Question

${lim}_{x\to \frac{\pi }{8}}\frac{cot4x-cos4x}{(\pi -8x{)}^3}$

Answer 1

${lim}_{x\to \frac{\pi }{8}}\frac{cot4x-cos4x}{(\pi -8x{)}^3}$

Let $x=\frac{\pi }{8}+y\Rightarrow y=x-\frac{\pi }{8}$

as $x\to \frac{\pi }{8},y\to 0$

$={lim}_{y\to 0}\frac{cot4(\frac{\pi }{8}+y)-cos4(\frac{\pi }{8}+y)}{[\pi -8(\frac{\pi }{8}+y){]}^3}$

$={lim}_{y\to 0}\frac{cot(\frac{\pi }{2}+4y)-cos(\frac{\pi }{2}+4y)}{(-8y{)}^3}$

$={lim}_{y\to 0}\frac{-tan4y+sin4y}{-(8{)}^3{y}^3}$

$={lim}_{y\to 0}\frac{\frac{sin4y}{cos4y}-sin4y}{8^3{y}^2}$

$={lim}_{y\to 0}\frac{sin4y-sin4ycos4y}{cos4y\times y^3\times 8^3}$

$={lim}_{y\to 0}\frac{sin4y(1-cos4y)}{cos4y\times y^3\times 8^3}$

$={lim}_{y\to 0}\frac{sin4y\left(2si{n}^{2}2y\right)}{cos4y\times y^3\times 8^3}$

$=\frac{2}{8^3}{lim}_{y\to 0}\frac{sin4y}{y}\times \frac{si{n}^{2}2y}{y}\times \frac{1}{cos4y}$

$=\frac{2}{8^3}({lim}_{y\to 0}\frac{sin4y}{4y}\times 4)\times {\left({lim}_{y\to 0}\frac{si{n}^{2}2y}{2y}\right)}^2\times 4\times \frac{1}{{lim}_{y\to 0}cos4y}$

$=\frac{2}{8^3}(1\times 4)\times \left(1\right)\times 4\times \frac{1}{1}$

$[\because {lim}_{\theta \to 0}\frac{sin\theta }{\theta }=1,{lim}_{\theta \to 0}cos\theta =1]$

$=\frac{2\times 4\times 4}{8\times 8\times 8}=\frac{1}{16}$