Find the limit of $\sqrt[n]{n^2+2n+3}$ using L'Hôpitals rule. Using the squeeze theorem $\lim\limits_{n \rightarrow \infty} (\sqrt[n]{n^2+2n+3}) = 1$:
$\sqrt[n]{3} \le \sqrt[n]{n^2+2n+3} \le \sqrt[n]{n}^2 \cdot \sqrt[n]{n} \cdot \sqrt[n]{2} \cdot \sqrt[n]{3}$
Now I'd like to see whether I could also use L'Hôpital here, but I can't figure how that'd work.
I've found one similar question with an answer that I don't understand (first one), that may help: limit of n-th root of polynomial
 A: HINT: write your term in the form
$$\exp\left(\frac{\ln(n^2+2n+3)}{n}\right)$$
A: Assuming $a_n>0$ for simplicity, we have
$$
\lim \sqrt[n]{a_n} = \lim \frac{a_{n+1}}{a_n}
$$
when both limits exist.
For $a_n = n^2+2n+3$ we have
$$
\lim \frac{a_{n+1}}{a_n} = 1
$$
Although this proof does not use L'Hôpital, it uses simpler results.
A: Hint: $X = (n^2 + 2n +3)^{1/n}$, $y = \ln(X) = \frac{\ln(n^2+2n+3)}{n}$.
Apply L'Hospital rule to find $\lim_{n\rightarrow \infty} y = 0$. Consequently, $\lim_{n\rightarrow\infty}x = e^0 =1$.
A: You know that to apply the L'Hospital rule you must have a rational function (you can take log or express as an exponent of $e$ given in previous answers). However, not always L'Hospital rule is efficient. It is easier to use comparison. You have already noted:
$$\lim_\limits{n\to\infty} \sqrt[n]{n^2}=1.$$
Now note:
$$\lim_\limits{n\to\infty} \frac{\sqrt[n]{n^2+2n+3}}{\sqrt[n]{n^2}}=\lim_\limits{n\to\infty} \left(1+\frac{2}{n}+\frac{3}{n^2}\right)^{\frac{1}{n}}=1.$$
Hence the original limit is also $1$.
A: Since
$\sqrt[n]{n}
\to 1$,
$\sqrt[n]{n^m}
=(\sqrt[n]{n})^m
\to 1$
for any $m$ > 0$.
Therefore
$\sqrt[n]{|P(n)|}
\to 1$
for any polynomial
$P(n)$.
