Computation of the succession $\lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right)$ For training I have decided to solve this limit of a succession
$$\lim\limits_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right).$$
My first attempt:
\begin{split}
\lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right) =\\
&=\lim_{n\to \infty } 3^{n+1}\left(1-\frac{3^{n\cdot \sqrt{1-\frac{1}{n^2}}}}{3\cdot 3^{n}}\right)\\
&=\ldots 
\end{split}
I have abandoned this approach because I wanted to use a notable limit (if $\{b_n\}\to 0$) that could probably be useful; but seeing the rounded brackets I have thought that I have occurred many times:
$$\lim_{n\to \infty} \frac{a^{b_n}-1}{b_n}$$
My second attempt:
\begin{split}
\lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right) =\\
&=\lim_{n\to \infty } 3^{\sqrt{n^2-1}}\left(\frac{3^{n+1}}{3^{\sqrt{n^2-1}}}-1\right)\\
&=\ldots 
\end{split}
Considering only
$$\frac{3^{n+1}}{3^{\sqrt{n^2-1}}}=3^{(n+1-\sqrt{n^2-1})}$$
Taking only the exponent I have:
$$n+1-\sqrt{n^2-1}=\frac{((n+1)-\sqrt{n^2-1})\cdot ((n+1)+\sqrt{n^2-1})}{(n+1)+\sqrt{n^2-1}}$$
$$=\frac{2n+2}{n+1+\sqrt{n^2+1}} \tag 1$$
If I take the limit of the $(1)$, easily:
$$\lim_{n\to \infty}\frac{2n+2}{n+1+\sqrt{n^2+1}}=1$$
i.e.
$$\lim_{n\to \infty}3^{(n+1-\sqrt{n^2-1})}=3$$
Definitely I think that this limit is $\infty\cdot 2=+\infty$.
I think that is correct. Do you users think there are better alternatives? Thanks in advance.

For $n>1$ you have
$$3^{n+1}-3^{\sqrt{n^2-1}}>3^{n+1}-3^{\sqrt{n^2}}=3^{n+1}-3^n=2\cdot3^n.$$
