Question

lim x approach to π/2 cotx-cosx/(π-2x)³

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Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\tt \:\lim_{x \: \to \: \frac{\pi}{2} } \:\dfrac{cotx - cosx }{{(\pi \: - 2x)}^{3}}$

☆ On substituting the value of x directly, we get indeterminant form.

☆ To evaluate this limit, the following formula have used :-

$\boxed{ \red{\tt \:\lim_{x\to0} \:\dfrac{sinx}{x} = 1}}$

☆ Now, Consider

$\tt \:\lim_{x \: \to \: \frac{\pi}{2} } \:\dfrac{cotx - cosx }{{(\pi \: - 2x)}^{3}}$

$\tt \ \: :  ⟼ Let \: x \: = \dfrac{\pi}{2} - y$

$\tt \ \: :  ⟼ \: As \: x \: \to \: \dfrac{\pi}{2} \: \: \: \tt\implies \:y \: \to \: 0$

$\tt \ \: :  ⟼ \tt \:\lim_{y\to0} \:\dfrac{cot(\dfrac{\pi}{2} - y) - cos(\dfrac{\pi}{2} - y) }{ { \bigg(\pi \: - 2(\dfrac{\pi}{2} - y) \bigg)}^{3} }$

$\tt \ \: :  ⟼ \tt \:\lim_{y\to0} \:\dfrac{tany - siny}{(\pi \: - \pi \: + 2y) ^{3} }$

$\tt \ \: :  ⟼ \tt \:\lim_{y\to0} \:\dfrac{\dfrac{siny}{cosy} - siny}{ {8y}^{3} }$

$\tt \ \: :  ⟼ \tt \:\dfrac{1}{8} \: \lim_{y\to0} \:\dfrac{siny \bigg(\dfrac{1}{cosy} - 1 \bigg) }{ {y}^{3} }$

$\tt \ \: :  ⟼ \dfrac{1}{8} \tt \:\lim_{y\to0} \:\dfrac{siny(1 - cosy)}{cosy \: \times ({y}^{3} )}$

$\tt \:  = \dfrac{1}{8} \tt \:\lim_{y\to0} \:\dfrac{siny}{y} \times \tt \:\lim_{y\to0} \:\dfrac{1}{cosy} \times \tt \:\lim_{y\to0} \:\dfrac{1 - cosy}{ {y}^{2} }$

$\tt \ \: :  ⟼ \dfrac{1}{8} \times 1 \times 1 \times \tt \:\lim_{y\to0} \:\dfrac{2 {sin}^{2}\dfrac{y}{2} }{ {y}^{2} }$

$\tt \ \: :  ⟼ \dfrac{1}{4} \: \tt \:\lim_{y\to0} \:\dfrac{ {sin}^{2} \dfrac{y}{2} }{\dfrac{ {y}^{2} }{4} \times 4}$

$\tt \ \: :  ⟼ \dfrac{1}{4} \times \dfrac{1}{4}$

$\tt \ \: :  ⟼ \dfrac{1}{16}$