Question

If $L_1=\underset{x\to 2020}{lim}\sum _{k=1}^{2020}\frac{\{x+k\}}{2020}$ and $L_2=\underset{x\to 2020}{lim}(\left[x\right]+[-x]),$ then which of the following option is correct ?

(where $\left[x\right]$ and $\left\{x\right\}$ represent the greatest integer function and fractional part of $x$ respectively)

• A. $L_1=0$ and $L_2$ does not exist
• B. $L_1=0$ and $L_2=-1$
• C. $L_1$ does not exist and $L_2=-1$
• D. $L_1=2020$ and $L_2$ does not exist

Answer 1

The correct option is C $L_1$ does not exist and $L_2=-1$

$L_1=\underset{x\to 2020}{lim}\sum _{k=1}^{2020}\frac{\{x+k\}}{2020}$

$=\underset{x\to 2020}{lim}\frac{1}{2020}(\{x+1\}+\{x+2\}+\cdots +\{x+2020\})$

$=\underset{x\to 2020}{lim}\frac{1}{2020}(\left\{x\right\}+\left\{x\right\}+\cdots 2020\text{terms})$

$=\underset{x\to 2020}{lim}\left\{x\right\}$
which does not exist $(\because \text{L.H.L}=1,\text{R.H.L}=0)$


$L_2=\underset{x\to 2020}{lim}(\left[x\right]+[-x])$

We know that $[-x]=-1-\left[x\right]$
$\Rightarrow L_2=\underset{x\to 2020}{lim}(\left[x\right]-1-\left[x\right])\Rightarrow L_2=\underset{x\to 2020}{lim}(-1)=-1$