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

$x$ is a variable and also represents the quotient of lengths of opposite side to hypotenuse of a right triangle. The inverse sine function is written as $\arcsin{(x)}$ or $\sin^{-1}{(x)}$ in inverse trigonometric mathematics.

In calculus, the limit of a function in the following form is often appeared. So, it is considered as a standard result and also used as a formula in calculus.

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

The limit of arcsin(x) by x as x approaches zero is equal to one. Let’s learn how to derive this limit rule before using it as a formula in calculus.

Transform the function in trigonometric form

Convert the inverse trigonometric sine function by the mathematical relationship between the trigonometric and inverse trigonometric functions.

Take $y \,=\, \sin^{-1}{x}$, then $x \,=\, \sin{y}$.

Therefore, the quotient of $\arcsin{x}$ by $x$ can be written as the ratio of $y$ to $\sin{y}$ mathematically.

$\implies$ $\dfrac{\sin^{-1}{x}}{x} \,=\, \dfrac{y}{\sin{y}}$

The limit of the function in terms of $x$ has to calculate as $x$ approaches zero but the inverse trigonometric function is now expressed in the form of trigonometric function and in terms of $y$.

We have taken that $y \,=\, \sin^{-1}{x}$

Therefore, if $x$ approaches zero, then $y$ tends to $\sin^{-1}{(0)}$

According to inverse trigonometry, the value of inverse sin of zero is equal to zero.

Therefore, if the value of $x$ closer to $0$, then the value of $y$ also approaches zero.

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

Transform the function in known limit rule

There is a trigonometric limit rule in calculus.

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

But our function is in reciprocal form.

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

Now, write the above algebraic trigonometric function in its reciprocal form.

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

According to quotient rule of limits, the limit of a quotient is equal to quotient of their limits.

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

Evaluate the Limit of the function

According to constant limit rule, The limit of one is always equal to one.

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

The limit of sinx/x as x approaches zero is equal to one. Therefore, the limit of $\dfrac{\sin{y}}{y}$ as $y$ tends to $0$ is also equal to one.

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

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

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

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

Therefore, it is proved that the limit of quotient of inverse sine by a variable as the variable approaches zero is equal to one.

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