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7.4 Q.II · Q55

Q.lim⁡x→0[1−cos⁡(nx)1−cos⁡(mx)]\displaystyle\lim_{x\to 0}\left[\frac{1-\cos(nx)}{1-\cos(mx)}\right]

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✓ Free question

Divide numerator and denominator by x2x^2: (1−cos⁡nx)/x2(1−cos⁡mx)/x2\dfrac{(1-\cos nx)/x^2}{(1-\cos mx)/x^2}. Using 1−cos⁡(kx)=2sin⁡2(kx/2)1-\cos(kx)=2\sin^2(kx/2), 1−cos⁡kxx2=2(sin⁡(kx/2)kx/2)2k24→k22\dfrac{1-\cos kx}{x^2}=2\left(\dfrac{\sin(kx/2)}{kx/2}\right)^2\dfrac{k^2}{4}\to\dfrac{k^2}{2}. So the ratio tends to n2/2m2/2=n2m2\dfrac{n^2/2}{m^2/2}=\dfrac{n^2}{m^2}.

✓Final answer

n2m2\dfrac{n^2}{m^2}

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