Can we prove $\lim\limits_{n\to \infty} \sqrt[n]{n^2+n}=1$ using Theorem 1 from Ch. 20 of Spivak's Calculus? In Spivak's Calculus there is the following theorem (I am paraphrasing)

Ch. 22, Theorem 1 If a function $f$ is defined in an open interval containing $c$ except perhaps at $c$, with $$\lim\limits_{x\to
> c}f(x)=l\tag{1}$$ then
For every sequence ${a_n}$ such that

*

*each $a_n$ is in the domain of $f$

*each $a_n\neq 0$

*$\lim\limits_{n\to\infty} a_n=c$
the sequence $f(a_n)$ satisfies
$$\lim\limits_{n\to\infty} f(a_n)=l\tag{2}$$
Conversely, if for every sequence $a_n$ satisfying the three
conditions above there is a sequence $f(a_n)$ satisfying $(2)$, then
$(1)$ is true.

Can apply this theorem directly to prove that
$$\lim\limits_{n\to \infty} \sqrt[n]{n^2+n}=1$$
?
Here is the proposed argument
$$\sqrt[n]{n^2+n}=e^{\frac{1}{n}\log{(n^2+n)}}$$
$$a_n=\frac{\log{(n^2+n)}}{n}$$
$$\lim\limits_{n\to \infty} a_n=0$$
Let $f(x)=e^x$.
Then $\lim\limits_{x\to 0} f(x)=1$.
Since each $a_n$ is in the domain of $f$, which is all numbers, each $a_n\neq 0$ and $$\lim\limits_{n\to \infty} a_n=0$$, we can infer that the sequence $f(a_n)$ satisfies
$$\lim\limits_{x\to\infty} f(a_n)=\lim\limits_{n\to \infty} \sqrt[n]{n^2+n}=1$$
 A: Without the use of theorems then:
\begin{align}
\sqrt[n]{n^2 + n} &= e^{\ln(n^2 + n)/n} = 1 + \frac{\ln(n^2 + n)}{n} + \frac{\ln^{2}(n^2 + n)}{2 \, n^2} + \mathcal{O}\left( \frac{\ln^3(n)}{n^3} \right) \\
&= 1 + \frac{2 \, \ln(n)}{n} + \frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right) + \mathcal{O}\left(\frac{1}{n^2}\right)
\end{align}
An alternate view is:
\begin{align}
\sqrt[n]{n^2 + n} &= e^{\ln(n^2 + n)/n} = e^{\ln\left(1 + \frac{1}{n}\right)/n} \times e^{2 \, \ln n/n} \\
&= e^{\ln\left(1 + \frac{1}{n}\right)/n} \, \left( 1 + \frac{2 \, \ln n}{n} + \frac{2^2 \, \ln^2 n}{2 \, n^2} + \mathcal{O}\left(\frac{\ln^3 n}{n^3}\right) \right)
\end{align}
Since both $\ln\left(1 + \frac{1}{n}\right)$ and $\frac{1}{n}$ tend to zero as $n \to \infty$ then $\text{exp}\left( \frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right)\right) \to 1$ as $n \to \infty$. This leaves
$$ \lim_{n \to \infty} \sqrt[n]{n^2 + n} = \lim_{n \to \infty} \left( 1 + \frac{2 \, \ln n}{n} + \frac{2^2 \, \ln^2 n}{2 \, n^2} + \mathcal{O}\left(\frac{\ln^3 n}{n^3}\right) \right). $$
Since $ \ln n < n$ then $\frac{\ln n}{n} \to 0$ as $n \to \infty$ then
$$ \lim_{n \to \infty} \sqrt[n]{n^2 + n} = 1.$$
Otherwise the problem proposer has provided a solution by use of a theorem which produces the same desired result.
