Continuity verification

Question

$$f\big(x\big) = \frac{\sin{x}}{x^{P}},\quad P \in Z$$

How to verify that $f(x)$ is contnuous at $x = 0$? I tried with $P = 1$, and then using $\sin(x)/x$ at limit $x \to 0 = 1$ rule. But if $P = 0$, then continuity depends on $\frac{\sin(x)}{x}$ alone, right? What could the values of $f(0)$ be?

What's the problem with $P=0$ ? $\sin$ is continuous everywhere ... — Hippalectryon, Sep 16 '15 at 20:03
Yeah, if x = 0, then can we apply limit x -> 0 $\frac{sin(x)}{x^P}$ = 1 rule?. — Mitty, Sep 16 '15 at 20:07
You should define at first place the value of $f$ at $x=0$. If $p=1$ and you define $f(0)=1$, then using the well known limit , $f(x)$ turns to be continuous. — Svetoslav, Sep 16 '15 at 20:11

score 0 · Answer 1 · answered Sep 16 '15 at 20:11

0

if $p>0$ then $0$ dosenot belong to domain $f$, thus Continuity cannot be defined

answered Sep 16 '15 at 20:11

R.N

4,318

MrYouMath · Accepted Answer · 2015-09-16T20:26:42.387

0

If you rewrite your problem to

$$\frac{\sin(x)}{x^P}=\frac{\sin(x)}{x\cdot x^{P-1}}=x^{1-P}\frac{\sin(x)}{x}$$

Then this will be the same as showing that, $x^{1-P}$ is continous/discontinous $x=0$. You can do this as $\frac{\sin(x)}{x}$ is continous at $x=0$ and the product of continous functions is again continous.

edited Sep 16 '15 at 20:26

answered Sep 16 '15 at 20:17

MrYouMath

15,833

Continuity verification

2 Answers2