Prove: from the definition of the limit if $a_n \to L$ then $|a_n| \to |L|$

What is the strategy for a proof of this nature? There are similar proof's of the type "given a sequence converges to $L$ show that some transform on the sequence implies the transform on the limit." Another example is proving: $a_n \to L$ implies $\dfrac{1}{a_n} \to \dfrac{1}{L}$.

  1. How do I begin with these proofs
  2. What is the overall 'proof strategy'
  3. Thanks

1) Limits are about control. In consideration of the definition of the limit, you are trying to control a certain quantity ($| |a_n| - |L| |$), by imposing suitable constraints (choosing $n$ large enough), and using known data ($a_n \to L$ .)

2) Recall the triangle inequality, valid for all pairs of real numbers $x,y$:

$|x+y| \le |x| + |y| $

An easy corollary is that $|x| - |y| \le |x-y|$, from which you can easily deduce that $||x| - |y| | \le |x-y|$ (*) Now consider how this inequality applies to your problem and lets you control the quantity of interest.

In the second problem, note that you ultimately need to figure out how to control $|\frac{1}{L} - \frac{1}{a_n}| = | \frac{a_n - L }{a_nL}|$ (note that the statement only makes sense for $L \ne 0.$ ) Since you know how to make the numerator small with the data given, you just need to make $|a_n L|$ large enough. Think about how you can use the definition of convergence and (*) to do this (choose the right $\epsilon .$)

3) You're welcome


The idea is to use the fact that $|a_n-L|$ can be made arbitrarily small to show $||a_n|-|L||$ can be made arbitrarily small. In this case, you can use that for each $x,y$, $$||x|-|y||\leq |x-y|$$

The second case is not always true: you need $L\neq 0$. Now note

$$\left| {\frac{1}{{{a_n}}} - \frac{1}{L}} \right| = \left| {\frac{{L - {a_n}}}{{{a_n}L}}} \right| = \frac{{\left| {{a_n} - L} \right|}}{{\left| {{a_n}L} \right|}}$$

Since $a_n\to L$, there is an $n_0$ such that $|a_n-L|<|L|/2$ whenever $n\geq n_0$. Then

$$|L|-|a_n|\leq |a_n-L|<|L|/2$$

so that $|a_n|>|L|/2$. Again, since $a_n\to L$, there is a $n_1$ for which

$$|a_n-L|<\epsilon |L|^2/2$$ whenever $n\geq n_1$. Then, for $n\geq \max\{n_0,n_1\}$, we must have

$$\left| {\frac{1}{{{a_n}}} - \frac{1}{L}} \right| = \frac{{\left| {{a_n} - L} \right|}}{{\left| {{a_n}L} \right|}} < \frac{2}{{\left| L \right|}}\frac{{\left| {{a_n} - L} \right|}}{{\left| L \right|}} < \frac{2}{{\left| L \right|}}\frac{{\varepsilon {{\left| L \right|}^2}}}{{2\left| L \right|}} = \varepsilon $$

Note there is some unusual steps involved, but you get the idea.

  • $\begingroup$ Also, for a more general case, it might be useful to mention that if $a_n \to L$, and $f$ is a function that is continuous in a neighborhood of $L$, then $f(a_n) \to f(L)$. $\endgroup$ – Joel Cohen Oct 1 '12 at 1:31
  • $\begingroup$ @JoelCohen Well, if that is related to $f(t)=t^{-1}$, then yes, but the proof is rather long. $\endgroup$ – Pedro Tamaroff Oct 1 '12 at 1:38
  • 1
    $\begingroup$ Your proof is perfectly fine. I just mentioned the general case as an aside because it covers a lot of cases (including $f(x) = |x|$ and $f(x) = \frac{1}{x}$) and isn't particularly difficult : Pick $\delta > 0$ such that $|x - L| \le \delta \Rightarrow |f(x) - f(L)| \le \epsilon$ (we can because $f$ is continuous at $L$). Pick $n_0 \ge 0$ such that for all $n \ge n_0$, we have $|a_n - L| \le \delta$ (we can because $a_n \to L$). Then for all $n \ge n_0$, we have $|f(a_n) - f(L)| \le \epsilon$. $\endgroup$ – Joel Cohen Oct 1 '12 at 2:46
  • $\begingroup$ @joel Yes. I don't know why I was thinking about a completely different result relating functions and limits. $\endgroup$ – Pedro Tamaroff Oct 1 '12 at 5:09
  • $\begingroup$ As a sidenote, proving $f(t)$ is cont. at $a$ is equivalent to proving $f(a_n)\to f(a)$ for every sequence that converges to $a$. $\endgroup$ – Pedro Tamaroff Oct 1 '12 at 5:15

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