Let $g \cdot \mathbb{R} \rightarrow \mathbb{R}$ with $|g(x)| \leq 1 $ for all $x \in \mathbb{R}$.

Prove that $f: \mathbb{R} \ni x \mapsto \left(x-x_{0}\right) \cdot g(x) \in \mathbb{R}$ is continuous.

Ok I have trouble because not any specific $x_{0}$ is given.

Also I think the inequality $|g(x)| \leq 1$ will cause trouble / no easy birth.

We will have to prove it for

  1. $|g(x)| = 1$
  2. $|g(x)| < 1$

is that right so far?

  1. $f(x) = (x-x_{0}) \cdot 1 = (x-x_{0})$

So in this case, the limit of function $f$ would be $x-x_{0}$.

Using squeeze theorem on this we get:

$$(x-x_{0}) \leq (x-x_{0}) \cdot 1 \leq (x-x_{0})$$

I think doesn't make much sense I wrote here but the limit should be $(x-x_{0})$ if $|g(x)| = 1$.

2.If $|g(x)| < 1$:

$$0 \leq (x-x_{0}) \cdot |g(x)| \leq (x-x_{0})$$

Hmm.. I don't know what to do don't know what $x_{0}$ is ... :(

I have task from exam 2005 published by someone who wrote it already, old student. Maybe he write wrong? I can't imagine how show continuity here, we don't know $x_{0}$ and not much info about $g(x)$ too?

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    $\begingroup$ You probably shall prove that $f$ is continuous at $x_0$. If $g$ isn't continuous at $x_1 \neq x_0$, then $f$ also is discontinuous at $x_1$. $\endgroup$ – Daniel Fischer Sep 7 '16 at 15:30
  • $\begingroup$ By the way, "$f: \mathbb{R} \ni x \mapsto \left(x-x_{0}\right) \cdot g(x) \in \mathbb{R}$" is really confusing notation. (Probably you didn't compose it yourself, but still.) "Define $f\colon\Bbb R\to\Bbb R$ by $f(x) = (x-x_0)\cdot g(x)$" is far better. $\endgroup$ – Greg Martin Sep 7 '16 at 16:05
  • $\begingroup$ But what theorem would you recommend to use for this? Epsilon-delta? Or just left and right side limit, compare, if same then continuous? $\endgroup$ – cnmesr Sep 7 '16 at 16:09
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    $\begingroup$ This question is almost the duplicate of your earlier question math.stackexchange.com/q/1916546/72031 where you had chosen a specific point $x_{0} = 3$ instead of a generic $x_{0}$. It appears that you have not understood the answer given there fully otherwise this new question would not have come into existence. $\endgroup$ – Paramanand Singh Sep 8 '16 at 8:21