1) Let's evaluate the limit of $\frac{\sin (-x)}{x}$ when $x\to 0$

ⓘ The function sin is continuous, its limit is its value at the limit of its argument

The limit of $\sin (-x)$ is $\sin \left(0\right)$ when $x\to 0$

Let's calculate: $\sin \left(0\right)$

ⓘ The sine is the y-coordinate (vertical axis) of the point reached on the unit circle after a rotation of $0$ radians from the horizontal axis pointing to the right, this is a remarkable value: $\sin \left(0\right)$ = $0$

$0$

For $\frac{\sin (-x)}{x}$, we encounter the indeterminate form 0/0 as $x$ tends to $0$

2) We look for a Taylor expansion of the expression: $\sin (-x)$

Apply the Taylor development $\sin \left(A\right)+\cos \left(A\right)·(X-A)$ of sin with X=$-x$ around A=$0$ (order N=1)

$\sin \left(0\right)+\cos \left(0\right)·(-x+0)$

ⓘ The sine is the y-coordinate (vertical axis) of the point reached on the unit circle after a rotation of $0$ radians from the horizontal axis pointing to the right, this is a remarkable value: $\sin \left(0\right)$ = $0$■The cosine is the x-coordinate (horizontal axis) of the point reached on the unit circle after a rotation of $0$ radians from the horizontal axis pointing to the right, this is a remarkable value: $\cos \left(0\right)$ = $1$■Adding 0 changes nothing: $-x+0$ = $-x$■Multiplying by 1 changes nothing: $1·(-x)$ = $-x$

$0-x$

3) We redo the calculation using the Taylor expansion: $-x$

$-\frac{x}{x}$

ⓘ The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor: $-\frac{x}{x}$ = $-1$

$-1$

So the limit of $\frac{\sin (-x)}{x}$ is $-1$ when $x\to 0$

▶Limit:   $-1$