Question

Discuss the continuity of \( f(x) \) at \( x = 0 \), where

Given: \[ f(x) = \begin{cases} \dfrac{1 - \cos(4x)}{x^2}, & x \neq 0 \\[6pt] 8, & x = 0 \end{cases} \]

Hint:

To check the continuity of a function at a point, compute the limit as \( x \to 0 \) and compare it to the value of the function at that point. If they match, the function is continuous at that point.

Answer 1

Step 1: Checking the limit as \( x \to 0 \).
To check the continuity of the function at \( x = 0 \), we need to verify if: \[ \lim_{x \to 0} f(x) = f(0) \] We already know that \( f(0) = 8 \). Now, let's compute the limit of \( f(x) \) as \( x \to 0 \).
For \( x \neq 0 \), \[ f(x) = \frac{1 - \cos 4x}{x^2} \] To find the limit, we use L'Hopital's Rule because the limit is of the indeterminate form \( \frac{0}{0} \). Differentiate the numerator and denominator:
Numerator: \[ \frac{d}{dx}(1 - \cos 4x) = 4 \sin 4x \] Denominator: \[ \frac{d}{dx}(x^2) = 2x \] Thus, by L'Hopital's Rule: \[ \lim_{x \to 0} \frac{1 - \cos 4x}{x^2} = \lim_{x \to 0} \frac{4 \sin 4x}{2x} \] Simplifying: \[ \lim_{x \to 0} \frac{4 \sin 4x}{2x} = 2 \lim_{x \to 0} \frac{\sin 4x}{x} \] Using the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we get: \[ 2 \cdot 4 = 8 \] Step 2: Conclusion.
Since \( \lim_{x \to 0} f(x) = 8 \) and \( f(0) = 8 \), the function is continuous at \( x = 0 \).