Math Doubts

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


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

If $x$ is angle of a right angled triangle then sine function is written as $\sin{x}$. The value of ratio of sin function to angle as the angle approaches zero is expressed in the following mathematical form.

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

The value of ratio of sin function to angle is equal to one as the angle approaches zero. It is used as a formula in calculus and the limit rule is called limit of ratio of sine function to angle rule as angle approaches zero.


There are two methods to prove this limit rule in calculus.

Relation of Sine function and angle

The limit of ratio of $\sin{x}$ to $x$ when $x$ tends to zero is derived in calculus on the basis of relation of sine function with angle of the right angled triangle.

Taylor or Maclaurin Series Method

The limit of ratio of $\sin{x}$ to $x$ as $x$ approaches to zero can also be derived in calculus as per Taylor or Maclaurin series.

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