continuty question :

Question

hysenberg · Answer

$$f(x)=\frac{ |x| }{ x};x \neq0$$
       $$1 ; x=0$$
discuss the continuty of this function

sidsiddhartha · Answer

first of all u need to know
$$|x|=x ; x \ge0$$
and
$$|x|=-x ;x<0$$
ok with this ?

amistre64 · Answer

$$f(x)=\begin{cases}\dfrac{|x|}{x}~;~x\ne0\\~~1~~~;~ x=0\end{cases}$$

hysenberg · Answer

yeah ok!!!

hysenberg · Answer

yeah @amistre64

amistre64 · Answer

the top is a switching function, almost. -1 to the left and 1 to the right

amistre64 · Answer

other than that, im not sure what there is to discuss :)

sidsiddhartha · Answer

so 
$$|x|/x=1 ;x \ge0$$
and 
$$|x|/x=-1 ;x<0$$
this is generally known as a signum functiion|dw:1402592843401:dw|

sidsiddhartha · Answer

so i think the only doubtful point where continuty is to be checked is at x=0

hysenberg · Answer

ok so i need to evaluate the right hand and left hand limits?

sidsiddhartha · Answer

yeah what are u getting?

hysenberg · Answer

im getting left limit -1 and right limit 1

sidsiddhartha · Answer

well 
$$\lim_{x \rightarrow (0-0)}f(x)=-1$$
and
$$\lim_{x \rightarrow ?(0+0)}f(x)=1$$
and value at x=0
f(0)=1
so they are not equal
so what can you conclude

hysenberg · Answer

ahhh so the function is discontinuos at x=0

sidsiddhartha · Answer

yeah good it is called discontinuity of first kind :)