Limit x tends to 0, quotient of arc sin x by x is equal to 1


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


$x$ is a literal and it represents the ratio of length of the opposite side to hypotenuse of a right angled triangle for an angle. The inverse of sine function for this value is written as $\sin^{-1} x$. The value of the ratio of $\sin^{-1} x$ to $x$ is the requisite result when the value of the $x$ approaches zero.

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


Transform the function in trigonometric form

Take $y = \sin^{-1} x$, then $x = \sin y$. Now, transform the entire function in terms of $y$ from $x$.

$x \,\to\, 0 \implies y \,\to\, \sin^{-1} 0$

$\therefore \,\,\,\,\,\, y \,\to\, 0$

If $x$ tends to $0$, then $y$ tends to $\sin^{-1} 0$. Therefore, $y$ is also tends to zero when the value of $x$ approaches to zero.

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


Evaluate it

The function is similar to the limit x tends to 0, sin x by x rule. On the basis of this formula, the value of the function can be evaluated mathematically.

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

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

$= \,\,$ $\dfrac{1}{1}$

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

Save (or) Share


Follow us
Email subscription
Copyright © 2012 - 2017 Math Doubts, All Rights Reserved