The value of lim_x→ 0≤ft((1-cos x)/(2x^2)) is :

The correct answer is Option (3) → \(\displaystyle\frac{1}{4}\)

Using \(1-\cos x = 2\sin^2\frac{x}{2}\):

    \( \displaystyle \frac{1-\cos x}{2x^2} = \frac{2\sin^2\frac{x}{2}}{2x^2} = \frac{\sin^2\frac{x}{2}}{x^2} \)

Writing \(x^2 = 4\left(\frac{x}{2}\right)^2\):

    \( \displaystyle = \frac{1}{4}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \)

As \(x \to 0\), \(\frac{x}{2} \to 0\), and by the standard limit \(\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1\):

    \( \displaystyle \lim_{x\to 0}\frac{1-\cos x}{2x^2} = \frac{1}{4}\times 1^2 = \frac{1}{4} \)

Tip: The result \(\lim_{x\to 0}\frac{1-\cos x}{x^2} = \frac{1}{2}\) is worth memorising. The extra factor 2 in the denominator here halves it to \(\frac{1}{4}\); forgetting it gives \(\frac{1}{2}\), Option (2).