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From an old exam:

The not-necessary-continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ is given. We know that $|f(x)| < 1$ for all $x \in \mathbb{R}$.

Let $a(x) = (x-3)f(x)$ and $b(x) = (x-3)^{2}f(x)$.

  1. Prove that $a$ is continuous at $x_{0}=3$
  2. Prove that $b$ is differentiable at $x_{0}=3$

1.We need to show that left and right side limit are the same. If that's the case, the function is continuous:

$$\lim_{x\rightarrow 3^{-}}\left((x-3)f(x)) \right )= (3-3)\cdot f(3)= 0 \cdot f(3)= 0$$

$$\lim_{x\rightarrow 3^{+}}\left((x-3)f(x)) \right )= (3-3)\cdot f(3)= 0 \cdot f(3)= 0$$

Thus the function $a$ is continuous at $x_{0}=3$.

2.I will use difference-quotient for proving differentiability:

$$\lim_{x\rightarrow 3^{-}}\left ( \frac{b(x)-b(x_{0})}{x-x_{0}} \right )= \lim_{x\rightarrow 3^{-}} \left ( \frac{(x-3)^{2}f(x)-(3-3)^{2}f(3)}{x-3} \right ) = \lim_{x\rightarrow 3^{-}} \frac{(x-3)^{2}f(x)-0 \cdot f(3)}{x-3}= \lim_{x\rightarrow 3^{-}}\frac{(x-3)^{2}f(x)}{x-3}= \lim_{x\rightarrow 3^{-}}(x-3)f(x) = 0$$

Apply same for the right side and get $0$ as well. Thus $b$ is differentiable at $x_{0} = 3$.

Did I do this task correctly or it's completely wrong..? :o

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    $\begingroup$ You can not assume that the limits of $f$ and $g$ exists, because they are not necessarily continuous! $\endgroup$ – Zestylemonzi Sep 6 '16 at 12:16
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    $\begingroup$ Try an $\epsilon - \delta$ proof - it should be easier. $\endgroup$ – Zestylemonzi Sep 6 '16 at 12:18
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    $\begingroup$ It doesn't make sense to take the limit $\lim_{x \rightarrow 3^-} f(x)$. It makes just as much sense to write $\lim_{x \rightarrow 3^-} (x-3)(\text{Avocado}^x) = 0 \cdot \text{Avocado}^3=0$ $\endgroup$ – Zestylemonzi Sep 6 '16 at 12:20
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    $\begingroup$ What @Zestylemonzi is true. You can't plug the limit into $f$ because $f$ is not continous. For 2. observe that your last equation before zero is actually the limes of $a(x)$ at $3$ and then use 1. $\endgroup$ – Maik Pickl Sep 6 '16 at 12:21
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    $\begingroup$ @cnmesr If you don't understand the $\epsilon-\delta$ proofs, you have no business using any other method. The $\epsilon-\delta$ is a definition, and you cannot truly understand continuity without understanding its definition. $\endgroup$ – 5xum Sep 6 '16 at 12:22
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You started off correctly but in flow used the wrong result that $\lim_{x \to 3}f(x) = f(3)$. This is not known in advance as we don't know whether $f$ is continuous or not. Another problem is that you are trying to deal with both $x \to 3^{-}$ and $x \to 3^{+}$ separately. This is required only when the definition of function concerned is different for cases $x < 3$ and $x > 3$.

You only to need to prove one result here and that is $$\lim_{x \to 3}(x - 3)f(x) = 0$$ and this is easily done by using Squeeze theorem. Let $a(x) = (x - 3)f(x)$ then we know that $$0 \leq |a(x)| = |(x - 3)||f(x)| \leq |x - 3|$$ because we know that $|f(x)| \leq 1$. Thus by Squeeze theorem we get $$\lim_{x \to 3}|a(x)| = 0$$ Further $$-|a(x)| \leq a(x)\leq |a(x)|$$ and again applying Squeeze theorem gives us $$\lim_{x \to 3}a(x) = 0$$ Now note that $a(3) = 0$ so $a$ is continuous at $3$.

The second part about $b(x)$ is easy if you know the first part.


As requested by OP here is an approach via $\epsilon-\delta$ definition. As I have mentioned in comments this definition can not be used to evaluate a limit of a function, but it can be used to check whether a number is a limit of the function or not. So in this method we need to guess the limit somehow.

For the current we are asked to prove that $a(x)$ is continuous at $x_{0} = 3$. By definition of continuity this is equivalent to proving that $\lim_{x \to 3}a(x) = a(3)$. Now $a(3) = 0$ so we have to prove that $\lim_{x \to 3}a(x) = 0$. So you see that the question itself has given you the limit $0$ so that the trouble of guessing the limit is not here. Lucky!!

Now to prove that $\lim_{x \to 3}a(x) = 0$ we need to ensure that for every $\epsilon > 0$ there is a number $\delta > 0$ such that $$|a(x) - 0| < \epsilon$$ whenever $0 < |x - 3| < \delta$. Thus we need to start with an $\epsilon > 0$ and somehow try to find a suitable $\delta > 0$ (depending on $\epsilon$) such that $|a(x)| < \epsilon$ whenever $0 < |x - 3| < \delta$.

Now let $\epsilon > 0$ be given. Our goal to satisfy the inequality $$|a(x)| < \epsilon$$ or $$|(x - 3)f(x)| < \epsilon$$ Now note that $|f(x)| < 1$ so we already know that $$|(x - 3)f(x)| = |x - 3| |f(x)| < |x - 3|$$ and hence if we can get $|x - 3| < \epsilon$ then automatically we will have $$|a(x)| < |x - 3| < \epsilon$$ and our goal will be achieved. Thus we can take $\delta = \epsilon$ here and then $0 < |x - 3| < \delta$ will imply $0 < |x - 3| < \epsilon$ and this will imply that $|a(x)| < \epsilon$. Our proof is now complete and $\lim_{x \to 3}a(x) = 0$.

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    $\begingroup$ Thank you very much currently learning this new theorem you have used here I've never heard about and it seems very powerful when normal limit tricks / epsilon delta cannot help me out :) Thank you for prividing your solution and especially thank you for showing this theorem! $\endgroup$ – cnmesr Sep 7 '16 at 11:24
  • $\begingroup$ @cnmesr Actually, the squeeze theorem is simply a way to skip over 4 or 5 steps from the $\epsilon-\delta$ proofs. I advise against using it before you know your basics! $\endgroup$ – 5xum Sep 7 '16 at 11:33
  • $\begingroup$ @cnmesr: Squeeze theorem also known as Sandwich Theorem is indeed a very powerful theorem but sadly it is not used much. The $\epsilon-\delta$ technique is actually a definition of limit and is not so much used to evaluate limits but rather to prove theorems on limits (eg Squeeze Theorem) and these theorems are then used to evaluate limits. $\endgroup$ – Paramanand Singh Sep 7 '16 at 11:34
  • $\begingroup$ Then show me how to do it with epsilon-delta please. I have learned it several hours yesterday and I could even solve some tasks of it correctly. But there were clear functions given and not as in this task. By unclear I mean something like $f(x) < 1$. So, what is $f(x)$? I don't know anything, only know it's smaller than 1... But I need to know the function to use epsilon delta. I don't know how to apply epsilon delta here because don't know function $f(x)$. $\endgroup$ – cnmesr Sep 7 '16 at 11:39
  • $\begingroup$ @cnmesr: I will add a solution based on $\epsilon-\delta$ but you should understand that $\epsilon-\delta$ method does not help you to evaluate a limit $\lim_{x \to a}f(x)$. It only helps you to check whether a given number $L$ is the limit of $f(x)$ or not as $x \to a$. So in order to use $\epsilon-\delta$ you must to be able to guess the limit by some means or perhaps your intuition. $\endgroup$ – Paramanand Singh Sep 7 '16 at 11:42
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$$\lim_{x\rightarrow 3^{-}}\left((x-3)f(x)) \right )= (3-3)\cdot f(3)= 0 \cdot f(3)= 0$$

Wrong. You implicitly assume several things here:

  1. That $\lim_{x\to 3^-} (x-3)f(x)=\lim_{x\to 3^-}(x-3)\cdot \lim_{x\to 3^-} f(x)$
  2. That $\lim_{x\to 3^-} f(x)$ exists
  3. That $\lim_{x\to 3^-}=f(3)$.

none of these things are neccesarily true from what you know!


A much easier approach to prove that $a(x)$ is continuous is to go from the $\epsilon-\delta$ definition. You can

  1. Simply calculate what $a(0)$ is
  2. Prove that for each $\epsilon$, taking a small enough $\delta $ will cause $|a(x)-a(3)|$ to be smaller than $\epsilon$ if $|x-3|<\delta$

Hint:

$|a(x)| = |(x-3)f(x)| = |x-3||f(x)|$

now, can you estimate one of those two factors? What do you know about $f$?

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  • $\begingroup$ Ouch, what would you recommend then? epsilon-delta.proof as Zestylemonzi has recommended? I don't know how to do it >.< No other ways? $\endgroup$ – cnmesr Sep 6 '16 at 12:20
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    $\begingroup$ @cnmesr There are always other ways. But my experience tells me that the $\epsilon-\delta$ proof should be the easiest. $\endgroup$ – 5xum Sep 6 '16 at 12:21
  • $\begingroup$ How do you know we start by making $a(0)$ btw? $\endgroup$ – cnmesr Sep 6 '16 at 12:25
  • $\begingroup$ @cnmesr What is $a(0)$, by definition? $\endgroup$ – 5xum Sep 6 '16 at 12:26
  • $\begingroup$ Ok nvm it. I skip this task and try to understand epsilon delta first... $\endgroup$ – cnmesr Sep 6 '16 at 12:29

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