Limit of $\sqrt[n]{2^n + 3^n} + \sqrt[n]{n^2 + 1}$ I'm learning calculus, specifically limit of sequences, and need with with the following exercise:

Find the limit of the following sequences
$$(a) \ \ a_n = \sqrt{2^n + 5^n} \quad \quad (b) \ \ b_n = \sqrt[n]{2^n + 3^n} + \sqrt[n]{n^2 + 1}.$$

Since I'm having difficulties for $(b)$, I'm going to show my work for $(a)$.
$(a)$ Using the Squeeze Theorem, we observe that
$$\sqrt{2}(2)^{\frac n2} = \sqrt{2^n + 2^n} \leq a_n \leq \sqrt{5^n + 5^n} = \sqrt{2}(5)^{\frac n2}.$$
Obviously
$$\sqrt{2}\lim_{n \to \infty} (2)^{\frac n2} = +\infty = \sqrt{5}\lim_{n \to \infty} (2)^{\frac n2}$$
so for the original sequence we have
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \sqrt{2^n + 5^n} = +\infty.$$

Is my work/method correct for $(a)$. I wanted to use a similar argument for $(b)$ but I was not able to do it. Is there a clever way to solve $(b)$? I also thought of calculating the limit by multiplying by the conjugate by I think it's not the right method here.
 A: Your work is correct for $a$, although I wouldn't bother bounding it above: the sequence diverges so it suffices to bound it from below by a divergent sequence. 
For b,notice that 
$$
\sqrt[n]{n^2 + 1}\sim \sqrt[n]{n^2}=n^{2/n}\rightarrow 1
$$
and for large enough $n$ 
$$
\sqrt[n]{2^n + 3^n}\sim \sqrt[n]{3^n}
$$
By the property seen here:
Show that $\lim_{n\to\infty} \sqrt[n]{a^{n} + b^{n}} = \max(a, b)$
Or just prove it by the squeeze theorem
$$
3=\sqrt[n]{3^n}\leq \sqrt[n]{2^n + 3^n}\leq \sqrt[n]{3^n + 3^n}=\sqrt[n]{2*3^n}\rightarrow 3
$$
yielding $b_n\rightarrow 1+3=4$
A: $$\sqrt[n]{2^n + 3^n} + \sqrt[n]{n^2 + 1}=3\sqrt[n]{1 + (\frac{2}{3})^n} + n^{\frac{1}{n}}\sqrt[n]{1 + \frac{1}{n^2}}=3\left(1 + (\frac{2}{3})^n\right)^{\frac{1}{n}}+n^{\frac{1}{n}}\left(1 + \frac{1}{n^2}\right)^{\frac{1}{n}}$$
we know that limit of 
$$\lim_{n\rightarrow \infty }n^{\frac{1}{n}}=1$$
see 
$$1 + (\frac{2}{3})^n<n$$
and
$$1 + \frac{1}{n^2}<n$$
so the
$$3\lim_{n\rightarrow \infty }\left(1 + (\frac{2}{3})^n\right)^{\frac{1}{n}}+\lim_{n\rightarrow \infty }n^{\frac{1}{n}}\lim_{n\rightarrow \infty }\left(1 + \frac{1}{n^2}\right)^{\frac{1}{n}}=3(1)+1(1)=4$$
A: By taking logs:
$$\log\sqrt[n]{2^n+3^n}-\log(3)=\frac 1n\log\left(1+\left(\frac 23\right)^n\right)\sim\frac 1n\left(\frac 23\right)^n\to 0$$
Similarly,
$$|\log\sqrt[n]{n^2+1}|\leq\frac 2n\log(n)\to 0.$$
