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Q. $\displaystyle\lim _{x \rightarrow c} f(x)$ doesn't exist
where $[ x ]$, denotes step up function and $\{ x \}$ fractional part function.

Limits and Derivatives

Solution:

$\displaystyle\lim _{x \rightarrow 3}[[x]]-[2 x-1]$
RHL $\displaystyle\lim _{h \rightarrow 0}[(3+h)]-[6+2 h-1]$
As $3<3+ h <4 $
$\Rightarrow [3+ h ]=3$ and $5+2 h <6$
$\Rightarrow [5+2 h]=5$
$\Rightarrow \displaystyle\lim _{h \rightarrow 0}(3-5)=-2$
LHL $\displaystyle\lim _{ h \rightarrow 0}[[3- h ]]-[6-2 h -1]$
As $2<3- h < 3 $
$\Rightarrow [3- h ]=2$ and $4<5-2 h <5$
$\Rightarrow [5-2 h ]=4$
$\Rightarrow \displaystyle\lim _{h \rightarrow 0}(2-4)=-2$
$\therefore \displaystyle\lim _{h \rightarrow c} f(x)$ exists, $c=3$
(B) $\displaystyle\lim _{ x \rightarrow 1}[ x ]- x \displaystyle\lim _{ x \rightarrow 1^{+}} f ( x )$
$=\displaystyle\lim _{ h \rightarrow 0}[1+ h ]-(1+ h )$
$=\displaystyle\lim _{ h \rightarrow 0}(1-1- h )=0 $
$\displaystyle\lim _{ x \rightarrow 1^{-}} f ( x )=\displaystyle\lim _{ h \rightarrow 0}[1- h ]-(1- h )$
$=\displaystyle\lim _{ h \rightarrow 0} 0-(1- h )=-1$
$\therefore \displaystyle\lim _{x \rightarrow 1} f(x)$ doesn't exists.
(C)$\displaystyle\lim _{ x \rightarrow 0}\{ x \}^{2}-\{- x \}^{2}$
$RHL \displaystyle\lim _{ h\rightarrow 0}\{ h \}^{2}-\{- h \}^{2}$
$\Rightarrow \displaystyle\lim _{ h \rightarrow 0}( h -[ h ])^{2}-((- h )-[- h ])^{2}$
$\Rightarrow \displaystyle\lim _{ h \rightarrow 0} h ^{2}-(- h +1)^{2} $
$\Rightarrow -1$
$LHL \displaystyle\lim _{ h \rightarrow 0}\{- h \}^{2}-\{ h \}^{2}$
$\Rightarrow \displaystyle\lim _{ h \rightarrow 0}((- h )-[- h ])^{2}-( h -[ H ])^{2}$
$\Rightarrow \displaystyle\lim _{ h \rightarrow 0}(- h +1)^{2}-( h -0)^{2} $
$\Rightarrow 1$
$\therefore \displaystyle\lim _{ x \rightarrow c } f ( x ) ; c =0$, doesnt' exist.
(d) $\displaystyle\lim _{x \rightarrow 0} \frac{\tan (\text{sgn} x)}{(\text{sgn} x)}$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0^{+}}, f(x)=\lim _{h \rightarrow 0} \frac{\tan 1}{1}=\tan 1$
$\displaystyle\lim _{x \rightarrow 0^{-}} f(x)=\displaystyle\lim _{h \rightarrow 0} \frac{\tan (-1)}{-1}=\tan 1$
$\therefore \displaystyle\lim _{x \rightarrow 0} f(x)$ exists.