Question

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\cot x-\cos x}{{\left(\pi -2x\right)}^3}=$

• A. $\frac{1}{20}$
• B. $\frac{1}{24}$
• C. $\frac{1}{29}$
• D. $\frac{1}{25}$

Answer 1

The correct option is B $\frac{1}{24}$

Consider the given function,

${lim}_{x\to \frac{\pi }{2}}\frac{\cot x-\cos x}{{(\pi -2x)}^3}$

Apply L-Hospital Roll, we get

${lim}_{x\to \frac{\pi }{2}}\frac{\frac{d}{dx}(\cot x-\cos x)}{\frac{d}{dx}{(\pi -2x)}^3}={lim}_{x\to \frac{\pi }{2}}\frac{-\cos e{c}^{2}x-(-\sin x)}{3.(-2){(\pi -2x)}^2}$

$={lim}_{x\to \frac{\pi }{2}}\frac{-\cos e{c}^{2}x+\sin x}{-6{(\pi -2x)}^2}\left(\frac{0}{0}\right)$

Again apply L-hospital Roll, we get

${lim}_{x\to \frac{\pi }{2}}\frac{\frac{d}{dx}(-\cos e{c}^{2}x+\sin x)}{-6\frac{d}{dx}{(\pi -2x)}^2}={lim}_{x\to \frac{\pi }{2}}\frac{-2\cos ecx.(-\cos ecx\cot x)+\cos x}{-\mathrm{6.2.}(-2)(\pi -2x)}$

$={lim}_{x\to \frac{\pi }{2}}\frac{2\cos e{c}^{2}x\cot x+\cos x}{24(\pi -2x)}$

Again apply L-Hospital Roll ,we get

$={lim}_{x\to \frac{\pi }{2}}\frac{2\frac{d}{dx}\cos e{c}^{2}x\cot x+\frac{d}{dx}\cos x}{24\frac{d}{dx}(\pi -2x)}$

$={lim}_{x\to \frac{\pi }{2}}\frac{2.\cos e{c}^{2}x(-\cos e{c}^{2}x)+2\cot x.2\cos ecx.\cos ecx\cot x}{24.(0-2)}$

$={lim}_{x\to \frac{\pi }{2}}\frac{-2.\cos e{c}^{4}x+2{\cot }^{2}x.\cos e{c}^{2}x.}{-48}$

$={lim}_{x\to \frac{\pi }{2}}\frac{-\cos e{c}^{4}x+{\cot }^{2}x.\cos e{c}^{2}x.}{-24}$

$=\frac{-\cos e{c}^4\frac{\pi }{4}+{\cot }^2\frac{\pi }{4}.\cos e{c}^2\frac{\pi }{4}}{-24}=\frac{-1^4+{0.1}^2}{-24}$

$=\frac{1}{24}$

Hence ,This is answer