$\lim _{x \rightarrow 0}\left(\csc 2 x-\frac{1}{2 x}\right)=$

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

Recognize and Rewrite Limit: We are given the limit to evaluate: $\newline$ $\lim_{x \to 0}(\csc 2x-\frac{1}{2x})$ $\newline$ First, we recognize that $\csc(2x)$ is the reciprocal of $\sin(2x)$ , so we can rewrite the limit as: $\newline$ $\lim_{x \to 0}(\frac{1}{\sin(2x)} - \frac{1}{2x})$
Find Common Denominator: We need to find a common denominator to combine the terms in the limit expression. The common denominator for $\sin(2x)$ and $2x$ is $2x\sin(2x)$ . So we rewrite the expression with the common denominator: $\newline$ $\lim_{x \rightarrow 0}\left(\frac{2x - \sin(2x)}{2x\sin(2x)}\right)$
Apply L'Hôpital's Rule: We can now apply L'Hôpital's Rule because the limit is in an indeterminate form $0/0$ . L'Hôpital's Rule states that if the limit of $f(x)/g(x)$ as $x$ approaches a value $c$ is in the form $0/0$ 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. $\newline$ So we take the derivative of the numerator and the denominator: $\newline$ Derivative of the numerator: $\frac{d}{dx}(2x - \sin(2x)) = 2 - 2\cos(2x)$ $\newline$ Derivative of the denominator: $\frac{d}{dx}(2x\sin(2x)) = 2\sin(2x) + 4x\cos(2x)$
Derivatives and Limit: Now we apply L'Hôpital's Rule by taking the limit of the derivatives: $\lim_{x \rightarrow 0}\left(\frac{2 - 2\cos(2x)}{2\sin(2x) + 4x\cos(2x)}\right)$
Apply L'Hôpital's Rule Again: We evaluate the limit of the new expression as $x$ approaches $0$ :
$\lim_{x \to 0}\left(\frac{2 - 2\cos(2x)}{2\sin(2x) + 4x\cos(2x)}\right) = \frac{2 - 2\cos(0)}{2\sin(0) + 4\cdot 0\cdot \cos(0)} = \frac{2 - 2\cdot 1}{2\cdot 0 + 4\cdot 0\cdot 1} = \frac{2 - 2}{0 + 0} = \frac{0}{0}$
We have reached another indeterminate form, so we must apply L'Hôpital's Rule again.
Derivatives and Limit: We take the derivative of the numerator and the denominator again: $\newline$ Derivative of the numerator: $\frac{d}{dx}(2 - 2\cos(2x)) = 4\sin(2x)$ $\newline$ Derivative of the denominator: $\frac{d}{dx}(2\sin(2x) + 4x\cos(2x)) = 4\cos(2x) - 8x\sin(2x)$
Derivatives and Limit: We take the derivative of the numerator and the denominator again: $\newline$ Derivative of the numerator: $\frac{d}{dx}(2 - 2\cos(2x)) = 4\sin(2x)$ $\newline$ Derivative of the denominator: $\frac{d}{dx}(2\sin(2x) + 4x\cos(2x)) = 4\cos(2x) - 8x\sin(2x)$ We apply L'Hôpital's Rule once more by taking the limit of the second derivatives: $\newline$ $\lim_{x \rightarrow 0}\left(\frac{4\sin(2x)}{4\cos(2x) - 8x\sin(2x)}\right)$
Derivatives and Limit: We take the derivative of the numerator and the denominator again: $\newline$ Derivative of the numerator: $\frac{d}{dx}(2 - 2\cos(2x)) = 4\sin(2x)$ $\newline$ Derivative of the denominator: $\frac{d}{dx}(2\sin(2x) + 4x\cos(2x)) = 4\cos(2x) - 8x\sin(2x)$ We apply L'Hôpital's Rule once more by taking the limit of the second derivatives: $\newline$ $\lim_{x \rightarrow 0}\left(\frac{4\sin(2x)}{4\cos(2x) - 8x\sin(2x)}\right)$ We evaluate the limit of the new expression as x approaches $0$ : $\newline$ $\lim_{x \rightarrow 0}\left(\frac{4\sin(2x)}{4\cos(2x) - 8x\sin(2x)}\right) = \frac{4\sin(0)}{4\cos(0) - 8\cdot 0 \cdot \sin(0)} = \frac{4\cdot 0}{4\cdot 1 - 0} = \frac{0}{4} = 0$