# Continuety of these functions

1. Dec 27, 2008

### transgalactic

describe the continuety of these functions
http://img233.imageshack.us/img233/6141/76604020ql2.gif [Broken]

regarding the [] functions i learned this fact

http://img184.imageshack.us/img184/2469/87823960yl2.gif [Broken]

Last edited by a moderator: May 3, 2017
2. Dec 27, 2008

### HallsofIvy

Staff Emeritus
Since you used "[x]" I was not sure if you meant "floor of x", $\floor(c)$ or "ceiling of x", [tex]\ceil(x)[/itex]. From the I'm second link it appears that you mean the "floor" function.
Yes, $\floor(x)$ is constant between intgers: n-1 from n-1 up to n and n from n to n+1 so it is really only necessary to look at what happens as you approach the integer n from below and above.

The last one appears to be "piecewise", defined by different formulas for x rational and irrational but there appears to be something written before "$\pi x$ for x rational. Is that "sin"?

Last edited: Dec 29, 2008
3. Dec 28, 2008

### transgalactic

yes its sin

4. Dec 29, 2008

### transgalactic

how to do a limit of this x*1/[x]

i get 0*+infinity
??

5. Dec 29, 2008

### transgalactic

how to solve the splitted
i cant imagine the graph
i cant do a limit
because rational and irrational are infinitely mixed
what to do?

6. Dec 29, 2008

### HallsofIvy

Staff Emeritus
For x any number other than an integer, if x is between n and n+ 1, [x]= n so
x/[x]= x/n and x[x]= xn. It should be obvious that the function is continuous there. For x slightly below n, say x= n-y, x/[x]= n-y/(n-1) and x[x]= (n-y)n-1= xn/xy. For x slightly larger than n, say x= n+ y, x/[x]= (n+y)/n and x[x]= (n+y)n. Are the limits of those, as y goes to 0, the same?

As for $sin(\pi x)$ for x rational, 0 for x irrational, use the fact that $\lim_{x\rightarrow a} f(x)= L$ if and only if $\lim_{n\rightarrow \infty} f(a_n)= L$ for any sequence ${a_n}$ converging to x. In particular, for any number x, there exist a sequence of irrational numbers converging to x so for such a sequence the limit of this sequence will be 0. That means that in order that the limit itself exist at x, we must also have $sin(a_n\pi)$ converge to 0 also. Since sine is a continuous function, it is easy to see that that limit is $sin(\pi x)$. That is, this function is continuous exactly for those x such that $sin(\pi x)= 0$.