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In a proof.

Prove that given:

$$\lim_{x \to a} f(x) = L$$ then

$$\lim_{x\to a} |f(x)| = |L|$$

We know that

$$|f(x) - L| < \epsilon \space \text{for} \space |x - a| < \delta_1$$

What is the objective then?

Do we prove there exists a $\delta_2$ such that $\displaystyle \lim_{x\to a} |f(x)| = |L|$

Or do we go from the fact thatit is true that: $|f(x) - L| < \epsilon \space \text{for} \space |x - a| < \delta_1$ and then somehow using $\delta_1$ derive that:

$| |f(x)| - |L| | < \epsilon$

Or do we find a $\delta$?

Thanks!

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  • $\begingroup$ Related. $\endgroup$ – Git Gud Dec 2 '14 at 18:40
  • $\begingroup$ Hint: $|x| = |(x-y) + y| \leq |x-y| + |y| \Rightarrow |x| - |y| \leq |x-y|$. Similarly, $|y| - |x| \leq |x-y|$. Hence $$||x| - |y|| \leq |x-y|$$ That inequality will be helpful. $\endgroup$ – Simon S Dec 2 '14 at 18:43
  • $\begingroup$ I do not want the proof, I just want to know the basics. Do we assume there is a $\delta$ already? $\endgroup$ – Amad27 Dec 2 '14 at 18:43
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Since you are given $$\lim_{x \to a} f(x) = L$$ then you already know anytime $|x-a|<\delta_1$ that $|f(x)-L|<\varepsilon$. By the reverse triangle inequality you also have $$\left||f(x)|-|L|\right|\leq |f(x)-L|$$ which means anytime $$|x-a|<\delta_1 \quad \text{then} \quad \left||f(x)|-|L|\right|<\varepsilon \\ \implies \lim_{x \to a}|f(x)|=|L|$$ so you don't need to find a $\delta_2$, as $\delta_1$ suffices.

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  • $\begingroup$ So we dont find a $\delta$ but when we prove that such and such is true, we use the original $\delta$ like the one you used as $\delta_1$? $\endgroup$ – Amad27 Dec 3 '14 at 18:14
  • $\begingroup$ You don't need to define something like $\delta = \min\{\delta_1, \delta_2\}$ if that's what you mean. We can just drop the subscript and refer to $\delta_1$ as $\delta$, since there is only one delta necessary in this proof. You don't need a second delta because of the reverse triangle inequality. We have $$|x-a|<\delta \\ \implies |f(x)-L|<\varepsilon \\ \implies ||f(x)|-|L||<\varepsilon$$ so the $\delta$ we already have is perfectly fine to use. $\endgroup$ – graydad Dec 3 '14 at 18:58
  • $\begingroup$ That is my point. you use the ORIGINAL GIVEN $\delta$ correct? You dont define a new one. $\endgroup$ – Amad27 Dec 4 '14 at 7:59
  • $\begingroup$ Yes. A second one is not necessary. $\endgroup$ – graydad Dec 4 '14 at 8:06
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You need to show that $\forall \epsilon >0, \ \exists \delta_2 > 0$, such that $$ | |f(x)| - |L|| \ < \epsilon \quad \forall |x-a| < \delta_2 $$

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  • $\begingroup$ Hi! But what is $\delta_2$? $\endgroup$ – Amad27 Dec 2 '14 at 18:42
  • $\begingroup$ @Amad27 Generally, to prove such things, you fix some arbitrary $\epsilon > 0$ and find the value of $\delta_2$, such that for any $x \in (a-\delta_2,a+\delta_2)$ you will have the desired inequality $||f(x)|-|L|| < \epsilon$. $\endgroup$ – gt6989b Dec 2 '14 at 18:45
  • $\begingroup$ So the ultimate goal is finding a $\delta$? which fits along with the inequality being $< \epsilon$? $\endgroup$ – Amad27 Dec 2 '14 at 18:47
  • $\begingroup$ Like using the reverse inequality we find that: $| |f(x)| - |L| | < |f(x) - L| < \epsilon$ but we cannot find a $\delta_2$? Or do we say $\delta_1 = \delta_2$ because the inequality works using $|f(x) - L|$? $\endgroup$ – Amad27 Dec 2 '14 at 18:49
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You always need to define the $\delta$ with respect to a given $\epsilon$, and you need to relate to $\delta_1$ as well.

In this case, the relationship should be easy once you notice the following fact: $$||a| - |b|| \leq |a - b| \quad \forall a,b\in \mathbb{R}$$

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By assumption the function $g$ obtained from $f$ by putting $g(x):=f(x)$ when $x\ne a$ and $g(a):=L$ is continuous at $a$. Therefore the function $|g|:={\rm abs}\circ g$ is continuous at $a$ as well.

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