Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{ x}}$
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{ x}}$
✓
Answer
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{ x}}$
$\Rightarrow\text{x}\rightarrow\frac{\pi}{4},\text{x}-\frac{\pi}{4}\rightarrow0,$ let $\text{x}-\frac{\pi}{4}=\text{y}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{x}}=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\frac{\Big(1-\sin2\big(\text{y}+\frac{\pi}{4}\big)\Big)}{1+\cos4\big(\text{y}+\frac{\pi}{4}\big)}$
$=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\Bigg(\frac{1-\sin\big(\frac\pi2+2\text{y}\big)}{1+\cos(\pi+4\text{y})}\Bigg)$
$=\lim\limits_{\text{x}-\frac\pi4\rightarrow{0}}\frac{1-\cos2\text{y}}{1-\cos4\text{y}}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\text{y}}{2\sin^22\text{y}}$
$=\frac{\lim\limits_{\text{y}\rightarrow{0}}\sin^2\text{y}}{\lim\limits_{\text{y}\rightarrow{0}}\sin^22\text{y}}$
$=\frac{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\text{y}}{\text{y}}\Big)^2\times\text{y}^2}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin2\text{y}}{2\text{y}}\Big)^2\times4\text{y}^2}$
$=\frac{1\times\text{y}^2}{1\times4\text{y}^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac14$
$\Rightarrow\text{x}\rightarrow\frac{\pi}{4},\text{x}-\frac{\pi}{4}\rightarrow0,$ let $\text{x}-\frac{\pi}{4}=\text{y}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{1-\sin2\text{x}}{1+\cos4\text{x}}=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\frac{\Big(1-\sin2\big(\text{y}+\frac{\pi}{4}\big)\Big)}{1+\cos4\big(\text{y}+\frac{\pi}{4}\big)}$
$=\lim\limits_{\text{x}-\frac{\pi}{4}\rightarrow{0}}\Bigg(\frac{1-\sin\big(\frac\pi2+2\text{y}\big)}{1+\cos(\pi+4\text{y})}\Bigg)$
$=\lim\limits_{\text{x}-\frac\pi4\rightarrow{0}}\frac{1-\cos2\text{y}}{1-\cos4\text{y}}$
$=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\text{y}}{2\sin^22\text{y}}$
$=\frac{\lim\limits_{\text{y}\rightarrow{0}}\sin^2\text{y}}{\lim\limits_{\text{y}\rightarrow{0}}\sin^22\text{y}}$
$=\frac{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\text{y}}{\text{y}}\Big)^2\times\text{y}^2}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin2\text{y}}{2\text{y}}\Big)^2\times4\text{y}^2}$
$=\frac{1\times\text{y}^2}{1\times4\text{y}^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\Big]$
$=\frac14$
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