Question

Evaluate: $$ \lim_{x \to \frac{\pi}{2}} \frac{\cot x - \cos x}{(\pi - 2x)^3} $$ 

• A. $\frac{1}{24}$
• B. $\frac{1}{16}$
• C. $\frac{1}{8}$
• D. $\frac{1}{4}$

Hint:

Logic Tip: L'Hopital's rule can be used here since it's a $0/0$ form, but differentiating $(\pi-2x)^3$ and the complex trigonometric numerator three times is highly prone to algebraic errors. The substitution $x = \pi/2 - h$ is much safer and elegantly brings the limit to standard forms.

Answer 1

The Correct Option is B

Concept:  
When dealing with limits involving trigonometric functions where $x \to a$, substituting $x = a - h$ (or $x = a + h$) where $h \to 0$ helps reduce the expression into standard limits like $\lim_{h \to 0} \frac{\sin h}{h} = 1$. 

Step 1: Apply substitution. 
Let $$ I = \lim_{x \to \frac{\pi}{2}} \frac{\cot x - \cos x}{(\pi - 2x)^3} $$ First simplify numerator: $$ \cot x - \cos x = \frac{\cos x}{\sin x} - \cos x = \frac{\cos x (1 - \sin x)}{\sin x} $$ So, $$ I = \lim_{x \to \frac{\pi}{2}} \frac{\cos x (1 - \sin x)}{\sin x (\pi - 2x)^3} $$ Now substitute: $$ x = \frac{\pi}{2} - h \quad (h \to 0) $$ Then, $$ \pi - 2x = 2h $$ So, $$ I = \lim_{h \to 0} \frac{\cos\left(\frac{\pi}{2} - h\right)\left(1 - \sin\left(\frac{\pi}{2} - h\right)\right)} {\sin\left(\frac{\pi}{2} - h\right)(2h)^3} $$ 

Step 2: Use identities. 
$$ \cos\left(\frac{\pi}{2} - h\right) = \sin h, \quad \sin\left(\frac{\pi}{2} - h\right) = \cos h $$ So, $$ I = \lim_{h \to 0} \frac{\sin h (1 - \cos h)}{\cos h \cdot 8h^3} $$ Using: $$ 1 - \cos h = 2\sin^2\left(\frac{h}{2}\right) $$ $$ I = \lim_{h \to 0} \frac{2 \sin h \sin^2\left(\frac{h}{2}\right)}{8h^3} $$ $$ I = \frac{1}{4} \lim_{h \to 0} \frac{\sin h}{h} \cdot \frac{1}{\cos h} \cdot \frac{\sin^2\left(\frac{h}{2}\right)}{h^2} $$ 

Step 3: Apply standard limits. 
Rewrite: $$ \frac{\sin^2\left(\frac{h}{2}\right)}{h^2} = \left(\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right)^2 \cdot \frac{1}{4} $$ So, $$ I = \frac{1}{4} \cdot \left(1\right) \cdot \left(\frac{1}{1}\right) \cdot (1)^2 \cdot \frac{1}{4} $$ $$ I = \frac{1}{16} $$ 

Final Answer: 
$$ \boxed{\frac{1}{16}} $$