If 0 < a < b, where a, b $\in\mathbb{R}$, determine $\lim \bigg(\dfrac{a^{n+1} + b^{n+1}}{a^{n} + b^{n}}\bigg)$ If 0 < a < b, where a, b $\in\mathbb{R}$, determine $\lim \bigg(\dfrac{a^{n+1} + b^{n+1}}{a^{n} + b^{n}}\bigg)$
The answer (from the back of the text) is $\lim \bigg(\dfrac{a^{n+1} + b^{n+1}}{a^{n} + b^{n}}\bigg) = b$ but I have no idea how to get there. The course is Real Analysis 1, so its a course on proofs. This chapter is on limit theorems for sequences and series. The squeeze theorem might be helpful. 
I can prove that $\lim \bigg(\dfrac{a^{n+1} + b^{n+1}}{a^{n} + b^{n}}\bigg) \le b$ but I can't find a way to also prove that it is larger than b
Thank you!
 A: Hint: Divide top and bottom by $b^n$.  
At the top  we get $b+a\left(\dfrac{a}{b}\right)^n$.  
Remarks: $1$.)  Presumably it has already been shown that if $|x|\lt 1$ then $\lim_{n\to\infty} x^n=0$.  If not, let $|x|=1+t$. By the Binomial Theorem, or the Bernoulli Inequality, $(1+t)^n \ge 1+nt$. From this one can prove quickly, even in full $\epsilon$-$N$ style, that $\lim_{n\to \infty}\frac{1}{(1+t)^n}=0$.
$2$) Suppose one does not think of the standard trick that we used. Knowing the answer is then helpful, for then it is natural to look at 
$$\frac{a^{n+1}+b^{n+1}}{a^n+b^n}-b.$$
After some simplification, this becomes 
$\dfrac{a^n-a^{n+1}}{a^n+b^n}$, and now what to do may be clearer.
A: HINT: Use the standard trick from first-year calculus, and divide top and bottom by the biggest thing in the denominator, $b^n$:
$$\lim_{n\to\infty}\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\lim_{n\to\infty}\frac{\dfrac{a^{n+1}}{b^n}+b}{\dfrac{a^n}{b^n}+1}\;.$$
There’s still some work to be done, but it’s easy work.
