Answer
$\frac{{\sqrt 2 }}{2}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \pi /4} \frac{{\sin x - \cos x}}{{\tan x - 1}} \cr
& {\text{Use the quotient property of limits}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \pi /4} \left( {\sin x - \cos x} \right)}}{{\mathop {\lim }\limits_{x \to \pi /4} \left( {\tan x - 1} \right)}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \pi /4} \left( {\sin x} \right) - \mathop {\lim }\limits_{x \to \pi /4} \left( {\cos x} \right)}}{{\mathop {\lim }\limits_{x \to \pi /4} \left( {\tan x} \right) - \mathop {\lim }\limits_{x \to \pi /4} \left( 1 \right)}} \cr
& {\text{Evaluate the limit}} \cr
& = \frac{{\sin \left( {\frac{\pi }{4}} \right) - \cos \left( {\frac{\pi }{4}} \right)}}{{\tan \left( {\frac{\pi }{4}} \right) - 1}}{\text{ = }}\frac{0}{0} \cr
& {\text{The limit is an indeterminate form type }}\frac{0}{0},{\text{ so we can apply}} \cr
& {\text{the L'Hospital's Rule}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to \pi /4} \frac{{\sin x - \cos x}}{{\tan x - 1}} = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\frac{d}{{dx}}\left[ {\sin x - \cos x} \right]}}{{\frac{d}{{dx}}\left[ {\tan x - 1} \right]}} \cr
& = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\cos x + \sin x}}{{{{\sec }^2}x}} \cr
& {\text{Evaluate the limit when }}x \to \pi /4 \cr
& = \frac{{\cos \left( {\pi /4} \right) + \sin \left( {\pi /4} \right)}}{{{{\sec }^2}\left( {\pi /4} \right)}} \cr
& = \frac{{\sqrt 2 /2 + \sqrt 2 /2}}{2} \cr
& = \frac{{\sqrt 2 }}{2} \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to \pi /4} \frac{{\sin x - \cos x}}{{\tan x - 1}} = \frac{{\sqrt 2 }}{2} \cr} $$
Another method
Rewrite the denominator:
$\displaystyle\lim_{x \to \pi /4} \frac{{\sin x - \cos x}}{{\tan x - 1}}= \displaystyle\lim_{x \to \pi /4} \frac{{\sin x - \cos x}}{{\frac{\sin x}{\cos x} - 1}}=\displaystyle\lim_{x \to \pi /4} \frac{{(\sin x - \cos x)\cos x}}{\sin x-\cos x}=\displaystyle\lim_{x \to \pi /4}\cos x=\cos\frac{\pi}{4}=\frac{{\sqrt 2 }}{2}$
