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$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{x}}{x}}$ formula


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

The limit of quotient of inverse sine function by a variable as the input approaches zero is equal to one. It is a standard result in calculus and used as a formula in mathematics.


Assume $x$ is a variable and represents the ratio of lengths of opposite side to hypotenuse in a right triangle. The inverse sine function in terms of $x$ is written as $\sin^{-1}{x}$ or $\arcsin{x}$ in inverse trigonometry.

The limit of the $\arcsin{x}$ by $x$ as $x$ approaches zero is written in mathematical form as follows.

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

In fact, the limit of $\arcsin{(x)}/x$ as $x$ tends to $0$ is equal to $1$. It is used often appeared in calculus. So, this standard inverse trigonometric function result is used as a formula in calculus.

Other forms

The limit rule of inverse trigonometric function can be written in several ways in calculus.

$(1) \,\,\,$ $\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{m}}{m}}$ $\,=\,$ $1$

$(2) \,\,\,$ $\displaystyle \large \lim_{p \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{p}}{p}}$ $\,=\,$ $1$

$(3) \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{y}}{y}}$ $\,=\,$ $1$


Learn how to prove that the limit of $\arcsin{(x)}/x$ as $x$ tends to zero is equal to one in calculus.

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