Browse Questions

# Show that the function defined by $g (x) = x – [x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$.

Toolbox:
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
Let $c$ be an integer.
$[c-h]=c-1$
$[c+h]=c$
$[c]=c$
$g(x)=x-[x]$
LHL:
At $x=c$
$\lim\limits_{\large x\to c}(x-[x])=\lim\limits_{\large h\to 0}[(c-h)-(c-1)]$
$\qquad\qquad\;\;\;\;=\lim\limits_{\large h\to 0}[c-h-c+1]$
$\qquad\qquad\;\;\;\;=1$
Step 2:
RHL:
At $x=c$
$\lim\limits_{\large x\to c}(x-[x])=\lim\limits_{\large h\to 0}[(c+h)-(c+h)]$
$\qquad\qquad\;\;\;\;=\lim\limits_{\large h\to 0}[c+h-c-h]$
$\qquad\qquad\;\;\;\;=0$
Step 3:
$f(c)=c-[c]=0$
$LHL \neq$ RHL =$f(c)$
$f$ is not continuous at integral points.
Step 4:
To find $LHL=\lim\limits_{\large x\to a}f(x)$,put $x=a-h$
As $x\to a,h\to 0$
LHL=$\lim\limits_{\large h\to 0}f(a-h)$
Similarly for RHL, put $x=a+h$
$\lim\limits_{\large x\to a}f(x)=\lim\limits_{\large h\to 0}f(a+h)$