Limit of Tan X over X at Zero/Proof 1

Theorem

$\ds \lim_{x \mathop \to 0} \frac {\tan x} x = 1$

Proof

By L'Hôpital's Rule:

\(\ds \lim_{x \mathop \to 0} \frac {\tan x} x\)

\(=\)

\(\ds \lim_{x \mathop \to 0} \frac {\sec^2 x} 1\)

Derivative of Tangent Function

\(\ds \)

\(=\)

\(\ds 1\)

Secant of Zero

$\blacksquare$