What is now this limit doing here?

$\begin{aligned} \lim_{x \to \frac{\pi}{2}} \left(\sin x - \cos x \right)^{\tan x} & = \lim_{x \to \frac{\pi}{2}} \sin x \left(1 - \frac{1}{\color{#3D99F6}{\tan x}} \right)^{\color{#3D99F6}{\tan x}} \quad \quad \small \color{#3D99F6}{\text{Let } n = \tan x}\\ & = \lim_{x \to \frac{\pi}{2}} \sin x \cdot{} \lim_{\color{#3D99F6}{n} \to \infty} \left(1 - \frac{1}{\color{#3D99F6}{n}} \right)^{\color{#3D99F6}{n}} \\ & = 1 \cdot{} \frac{1}{e} \approx \boxed{0.36787944} \end{aligned}$

By inserting the limit, we can clearly see that the expression takes the form ${ 1 }^{ \infty }$ , which can be easily solved by using the exponential conversion. So, the following limit can be written as:

${ e }^{ \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ (\sin { x-\cos { x } -1) } \tan { x } } }$

Which can be further simplified into

${ e }^{ \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { (\sin { x-\cos { x } -1) } }{ \cot { x } } } }$

Now, as we can clearly see that the limit in the exponent tends to a $\frac {0} {0}$ form, so we'll use the L'Hôpital's Rule to simplify it further.

Differentiating the numerator and denominator in the power, with respect to $x$ , we get:

${ e }^{ \lim _{ x\rightarrow \frac { \pi }{ 2 } }{ \frac { \cos { x } +\sin { x } }{ -\csc ^{ 2 }{ x } } } }$

Now, inserting the limit into the formed expression:

${ e }^{ \frac { 0+1 }{ -1 } }={ e }^{ -1 }=0.36787944$

Cheers! :)