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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 17 of 72
Marks: +1, -0
lim⁡x→0cos⁡2x−1cos⁡x−1\lim\limits_{x\to 0}\frac{\cos 2x-1}{\cos x-1}
Solution:  
We have, lim⁡x→0cos⁡2x−1cos⁡x−1\lim\limits_{x\to 0}\frac{\cos 2x-1}{\cos x-1}
= lim⁡x→0cos⁡2x−1−(−1−cos⁡x)\lim\limits_{x\to 0}\frac{\cos 2x-1}{-(-1-\cos x)} = lim⁡x→01−cos⁡2x1−cos⁡x×1+cos⁡x1+cos⁡x\lim\limits_{x\to 0}\frac{1-\cos 2x}{1-\cos x} \times \frac{1+\cos x}{1+\cos x}
= lim⁡x→0(2sin⁡2x)(1+cos⁡x)1−cos⁡2x\lim\limits_{x\to 0}\frac{(2\sin^2 x)(1+\cos x)}{1-\cos^2 x} = lim⁡x→02(1+cos⁡x)\lim\limits_{x\to 0}2(1+\cos x)
= 2 (1 + cos 0) = 2 (1 + 1) = 2 × 2 = 4.
Alternative solution :
lim⁡x→0cos⁡2x−1cos⁡x−1\lim\limits_{x\to 0}\frac{\cos 2x-1}{\cos x-1} = lim⁡x→01−2sin⁡2x−11−2sin⁡2(x2)−1\lim\limits_{x\to 0}\frac{1-2\sin^2 x-1}{1-2\sin^2\left(\frac{x}{2}\right)-1}
= lim⁡x→0sin⁡2xsin⁡2(x2)\lim\limits_{x\to 0}\frac{\sin^2 x}{\sin^2\left(\frac{x}{2}\right)} = lim⁡x→0sin⁡2xx2×x2×(x2)2sin⁡2(x2)×1(x2)2\lim\limits_{x\to 0}\frac{\sin^2 x}{x^2} \times x^2 \times \frac{\left(\frac{x}{2}\right)^2}{\sin^2\left(\frac{x}{2}\right)} \times \frac{1}{\left(\frac{x}{2}\right)^2}
= lim⁡x→0sin⁡2xx2\lim\limits_{x\to 0}\frac{\sin^2 x}{x^2} . lim⁡x→0x2sin⁡2(x2)×x2×4x2\lim\limits_{x\to 0}\frac{\frac{x}{2}}{\sin^2\left(\frac{x}{2}\right)} \times x^2 \times \frac{4}{x^2} = 4
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