1) Compute the limit of $\frac{\sin \left(2x\right)}{x}$ as $x\to 0$

ⓘ The limit of a product of finite limits is the product of the limits

The limit of $2x$ is $2·0$ when $x\to 0$

⇥1.1) Let's calculate: $2·0$

⇥ⓘ A product with a zero factor is zero: $2·0$ = $0$

⇥ $0$

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

The limit of $\sin \left(2x\right)$ is $\sin \left(0\right)$ when $x\to 0$

⇥1.2) Let's evaluate: $\sin \left(0\right)$

⇥ⓘ 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 $\frac{\sin \left(2x\right)}{x}$, we encounter the indeterminate form 0/0 as $x$ tends to $0$

2) We need a Taylor expansion for: $\sin \left(2x\right)$

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=$2x$ at the point A=$0$ of order N=1

$\sin \left(0\right)+\cos \left(0\right)·(2x+0)$

ⓘ 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: $2x+0$ = $2x$■Multiplying by 1 changes nothing: $1·2x$ = $2x$

$0+2x$

3) Let's restart using the Taylor expansion: $2x$

$\frac{2x}{x}$

ⓘ The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor $x$: $\frac{2x}{x}$ = $2$

$2$

So the limit of $\frac{\sin \left(2x\right)}{x}$ is $2$ when $x\to 0$

▶Limit:   $2$