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I have the following hypotheses: $\alpha_n \to 0, \beta_n \to 0, \alpha_n < 0 < \beta_n$. I need to show that the following two limits exist so that I may add them:

$$\lim \limits_{n \to \infty} \frac{-\alpha_n}{\beta_n -\alpha_n}, \lim \limits_{n \to \infty} \frac{\beta_n}{\beta_n - \alpha_n}$$

Well, I know that each limit is between $0$ and $1$. But I'm not sure how to show that these limits actually converge and don't fluctuate? (Or, it's also a possibility that I don't need the fact that the limits exist to add them, but I'm not sure if that's true)

This question is motivated by Exercise 5.19 in Baby Rudin.

Edit: So either I'm doing something wrong by wanting to add the limits, or we don't need existence of the limits because I found a counter example: take $\beta_n = \frac{1}{n} \cos^2 n, \alpha_n = - \frac{1}{n} \sin^2 n$. Then $ \beta_n < 0 < \alpha_n$ (strict because $\pi$ is irrational), and the quotients do not converge (they are equal to $\cos^2 n$ and $\sin^2 n$).

Thus, here is my whole solution so someone can point out where I'm going wrong. I need to prove that for $D_n = \frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n}$, $f$ defined in $(-1,1)$, $f'$ exists at $0$, and $-1 < \alpha_n < 0 < \beta_n < 1$, $\alpha_n, \beta_n \to 0$, we have $\lim D_n = f'(0)$.

\begin{align} \lim D_n &= \lim \frac{f(\beta_n) - f(0) + f(0) - f(\alpha_n)}{\beta_n - \alpha_n}\\ &= \lim \frac{f(\beta_n) - f(0)}{\beta_n - \alpha_n} + \lim \frac{f(0) - f(\alpha_n)}{\beta_n - \alpha_n}\\ &= \lim \frac{f(\beta_n) - f(0)}{\beta_n - 0} \lim \frac{\beta_n}{\beta_n - \alpha_n} + \lim \frac{f(0) - f(\alpha_n)}{0 - \alpha_n} \lim \frac{-\alpha_n}{\beta_n - \alpha_n}\\ &= \lim f'(0) \lim \frac{\beta_n - \alpha_n}{\beta_n - \alpha_n} = f'(0) \end{align}

MT_
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3 Answers3

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If $\alpha_n = -1/n$ and $\beta_n = \left\{ \begin{array}{ll}1/n & n \textrm{ even} \\ 1/2^{n} & n \textrm{ odd} \end{array}\right.$, then we see that $$\dfrac{-\alpha_n}{\beta_n - \alpha_n} = \left\{ \begin{array}{ll}\frac{1}{2} & n \textrm{ even} \\[5pt] \frac{2^n}{n+2^n} & n \textrm{ odd} \end{array}\right.$$

This has two convergent subsequences with different limits, so the limit does not necessarily exist in general.

Rolf Hoyer
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  • Yeah, I updated my post with a different counterexample. So now I don't know what I'm doing wrong in this problem - I'm going to update the post with my entire solution to see where I'm going wrong. – MT_ Apr 09 '15 at 18:07
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Some ideas: 1. If $a_n, b_n \to L$ and $t_n$ is any sequence in $[0,1],$ then $t_na_n+(1-t_n)b_n \to L.$

2. $$D_n = \frac{\beta_n}{\beta_n-\alpha_n}\cdot \frac{f(\beta_n) - f(0)}{\beta_n}+\frac{-\alpha_n}{\beta_n-\alpha_n}\cdot\frac{f(0)-f(a_n)}{0-a_n}.$$

zhw.
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  • The "1." is exactly what I need. Do you have a proof? – MT_ Apr 09 '15 at 18:41
  • I do, but it's better if you wrestle with it for a while. – zhw. Apr 09 '15 at 18:45
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    Fair enough. It makes sense intuitively, and it shouldn't be too hard to prove. – MT_ Apr 09 '15 at 18:47
  • Is it necessary that $t_n$ in $[0,1]$, or just that it is bounded? Or is bounded even necessary? – MT_ Apr 09 '15 at 18:48
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    If we are only given $t_n$ bounded, I have shown that the statement still holds. Bounded, however, is a necessary condition. – MT_ Apr 09 '15 at 19:01
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Answering my own question: I should have just not split up the limit, and used the following fact:

If $a_n, b_n \to L$ and $t_n$ is bounded, then $t_n a_n + (1 - t_n) b_n \to L$.

Proof. Denote $M \geq |t_n|$. The main idea is to use the following inequality:

$|t_n a_n + (1-t_n) b_n - L| \leq |t_n||a_n - L| + |1 - t_n||b_n - L| \leq M |a_n - L| + (1+M)|b_n - L| < \varepsilon$

for $n > N$, with $N$ chosen sufficiently large.

MT_
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