CUET UG 2025 — Mathematics Calculus

For the function to be continuous at $x = 0$x=0:

$$\lim_{x \to 0} \frac{1-\cos 8x}{16x^2} = k$$x→0lim​16x21−cos8x​=k

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Using the trigonometric identity $1 - \cos \theta = 2\sin^2\left(\frac{\theta}{2}\right)$1−cosθ=2sin2(2θ​) with $\theta = 8x$θ=8x:

$$1 - \cos 8x = 2\sin^2(4x)$$1−cos8x=2sin2(4x)

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$$\lim_{x \to 0} \frac{1-\cos 8x}{16x^2}$$x→0lim​16x21−cos8x​

$$= \lim_{x \to 0} \frac{2\sin^2(4x)}{16x^2}$$=x→0lim​16x22sin2(4x)​

$$= \lim_{x \to 0} \frac{\sin^2(4x)}{8x^2}$$=x→0lim​8x2sin2(4x)​

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Rewriting the expression:

$$\frac{\sin^2(4x)}{8x^2}$$8x2sin2(4x)​

$$= \frac{\sin^2(4x)}{(4x)^2} \times \frac{(4x)^2}{8x^2}$$=(4x)2sin2(4x)​×8x2(4x)2​

$$= \left[\frac{\sin(4x)}{4x}\right]^2 \times \frac{16x^2}{8x^2}$$=[4xsin(4x)​]2×8x216x2​

$$= \left[\frac{\sin(4x)}{4x}\right]^2 \times 2$$=[4xsin(4x)​]2×2

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Using the standard limit $\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1$x→0lim​xsinx​=1:

$$\lim_{x \to 0} \frac{\sin(4x)}{4x} = 1$$x→0lim​4xsin(4x)​=1

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$$\lim_{x \to 0} \left[\frac{\sin(4x)}{4x}\right]^2 \times 2$$x→0lim​[4xsin(4x)​]2×2

$$= (1)^2 \times 2$$=(1)2×2

$$= 2$$=2

Therefore, $k = 2$k=2