limit as x approaches 0 of ((tan(5x)))/x

1) Let's evaluate the limit of

\frac{\tan (5 x)}{x}

when

x \to 0

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

The limit of

5 x

5 \cdot 0

when

x \to 0

⇥1.1) Let's evaluate:

5 \cdot 0

⇥ⓘ A product with a zero factor is zero:

5 \cdot 0

0

⇥

0

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

The limit of

\tan (5 x)

\tan (0)

when

x \to 0

⇥1.2) Let's evaluate:

\tan (0)

⇥ⓘ The tangent is the y-coordinate (vertical axis) divided by 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, provided the x-coordinate is not zero, this is a remarkable value:

\tan (0)

0

⇥

0

For

\frac{\tan (5 x)}{x}

, we encounter the indeterminate form 0/0 as

x

tends to

0

2) We need a Taylor expansion for:

\tan (5 x)

Using Taylor's formula:

\tan (A) + (1 + {\tan (A)}^{2}) \cdot (X - A)

for the function tan, we compute the Taylor expansion for X=

5 x

at the point A=

0

of order N=1

\tan (0) + (1 + {\tan (0)}^{2}) \cdot (5 x + 0)

ⓘ The tangent is the y-coordinate (vertical axis) divided by 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, provided the x-coordinate is not zero, this is a remarkable value:

\tan (0)

0

■Adding 0 changes nothing:

5 x + 0

5 x

0 + (1 + 0^{2}) \cdot 5 x

3) Let's restart using the Taylor expansion:

5 x

\frac{5 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{5 x}{x}

5

5

So the limit of

\frac{\tan (5 x)}{x}

5

when

x \to 0

▶Limit:

5