Question

If [x] is the greatest integer function and \( f(x) = \begin{cases} \frac{2[x]-x}{|x|} & x \neq 0 \\ 1 & x=0 \end{cases} \) is a real valued function, then f is

• A. continuous at x=0
• B. continuous at x=1
• C. left continuous at x=0
• D. right continuous at x=1

Hint:

For a function to be continuous at a point \(c\), the left-hand limit, the right-hand limit, and the function's value at that point must all be equal and finite. For greatest integer functions, always check the limits from both sides, as the value of \([x]\) changes at integer points.

Answer 1

The Correct Option is D

Let's analyze the continuity at the points of interest.
Continuity at x=0:
The function value is \(f(0) = 1\).
Left-hand limit (LHL): \( \lim_{x \to 0^-} f(x) \). For \(x\) slightly less than 0 (e.g., -0.1), \([x] = -1\) and \(|x| = -x\).
LHL = \( \lim_{x \to 0^-} \frac{2(-1)-x}{-x} = \lim_{x \to 0^-} \frac{-2-x}{-x} = \infty \).
Since the LHL is not finite, the function is not continuous at x=0.
Continuity at x=1:
The function value at x=1 is \( f(1) = \frac{2[1]-1}{|1|} = \frac{2(1)-1}{1} = 1 \).
Left-hand limit (LHL) at x=1: \( \lim_{x \to 1^-} f(x) \). For \(x\) slightly less than 1 (e.g., 0.9), \([x] = 0\) and \(|x|=x\).
LHL = \( \lim_{x \to 1^-} \frac{2(0)-x}{x} = \lim_{x \to 1^-} \frac{-x}{x} = -1 \).
Since LHL \( \neq f(1) \), the function is not left-continuous at x=1.
Right-hand limit (RHL) at x=1: \( \lim_{x \to 1^+} f(x) \). For \(x\) slightly greater than 1 (e.g., 1.1), \([x]=1\) and \(|x|=x\).
RHL = \( \lim_{x \to 1^+} \frac{2(1)-x}{x} = \frac{2-1}{1} = 1 \).
Since the Right-hand limit (RHL) is equal to the function value \(f(1)\), the function is right-continuous at x=1.