Continuety of these functions

[transgalactic]

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


regarding the [] functions i learned this fact

http://img184.imageshack.us/img184/2469/87823960yl2.gif

 

[HallsofIvy]

Since you used "[x]" I was not sure if you meant "floor of x", \floor(c) or "ceiling of x", \ceil(x)[/itex]. From the I&#039;m second link it appears that you mean the &quot;floor&quot; function.<br /> 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.<br /> <br /> The last one appears to be &quot;piecewise&quot;, defined by different formulas for x rational and irrational but there appears to be something written before &quot;\pi x for x rational. Is that &quot;sin&quot;?

 

[transgalactic]

yes its sin

does it change your answer?

 

[transgalactic]

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

i get 0*+infinity
??

 

[transgalactic]

how to solve the splitted
i can't imagine the graph
i can't do a limit
because rational and irrational are infinitely mixed
what to do?

 

[HallsofIvy]

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= x<sup>n/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.</sup>