Question

If \( f(x) = \begin{cases} x \left( 1 + \frac{1}{2} \sin(\log x^2) \right), & x \neq 0 \\ 0, & x = 0 \end{cases} \), then \( \lim_{x \to 0} \frac{f(x) - f(0)}{x} \)

• A. is equal to \( f(0) \)
• B. does not exist
• C. is equal to \( \frac{1}{2} \)
• D. is equal to \( f(1) \)

Hint:

To evaluate the limit \( \lim_{x \to 0} \frac{f(x) - f(0)}{x} \), substitute the given piecewise definition of \( f(x) \). Simplify the expression and analyze the behavior of the trigonometric function as \( x \) approaches 0. If the argument of the sine function tends to infinity, and the sine function oscillates, the overall limit might not exist.

Answer 1

The Correct Option is B

We need to find \( \lim_{x \to 0} \frac{f(x) - f(0)}{x} \).
Given \( f(0) = 0 \), so the limit becomes \( \lim_{x \to 0} \frac{f(x)}{x} \).
For \( x \neq 0 \), \( f(x) = x \left( 1 + \frac{1}{2} \sin(\log x^2) \right) \).
So, \( \frac{f(x)}{x} = 1 + \frac{1}{2} \sin(\log x^2) \).
We need to evaluate \( \lim_{x \to 0} \left( 1 + \frac{1}{2} \sin(\log x^2) \right) \).
Let \( t = \log x^2 = 2 \log |x| \).
As \( x \to 0 \), \( |x| \to 0^+ \), so \( \log |x| \to -\infty \), and \( t \to -\infty \).
The limit becomes \( \lim_{t \to -\infty} \left( 1 + \frac{1}{2} \sin(t) \right) \).
The function \( \sin(t) \) oscillates between -1 and 1 as \( t \to -\infty \).
Therefore, \( 1 + \frac{1}{2} \sin(t) \) oscillates between \( 1 + \frac{1}{2}(-1) = \frac{1}{2} \) and \( 1 + \frac{1}{2}(1) = \frac{3}{2} \).
Since the limit oscillates between two different values, the limit does not exist.