Math Doubts

$\displaystyle \lim_{x \,\to\, 0} \dfrac{\sin{x}}{x}$ formula


$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}} \,=\, 1$

The limit of ratio of sin of angle to angle as the angle approaches zero is equal to one. This standard result is used as a rule to evaluate the limit of a function in which sine is involved.


$x$ is a variable and represents angle of a right triangle. The sine function is written as $\sin{x}$ as per trigonometry. The limit of quotient of $\sin{x}$ by $x$ as $x$ approaches zero is often appeared in calculus.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$

Actually, the limit of $\sin{(x)}/x$ as $x$ tends to $0$ is equal to $1$ and this standard trigonometric function result is used as a formula everywhere in calculus.


There are two ways to prove this limit of trigonometric function property in mathematics.

Relation between Sine function and angle

It is derived on the basis of close relation between $\sin{x}$ function and angle $x$ as the angle $x$ closer to zero.

Taylor (or) Maclaurin Series Method

It can also be derived by the expansion of $\sin{x}$ function as per Taylor (or) Maclaurin series.

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