If [x] denotes the greatest integer less than or equal to x; then the $lim x \to 1 (1 - x + [x - 1] + [1 - x])$ is:

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 0
$L = {lim}_{x \to 1} (1 - x + [x - 1] + [1 - x])$
$\Rightarrow [x - 1] = [x] - 1$
and $[1 - x] = 1 + [- x]$
$= 1 - 1 - [x]$
$∴ L = {lim}_{x \to 1} (x - x + [x] - 1 + 1 - 1 - [x])$
$= 0$
Hence, the answer is $0.$

Suggest Corrections

Similar questions

[x]

x

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

Let $f (x) = x {(- 1)}^{[\frac{1}{x}]} . x \neq 0$ , where [x] denotes the greatest integer less than or equal to x. then $lim x \to 0 f (x)$

n \in N,

[x]

{lim}_{x \to n} (- 1)^{[x]}

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{sin [x]}{[x]}, & [x] \neq 0 0, & [x] = 0 \end{matrix}

[x]

x

lim x \to 0 f (x)

Join BYJU'S Learning Program