Rationalise?

This could well be done in one line by L'hopital but I'll use a pure trigonometric approach. $\small{\color{#3D99F6}{(Using~1-\cos x=2 \sin^2 \frac{x}{2} ,\text{ sinx=2 sin}\frac{x}{2} \cos \frac{x}{2})}}$ $\dfrac{\tan x + \cos x - 1}{\tan x - \cos x + 1} =\dfrac{2\frac{\sin(x/2) \cos(\frac{x}{2})}{\cos x}-2 \sin^2 (\frac{x}{2})}{2\frac{\sin(x/2) \cos(\frac{x}{2})}{\cos x}+2 \sin^2 (\frac{x}{2})}$ $=\dfrac{\frac{\cos (\frac{x}{2})}{\cos x}-\sin (\frac{x}{2})}{\frac{\cos (\frac{x}{2})}{\cos x}+\sin (\frac{x}{2})}$ Applying limits, we get: $\lim_{x \rightarrow 0} \dfrac{\frac{\cos (\frac{x}{2})}{\cos x}-\sin (\frac{x}{2})}{\frac{\cos (\frac{x}{2})}{\cos x}+\sin (\frac{x}{2})}=\large \dfrac{1-0}{1+0}=\boxed1$

As $x\rightarrow 0$ , we see that the expression $\frac{\tan x + \cos x - 1}{\tan x - \cos x + 1}$ becomes $\frac{0}{0}$ , hence we can use L'Hopital's rule:

$\lim_{x \rightarrow 0} \dfrac{\tan x + \cos x - 1}{\tan x - \cos x + 1} = \lim_{x \rightarrow 0} \dfrac{\sec^2 x -\sin x}{\sec^2 x +\sin x} = \dfrac{1-0}{1+0}=\boxed{1}$