &amp; 0.999999992549419 
\end{array}
\right)$$
We can use that
$$\frac{1}{n}\sqrt[n]{n^n-1}=\frac{1}{n}\sqrt[n]{n^n}\sqrt[n]{1-\frac1{n^n}}=\sqrt[n]{1-\frac1{n^n}}=e^{\frac{\log\left(1-\frac1{n^n}\right)}{n}} \to 1$$
indeed by standard limits
$$\frac{\log\left(1-\frac1{n^n}\right)}{n}=\frac{\log\left(1-\frac1{n^n}\right)}{\frac1{n^n}}\frac1{n^{n+1}} \to -1\cdot 0=0$$