Question

For which of the following fucntions $f\left(x\right)$, does the $\underset{x\to c}{lim}f\left(x\right)$ does not exist?

• A. $f\left(x\right)=\left[\right[x\left]\right]-[2x-1]$, $c=3$
• B. $f\left(x\right)=\left[x\right]-x$, $c=1$
• C. $f\left(x\right)={\left\{x\right\}}^2-{\{-x\}}^2$, $c=0$
• D. $f\left(x\right)=\frac{\tan \left(sgnx\right)}{sgnx}$, $c=0$

Answer 1

Answer: (B), (C)

The correct options are
B $f\left(x\right)=\left[x\right]-x$, $c=1$
C $f\left(x\right)={\left\{x\right\}}^2-{\{-x\}}^2$, $c=0$

Option A, $f\left(x\right)=\left[\right[x\left]\right]-[2x-1]$, $c=3$

$LHL={lim}_{x\to 3^{-}}f\left(x\right)={lim}_{h\to 0}f(3-h)={lim}_{h\to 0}\left[\right[3-h\left]\right]-\left[2\right(3-h)-1]={lim}_{h\to 0}\left[2\right]-[5-2h]=2-4=-2$

$RHL={lim}_{x\to 3^{+}}f\left(x\right)={lim}_{h\to 0}f(3+h)={lim}_{h\to 0}\left[\right[3+h\left]\right]-\left[2\right(3+h)-1]={lim}_{h\to 0}\left[3\right]-[5+2h]=3-5=-2$

Here, LHL=RHL, so limit exists.

Option B, $f\left(x\right)=\left[x\right]-x$, $c=1$

$LHL={lim}_{x\to 1^{-}}f\left(x\right)={lim}_{h\to 0}f(1-h)={lim}_{h\to 0}[1-h]-(1-h)={lim}_{h\to 0}0-1=-1$

$RHL={lim}_{x\to 1^{+}}f\left(x\right)={lim}_{h\to 0}f(1+h)={lim}_{h\to 0}[1+h]-(1+h)={lim}_{h\to 0}1-1=0$

Here,$LHL\ne RHL$

Hence, limit does not exists.

Option C,$f\left(x\right)={\left\{x\right\}}^2-{\{-x\}}^2$, $c=0$

$LHL={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{h\to 0}f(-h)={lim}_{h\to 0}{\{-h\}}^2-{\{-(-h\left)\right\}}^2={lim}_{h\to 0}0-1=-1$

$RHL={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{h\to 0}f\left(h\right)={lim}_{h\to 0}{\left\{h\right\}}^2-{\{-h\}}^2={lim}_{h\to 0}1-0=1$

Here,$LHL\ne RHL$

Hence, the limit does not exist.

Option D, $f\left(x\right)=\frac{\tan \left(sgnx\right)}{sgnx}$

We know $sgn\left\{x\right\}=\{\begin{array}{c}-1x<0\\ 1x\ge 0\end{array}$

$RHL={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}\frac{\tan \left(sgnx\right)}{sgnx}=\frac{\tan \left(1\right)}{1}$

$LHL={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}\frac{\tan \left(sgnx\right)}{sgnx}=\frac{\tan (-1)}{(-1)}=\frac{\tan \left(1\right)}{1}$

Here, LHL=RHL, so limit exists.