# 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!

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Though it's obvious, you should indicate what variable you are taking the limit with respect to, at to what point. – David Mitra Dec 9 '12 at 0:25

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.

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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.

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