Two similar limit problems

Question

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

1. $((4^{10}+2^{n})^{\frac{1}{n}})$
2. $((3n^{2}+n)^{\frac{1}{n}})$

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.

Answer 1

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.

Answer 2

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$$\lim_{n\to\infty}(3n^2)^{1/n}=\lim_{n\to\infty}(4n^2)^{1/n}=1.$$