Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 47

Find the limit $\displaystyle \lim\limits_{x \to \pi/4} \left( \frac{1 - \tan x}{\sin x - \cos x}\right)$

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\begin{equation}
\begin{aligned}
\lim\limits_{x \to \pi/4} \left( \frac{1 - \tan x}{\sin x - \cos x}\right) &= \lim\limits_{x \to \pi/4} \left( \frac{1-\frac{\sin x}{\cos x}}{\sin x - \cos x}\right)\\
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&= \lim\limits_{x \to \pi/4} \frac{\cos x - \sin x}{\sin x - \cos x} \left( \frac{1}{\cos x} \right)\\
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&= \lim\limits_{x \to \pi/4} \frac{-\cancel{(\sin x - \cos x)}}{\cancel{(\sin x - \cos x)}} \left( \frac{1}{\cos x}\right)\\
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& = \lim\limits_{x \to \pi/4} \frac{-1}{\cos x}\\
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& = \frac{-1}{\cos \frac{\pi}{4}}\\
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& = \frac{-1}{\frac{\sqrt{2}}{2}} = \frac{-2}{\sqrt{2}} && \text{(By rationalizing the denominator)}\\
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& = \frac{-2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\
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& = \frac{-\cancel{2}\sqrt{2}}{\cancel{2}} \\
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& = - \sqrt{2}
\end{aligned}
\end{equation}
$