Using Squeeze Theorem, find limit $(5^{n} + 6^{n})^{\frac{1}{n}}$. Find limit of $(a_{n}) = (5^{n} + 6^{n})^{\frac{1}{n}}$ given $b^{\frac{1}{n}} \to 1, b \in R^{+}$

Is it too cheeky to say let $b = (5^{n} + 6^{n})$ and say by definition $(5^{n} + 6^{n})^{\frac{1}{n}} \to 1$?
As pointed out in the comments, we have $6^n\leq5^n+6^n\leq2\cdot6^n$ so the desired limit is inbetween $\lim_{n\to\infty}\left(6^n\right)^{1/n}=6$ and $\lim_{n\to\infty}\left(2\cdot6^n\right)^{1/n}=6\cdot\lim_{n\to\infty}2^{1/n}=6\cdot1=6$. Both limits equal $6$ so the final answer is $6$ as well.
$$(5^n+6^n)^{1/n}=6\left(1+\left(\frac56\right)^n\right)^{1/n}.$$

Now as

$$1&lt;1+\left(\frac56\right)^n&lt;2,$$ 

$$1&lt;\left(1+\left(\frac56\right)^n\right)^{1/n}&lt;2^{1/n},$$ 
you can use the hint given.