Evaluate the limit: $\newline$ $\lim _{x \rightarrow 0}\left(\frac{1}{x}-\cot x\right)=$

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Evaluate definite integrals using the power rule

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Q. Evaluate the limit:

\newline

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

Recognize singularity: We are given the limit to evaluate: $\newline$ $\lim_{x \to 0}\left(\frac{1}{x} - \cot(x)\right)$ $\newline$ First, we recognize that both $\frac{1}{x}$ and $\cot(x)$ have singularities at $x = 0$ , which means we need to find a way to combine these terms to eliminate the singularity. We can do this by finding a common denominator.
Find common denominator: To find a common denominator, we need to express $\cot(x)$ in terms of sine and cosine: $\newline$ $\cot(x) = \frac{\cos(x)}{\sin(x)}$ $\newline$ Now we can rewrite the limit as: $\newline$ $\lim_{x \to 0}\left(\frac{1}{x} - \frac{\cos(x)}{\sin(x)}\right)$
Combine fractions: The common denominator between $\frac{1}{x}$ and $\frac{\cos(x)}{\sin(x)}$ is $x\sin(x)$ . We multiply the numerator and denominator of $\frac{1}{x}$ by $\sin(x)$ to get: $\newline$ $\lim_{x \to 0}\left(\frac{\sin(x)}{x\sin(x)} - \frac{\cos(x)}{\sin(x)}\right)$
Apply L'Hôpital's Rule: Now that we have a common denominator, we can combine the fractions: $\lim_{x \to 0}\left(\frac{\sin(x) - x\cos(x)}{x\sin(x)}\right)$
Differentiate numerator and denominator: We can now apply L'Hôpital's Rule because the limit is in an indeterminate form $(0/0)$ . We differentiate the numerator and the denominator with respect to $x$ :
Numerator's derivative: $\frac{d}{dx}(\sin(x) - x\cos(x)) = \cos(x) - (\cos(x) - x\sin(x))$
Denominator's derivative: $\frac{d}{dx}(x\sin(x)) = \sin(x) + x\cos(x)$
Evaluate limit: We simplify the derivatives: $\newline$ Numerator's derivative simplifies to: $\cos(x) - \cos(x) + x\sin(x) = x\sin(x)$ $\newline$ Denominator's derivative is already simplified: $\sin(x) + x\cos(x)$ $\newline$ Now we can rewrite the limit using these derivatives: $\newline$ $\lim_{x \to 0}\left(\frac{x\sin(x)}{\sin(x) + x\cos(x)}\right)$
Apply L'Hôpital's Rule again: We can now evaluate the limit by substituting $x = 0$ into the simplified expression: $\lim_{x \to 0}\left(\frac{x\sin(x)}{\sin(x) + x\cos(x)}\right) = \frac{0\sin(0)}{\sin(0) + 0\cos(0)} = \frac{0}{0}$ However, this is still an indeterminate form, which means we need to apply L'Hôpital's Rule again.
Differentiate second time: We differentiate the numerator and denominator again: $\newline$ Numerator's second derivative: $\frac{d}{dx}(x\sin(x)) = \sin(x) + x\cos(x)$ $\newline$ Denominator's second derivative: $\frac{d}{dx}(\sin(x) + x\cos(x)) = \cos(x) + \cos(x) - x\sin(x) = 2\cos(x) - x\sin(x)$
Evaluate final limit: We simplify the second derivatives: $\newline$ Numerator's second derivative simplifies to: $\sin(x) + x\cos(x)$ $\newline$ Denominator's second derivative simplifies to: $2\cos(x) - x\sin(x)$ $\newline$ Now we can rewrite the limit using these second derivatives: $\newline$ $\lim_{x \to 0}\left(\frac{\sin(x) + x\cos(x)}{2\cos(x) - x\sin(x)}\right)$
Evaluate final limit: We simplify the second derivatives: $\newline$ Numerator's second derivative simplifies to: $\sin(x) + x\cos(x)$ $\newline$ Denominator's second derivative simplifies to: $2\cos(x) - x\sin(x)$ $\newline$ Now we can rewrite the limit using these second derivatives: $\newline$ $\lim_{x \to 0}\left(\frac{\sin(x) + x\cos(x)}{2\cos(x) - x\sin(x)}\right)$ We can now evaluate the limit by substituting $x = 0$ into the second derivatives: $\newline$ $\lim_{x \to 0}\left(\frac{\sin(x) + x\cos(x)}{2\cos(x) - x\sin(x)}\right) = \frac{\sin(0) + 0\cos(0)}{2\cos(0) - 0\sin(0)} = \frac{0}{2} = 0$