Question

Find the limit: \[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \]

Hint:

When dealing with indeterminate forms like \( \frac{0}{0} \), use L'Hopital's rule by differentiating the numerator and denominator separately. Also, remember the standard limit \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \).

Answer 1

We are tasked with finding the limit: \[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x^2} \] Step 1: Apply L'Hopital's Rule
We see that the limit is of the indeterminate form \( \frac{0}{0} \), so we apply L'Hopital's rule, which states that if the limit is indeterminate, we can differentiate the numerator and denominator separately and then evaluate the limit.
Differentiate the numerator: \[ \frac{d}{dx} \left( 1 - \cos(2x) \right) = 2 \sin(2x) \] Differentiate the denominator: \[ \frac{d}{dx} \left( x^2 \right) = 2x \] Now, the limit becomes: \[ \lim_{x \to 0} \frac{2 \sin(2x)}{2x} \] Step 2: Simplify and Apply the Standard Limit
Simplify the expression: \[ \lim_{x \to 0} \frac{\sin(2x)}{x} \] We can use the standard limit: \[ \lim_{x \to 0} \frac{\sin(kx)}{x} = k \] For \( k = 2 \), this becomes: \[ \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \] Thus, the value of the limit is: \[ \boxed{2} \]