Evaluate the following limit: _ x 8 4x- 4x ( -8x)^3

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{8}}\frac{\cot4\text{x}-\cos4\text{x}}{(\pi-8\text{x})^3}$

✓

Answer

$\lim\limits_{\text{x}\rightarrow\frac{\pi}{8}}\frac{\cot4\text{x}-\cos4\text{x}}{(\pi-8\text{x})^3}$
When $\text{x}\rightarrow\frac\pi8,\frac\pi8-\text{x}\rightarrow0,$ let $\frac{\pi}{8}-\text{x}=\text{y}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cot4\big(\frac\pi8-\text{y}\big)-\cos4\big(\frac\pi8-\text{y}\big)}{(8)^3\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\cot4\big(\frac\pi2-4\text{y}\big)-\cos4\big(\frac\pi2-4\text{y}\big)}{(8)^3\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan4\text{y}-\sin4\text{y}}{(8)^2\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\frac{\sin4\text{y}}{\cos4\text{y}}-\sin4\text{y}}{8^3\text{y}^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}-\sin4\text{y}\cos4\text{y}}{\cos4\text{y}\times\text{y}^3\times8^3}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}\big(2\sin^22\text{y}\big)}{\cos4\text{y}\times\text{y}^3\times8^3}$
$=\frac{2}{8^3}\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}}{\text{y}}\times\frac{\sin^22\text{y}}{\text{y}^2}\times\frac{1}{\cos4\text{y}}$
$=\frac{2}{8^3}\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin4\text{y}}{4\text{y}}\times4\Big)\times\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin2\text{y}}{2\text{y}}\Big)^2\times4\times\frac{1}{\lim\limits_{\text{y}\rightarrow0}\cos4\text{y}}$
$=\frac{2}{8^2}(1\times4)\times(1)\times4\times\frac11$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\cos\theta=1\Big]$
$=\frac{2\times4\times4}{8\times8\times8}$
$\frac{1}{16}$