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limx0(ln(cos3x)ln(cos2x))\lim_{x \to 0}\left(\frac{\ln(\cos 3x)}{\ln(\cos 2x)}\right)

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Q. limx0(ln(cos3x)ln(cos2x))\lim_{x \to 0}\left(\frac{\ln(\cos 3x)}{\ln(\cos 2x)}\right)
  1. Identify Limit Form: Identify the form of the limit to determine the method of solution.\newlineAs xx approaches 00, both cos(3x)\cos(3x) and cos(2x)\cos(2x) approach 11. Therefore, ln(cos(3x))\ln(\cos(3x)) and ln(cos(2x))\ln(\cos(2x)) both approach ln(1)\ln(1), which is 00. This gives us a 0/00/0 indeterminate form, which suggests that we can use L'Hôpital's Rule to evaluate the limit.
  2. Apply L'Hôpital's Rule: Apply L'Hôpital's Rule. L'Hôpital's Rule states that if the limit of f(x)/g(x)f(x)/g(x) as xx approaches a value cc is in the indeterminate form 0/00/0 or /\infty/\infty, then the limit is the same as the limit of f(x)/g(x)f'(x)/g'(x) as xx approaches cc, provided that the derivatives exist and the limit of f(x)/g(x)f'(x)/g'(x) is determinate.
  3. Differentiate Numerator and Denominator: Differentiate the numerator and denominator separately.\newlineThe derivative of ln(cos(3x))\ln(\cos(3x)) with respect to xx is 1cos(3x)(sin(3x))3\frac{1}{\cos(3x)} \cdot (-\sin(3x)) \cdot 3, using the chain rule.\newlineThe derivative of ln(cos(2x))\ln(\cos(2x)) with respect to xx is 1cos(2x)(sin(2x))2\frac{1}{\cos(2x)} \cdot (-\sin(2x)) \cdot 2, also using the chain rule.
  4. Simplify Derivatives: Simplify the derivatives.\newlineThe derivative of ln(cos(3x))\ln(\cos(3x)) simplifies to 3tan(3x)-3\tan(3x).\newlineThe derivative of ln(cos(2x))\ln(\cos(2x)) simplifies to 2tan(2x)-2\tan(2x).
  5. Apply L'Hôpital's Rule: Apply L'Hôpital's Rule by taking the limit of the derivatives.\newlinelimx0(3tan(3x)2tan(2x))=limx0(3tan(3x)2tan(2x))\lim_{x \to 0}(\frac{-3\tan(3x)}{-2\tan(2x)}) = \lim_{x \to 0}(\frac{3\tan(3x)}{2\tan(2x)})
  6. Evaluate Simplified Expression: Evaluate the limit of the simplified expression.\newlineAs xx approaches 00, tan(3x)\tan(3x) approaches 3x3x and tan(2x)\tan(2x) approaches 2x2x because tan(x)\tan(x) is approximately equal to xx for small values of xx.\newlineTherefore, the limit becomes limx0(3×3x/2×2x)=limx0(9x/4x)\lim_{x \to 0}(3 \times 3x / 2 \times 2x) = \lim_{x \to 0}(9x / 4x).
  7. Simplify and Evaluate Limit: Simplify the expression and evaluate the limit.\newlineThe xx's cancel out, and we are left with 94\frac{9}{4}. Since there are no xx terms left, the limit is simply 94\frac{9}{4}.

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