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lim_(x rarr0)(csc 2x-(1)/(2x))=

lim⁡x→0(csc⁡2x−12x)= \lim _{x \rightarrow 0}\left(\csc 2 x-\frac{1}{2 x}\right)=

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Full solution

Q. lim⁡x→0(csc⁡2x−12x)= \lim _{x \rightarrow 0}\left(\csc 2 x-\frac{1}{2 x}\right)=
  1. Recognize and Rewrite Limit: We are given the limit to evaluate: \newlinelim⁡x→0(csc⁡2x−12x)\lim_{x \to 0}(\csc 2x-\frac{1}{2x})\newlineFirst, we recognize that csc⁡(2x)\csc(2x) is the reciprocal of sin⁡(2x)\sin(2x), so we can rewrite the limit as:\newlinelim⁡x→0(1sin⁡(2x)−12x)\lim_{x \to 0}(\frac{1}{\sin(2x)} - \frac{1}{2x})
  2. Find Common Denominator: We need to find a common denominator to combine the terms in the limit expression. The common denominator for sin⁡(2x)\sin(2x) and 2x2x is 2xsin⁡(2x)2x\sin(2x). So we rewrite the expression with the common denominator:\newlinelim⁡x→0(2x−sin⁡(2x)2xsin⁡(2x))\lim_{x \rightarrow 0}\left(\frac{2x - \sin(2x)}{2x\sin(2x)}\right)
  3. Apply L'Hôpital's Rule: We can now apply L'Hôpital's Rule because the limit is in an indeterminate form 0/00/0. 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 form 0/00/0 or ∞/∞\infty/\infty, then the limit is the same as the limit of the derivatives of the numerator and the denominator, provided that the limit of the derivatives exists.\newlineSo we take the derivative of the numerator and the denominator:\newlineDerivative of the numerator: ddx(2x−sin⁡(2x))=2−2cos⁡(2x)\frac{d}{dx}(2x - \sin(2x)) = 2 - 2\cos(2x)\newlineDerivative of the denominator: ddx(2xsin⁡(2x))=2sin⁡(2x)+4xcos⁡(2x)\frac{d}{dx}(2x\sin(2x)) = 2\sin(2x) + 4x\cos(2x)
  4. Derivatives and Limit: Now we apply L'Hôpital's Rule by taking the limit of the derivatives: lim⁡x→0(2−2cos⁡(2x)2sin⁡(2x)+4xcos⁡(2x))\lim_{x \rightarrow 0}\left(\frac{2 - 2\cos(2x)}{2\sin(2x) + 4x\cos(2x)}\right)
  5. Apply L'Hôpital's Rule Again: We evaluate the limit of the new expression as xx approaches 00:
    lim⁡x→0(2−2cos⁡(2x)2sin⁡(2x)+4xcos⁡(2x))=2−2cos⁡(0)2sin⁡(0)+4⋅0⋅cos⁡(0)=2−2⋅12⋅0+4⋅0⋅1=2−20+0=00\lim_{x \to 0}\left(\frac{2 - 2\cos(2x)}{2\sin(2x) + 4x\cos(2x)}\right) = \frac{2 - 2\cos(0)}{2\sin(0) + 4\cdot 0\cdot \cos(0)} = \frac{2 - 2\cdot 1}{2\cdot 0 + 4\cdot 0\cdot 1} = \frac{2 - 2}{0 + 0} = \frac{0}{0}
    We have reached another indeterminate form, so we must apply L'Hôpital's Rule again.
  6. Derivatives and Limit: We take the derivative of the numerator and the denominator again:\newlineDerivative of the numerator: ddx(2−2cos⁡(2x))=4sin⁡(2x)\frac{d}{dx}(2 - 2\cos(2x)) = 4\sin(2x)\newlineDerivative of the denominator: ddx(2sin⁡(2x)+4xcos⁡(2x))=4cos⁡(2x)−8xsin⁡(2x)\frac{d}{dx}(2\sin(2x) + 4x\cos(2x)) = 4\cos(2x) - 8x\sin(2x)
  7. Derivatives and Limit: We take the derivative of the numerator and the denominator again:\newlineDerivative of the numerator: ddx(2−2cos⁡(2x))=4sin⁡(2x)\frac{d}{dx}(2 - 2\cos(2x)) = 4\sin(2x)\newlineDerivative of the denominator: ddx(2sin⁡(2x)+4xcos⁡(2x))=4cos⁡(2x)−8xsin⁡(2x)\frac{d}{dx}(2\sin(2x) + 4x\cos(2x)) = 4\cos(2x) - 8x\sin(2x) We apply L'Hôpital's Rule once more by taking the limit of the second derivatives:\newlinelim⁡x→0(4sin⁡(2x)4cos⁡(2x)−8xsin⁡(2x))\lim_{x \rightarrow 0}\left(\frac{4\sin(2x)}{4\cos(2x) - 8x\sin(2x)}\right)
  8. Derivatives and Limit: We take the derivative of the numerator and the denominator again:\newlineDerivative of the numerator: ddx(2−2cos⁡(2x))=4sin⁡(2x)\frac{d}{dx}(2 - 2\cos(2x)) = 4\sin(2x)\newlineDerivative of the denominator: ddx(2sin⁡(2x)+4xcos⁡(2x))=4cos⁡(2x)−8xsin⁡(2x)\frac{d}{dx}(2\sin(2x) + 4x\cos(2x)) = 4\cos(2x) - 8x\sin(2x) We apply L'Hôpital's Rule once more by taking the limit of the second derivatives:\newlinelim⁡x→0(4sin⁡(2x)4cos⁡(2x)−8xsin⁡(2x))\lim_{x \rightarrow 0}\left(\frac{4\sin(2x)}{4\cos(2x) - 8x\sin(2x)}\right) We evaluate the limit of the new expression as x approaches 00:\newlinelim⁡x→0(4sin⁡(2x)4cos⁡(2x)−8xsin⁡(2x))=4sin⁡(0)4cos⁡(0)−8⋅0⋅sin⁡(0)=4⋅04⋅1−0=04=0\lim_{x \rightarrow 0}\left(\frac{4\sin(2x)}{4\cos(2x) - 8x\sin(2x)}\right) = \frac{4\sin(0)}{4\cos(0) - 8\cdot 0 \cdot \sin(0)} = \frac{4\cdot 0}{4\cdot 1 - 0} = \frac{0}{4} = 0

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