Step 1: Consider the given limit \( \lim_{x \to 0} \frac{| \sin x |}{x} \).
For \( x \to 0^+ \) (approaching 0 from the right), \( \sin x \) is positive, so \( | \sin x | = \sin x \).
The limit becomes: \[ \lim_{x \to 0^+} \frac{\sin x}{x} = 1. \] For \( x \to 0^- \) (approaching 0 from the left), \( \sin x \) is negative, so \( | \sin x | = -\sin x \). The limit becomes: \[ \lim_{x \to 0^-} \frac{-\sin x}{x} = -1. \]
Since the limit from the right is 1 and the limit from the left is -1, the two one-sided limits are not equal.
Hence, the limit does not exist.
Match List-I with List-II
List-I | List-II |
---|---|
(A) \( f(x) = |x| \) | (I) Not differentiable at \( x = -2 \) only |
(B) \( f(x) = |x + 2| \) | (II) Not differentiable at \( x = 0 \) only |
(C) \( f(x) = |x^2 - 4| \) | (III) Not differentiable at \( x = 2 \) only |
(D) \( f(x) = |x - 2| \) | (IV) Not differentiable at \( x = 2, -2 \) only |
Choose the correct answer from the options given below:
Match List-I with List-II
List-I | List-II |
---|---|
(A) \( f(x) = |x| \) | (I) Not differentiable at \( x = -2 \) only |
(B) \( f(x) = |x + 2| \) | (II) Not differentiable at \( x = 0 \) only |
(C) \( f(x) = |x^2 - 4| \) | (III) Not differentiable at \( x = 2 \) only |
(D) \( f(x) = |x - 2| \) | (IV) Not differentiable at \( x = 2, -2 \) only |
Choose the correct answer from the options given below: