# Thread: show f not continous on a

1. ## show f not continous on a

For x exsits in real numbers, there is unique n exsists in integers s.t n<=x<n+1. denote this n as [x]. Thus, we obtain a function F: real numbers arrow real numbers, x arrow [x]. e.g. [-2.5]= -3.

Let a exsist in Integers show f is not continuous at a. ( use suitable sequences which converge to a )??????

Hope this makes sense thanks.

For x exsits in real numbers, there is unique n exsists in integers s.t n<=x<n+1. denote this n as [x]. Thus, we obtain a function F: real numbers arrow real numbers, x arrow [x]. e.g. [-2.5]= -3.

Let a exsist in Integers show f is not continuous at a. ( use suitable sequences which converge to a )??????

Hope this makes sense thanks.
Condiser the sequnces for $a \in \mathbb{Z}$ $x_n=a-\frac{1}{n}$ and $y_n=a+\frac{1}{n}$ ; $n=1,2,3...$

what would $f(x_n)$ equal and $f(y_n)$

I hope this helps.

Good luck.

3. Originally Posted by TheEmptySet
Condiser the sequnces for $a \in \mathbb{Z}$ $x_n=a-\frac{1}{n}$ and $y_n=a+\frac{1}{n}$ ; $n=1,2,3...$

what would $f(x_n)$ equal and $f(y_n)$

I hope this helps.

Good luck.

em.. im sorry any chance a bit more in depth here?

4. surely as n gets infinitely large $x_n=a-\frac{1}{n}$ tends to $a$
$y_n=a+\frac{1}{n}$ tends to $a$

as a result $f(x_n)=a-1$ ???
$f(y_n)=a-1$

surely as n gets infinitely large $x_n=a-\frac{1}{n}$ tends to $a$
$y_n=a+\frac{1}{n}$ tends to $a$

as a result $f(x_n)=a-1$ ???
$f(y_n)=a-1$
note that $x_n < a$ for all n and

$y_n > a$ for all n

so $f(x_n) = a-1$ for all n

$f(a)=a$

$f(y_n) =a$ for all n

so both sequences converge to the same value, but their immages go to two different values.

Therefore the function is not continous at a when a is an integer.

6. $f(y_n) =a$ for all n

thanks for the help!
can you run this by me, understand the rest and looks like was on the right line but cant see this yet?

7. Also how would i show f is R-integrable on any interval [a,b] ???