Two similar limit problems I have two limit problems which are quite similar, so I've put them both into this post.


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*$((4^{10}+2^{n})^{\frac{1}{n}})$

*$((3n^{2}+n)^{\frac{1}{n}})$
Attempt:
For 1, I'm not sure if this is permissible. I know that $\lim_{n \to \infty}(x^{n}+y^{n})^{\frac{1}{n}} = \max\{x,y\}$. Obviously $4^{10}$ is a constant so it's not quite in the same form, but I believe it is still okay to use this result, and so the limit is $2$.
For 2, I've got $((3n^{2}+n)^{\frac{1}{n}}) = (n(3n+1)^{\frac{1}{n}})$, but then I'm not too sure how to proceed. 
I also thought of taking the log and proceeding that way, however in the notes I'm learning from they haven't covered l'hopital yet, so I'm trying to not use it.
Thanks.
 A: For the first one we have
$$(4^{10}+2^{n})^{\frac{1}{n}}=2\left(\frac{4^{10}}{2^n}+1\right)^{\frac{1}{n}}\to 2\cdot 1^0 =2$$
and
$$(3n^{2}+n)^{\frac{1}{n}}=e^{\log(3n^{2}+n)^{\frac{1}{n}}}=e^{\frac{\log (3n^2+n) }{n}}\to e^0= 1$$
recall indeed that $\frac{\log n}n \to 0$ and therefore
$$\frac{\log (3n^2+n) }{n}\le \frac{\log (3n^2+6n+3) }{n}=\frac{\log (3(n+1)^2) }{n}=\frac{2\log (n+1)+\log 3 }{n}=$$
$$=2\frac{n+1}n\frac{\log (n+1) }{n+1}+\frac{\log 3}n\to 2 \cdot 1 \cdot 0 + 0 =0$$
As an alternative for the second we can also use that
$$n^\frac1n \le(3n^{2}+n)^{\frac{1}{n}} \le (3n^{2}+n^2)^{\frac{1}{n}}=(4n^{2})^{\frac{1}{n}}$$
and conclude in a similar way by squeeze theorem.
A: Take $\varepsilon>0$. Then $\lim_{n\to\infty}(2+\varepsilon)^n=\infty$ and therefroe $(2+\varepsilon)^n>4^{10}$, if $n\gg1$. But then, if $n$ is large enough,$$(2^n)^{1/n}<(4^{10}+2^n)^{1/n}<\bigl((2+\varepsilon)^n+2^n\bigr)^{1/n}.$$Since $\lim_{n\to\infty}(2^n)^{1/n}=2$ and$$\lim_{n\to\infty}\bigl((2+\varepsilon)^n+2^n\bigr)^{1/n}=\max\{2+\varepsilon,2\}=2+\varepsilon,$$it is easy now to prove that the limit of your sequence is $2$.
For the other sequence, you can use the fact that $3n^2<3n^2+n<4n^2$ and that$$\lim_{n\to\infty}(3n^2)^{1/n}=\lim_{n\to\infty}(4n^2)^{1/n}=1.$$
