Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
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

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.