Question

${lim}_{x\to 1}(1-x+[x-1]+[1-x\left]\right)$ is equal to (where [.] denotes greatest interger function):

Answer 1

We have,

${lim}_{x\to 1}(1-x+[x-1]+[1-x])$

To do this problem, The following two points can be noted.

Then,

$[u+k]=\left[u\right]+kifk\in Z$

$\left[u\right]+[-u]=-1,ifk\notin Z$

Now,

$x\to 1\Rightarrow x\in (1-\delta ,1)\cup (1,1+\delta )$

Where $\delta$ is very small, So $x$ can be assumed as a non integer.

Now, Given that,

${lim}_{x\to 1}(1-x+[x-1]+[1-x])$

$={lim}_{x\to 1}(1-x+\left[x\right]-1+[-x]+1)$

$={lim}_{x\to 1}(-x+\left[x\right]+[-x]+1)\because (\left[x\right]+[-x]=-1)$

$={lim}_{x\to 1}(-x-1+1)$

$={lim}_{x\to 1}(-x)$

$=-1$

Hence, this is the answer.