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Miscellaneous 7 II · Q122

Q.lim⁡x→0[x(6x−3x)cos⁡(6x)−cos⁡(4x)]\displaystyle\lim_{x\to 0}\left[\frac{x(6^x-3^x)}{\cos(6x)-\cos(4x)}\right]

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6x−3x=3x(2x−1)∼1×xlog⁡2=xlog⁡26^x-3^x=3^x(2^x-1)\sim1\times x\log2=x\log2, so the numerator x(6x−3x)∼x2log⁡2x(6^x-3^x)\sim x^2\log2. Using the sum-to-product identity, cos⁡6x−cos⁡4x=−2sin⁡5xsin⁡x∼−2(5x)(x)=−10x2\cos6x-\cos4x=-2\sin5x\sin x\sim-2(5x)(x)=-10x^2. The limit i …

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