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

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

The limit of $\sin \left(x\right)$ 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 $\frac{\sin \left(x\right)}{x}$, the limit takes the indeterminate form 0/0 when $x$ → $0$

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

Using Taylor's formula: $\sin \left(A\right)+\cos \left(A\right)·(X-A)$ for the function sin, we compute the Taylor expansion for X=$x$ at the point A=$0$ of 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) Let's restart 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 $x$: $\frac{x}{x}$ = $1$

$1$

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

▶Limit:   $1$