If $f(x)=[x]-\left[\frac{x}{4}\right], x \in R$, where $[x]$ denotes the greatest integer function, then:

(A) Both $\lim_{x \rightarrow 4^-} f(x)$ and $\lim_{x \rightarrow 4^+} f(x)$ exist but are not equal

(B) $\lim_{x \rightarrow 4^-} f(x)$ exists but $\lim_{x \rightarrow 4^+} f(x)$ does not exist

(C) $\lim_{x \rightarrow 4^+} f(x)$ exists but $\lim_{x \rightarrow 4^-} f(x)$ does not exist

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Limits

f (x) = [x] - [\frac{x}{4}], x \in R

, where

[x]

denotes the greatest integer function, then:

(A)Both $lim_{x \to 4^{-}} f (x)$ and $lim_{x \to 4^{+}} f (x)$ exist but are not equal
(B) $lim_{x \to 4^{-}} f (x)$ exists but $lim_{x \to 4^{+}} f (x)$ does not exist
(C) $lim_{x \to 4^{+}} f (x)$ exists but $lim_{x \to 4^{-}} f (x)$ does not exist
(D) $f$ is continuous at $x = 4$ ✓ Correct

Answer: (D)

Approaching from the right, just above

x = 4

, the greatest-integer parts are

[x] = 4

and

[\frac{x}{4}] = 1

, so

f \to 4 - 1 = 3

. Approaching from the left, just below

4

[x] = 3

and

[\frac{x}{4}] = 0

, so

f \to 3 - 0 = 3

. At the point itself,

f (4) = [4] - [1] = 4 - 1 = 3

. The left limit, the right limit and the value all equal

3

, so

f

is continuous at

x = 4

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option