Question

Limit $$ \lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{\csc x - 2} $$

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Answer


To find the limit: $$ \lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{\csc x - 2} $$

First, evaluate $\cot^2 x$ and $\csc x$ at $x = \frac{\pi}{6}$:

  1. For $\cot \left( \frac{\pi}{6} \right)$: $$ \cot \left( \frac{\pi}{6} \right) = \frac{1}{\tan \left( \frac{\pi}{6} \right)} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} $$ Thus, $$ \cot^2 \left( \frac{\pi}{6} \right) = (\sqrt{3})^2 = 3 $$

  2. For $\csc \left( \frac{\pi}{6} \right)$: $$ \csc \left( \frac{\pi}{6} \right) = \frac{1}{\sin \left( \frac{\pi}{6} \right)} = \frac{1}{\frac{1}{2}} = 2 $$

Now, substitute these values into the limit expression: $$ \lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{\csc x - 2} = \frac{3 - 3}{2 - 2} = \frac{0}{0} $$

The $\frac{0}{0}$ form indicates that we should employ L'Hôpital's Rule. Apply L'Hôpital's Rule by differentiating the numerator and the denominator:

  • Differentiate the numerator: $$ \frac{d}{dx} (\cot^2 x - 3) = \frac{d}{dx} (\cot^2 x) = 2 \cot x \cdot (-\csc^2 x) $$
  • Differentiate the denominator: $$ \frac{d}{dx} (\csc x - 2) = \frac{d}{dx} (\csc x) = -\csc x \cot x $$

Applying L'Hôpital's Rule: $$ \lim_{x \to \frac{\pi}{6}} \frac{2 \cot x \cdot (-\csc^2 x)}{-\csc x \cot x} = \lim_{x \to \frac{\pi}{6}} 2 \csc x $$

Evaluate at $x = \frac{\pi}{6}$: $$ 2 \csc \left( \frac{\pi}{6} \right) = 2 \cdot 2 = 4 $$

Thus, the limit is: $$ \boxed{4} $$


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