# Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$

The limit of $x$ minus sine of angle $x$ divided by $x$ cube should be evaluated in this limit problem as the value of $x$ approaches zero. Firstly, let us try to evaluate the limit by direct substitution. Now, substitute $x$ is equal to zero in the rational function.

$=\,\,$ $\dfrac{0-\sin{0}}{0^3}$

The sine of zero radian is equal to zero as per the trigonometric mathematics.

$=\,\,$ $\dfrac{0-0}{0}$

$=\,\,$ $\dfrac{0}{0}$

It is evaluated as per the direct substitution method that the limit of variable $x$ minus sine of angle $x$ divided by cube of $x$ is indeterminate as the value of $x$ tends to zero. It clears to us that the direct substitution method is not useful to find the limit initially.

### Change of variables

Learn how to calculate the limit of $x$ minus sine of angle $x$ divided by the cube of $x$ as $x$ tends to zero by using change of variables.

### L’Hospital’s rule

Learn how to use the l’hôpital’s rule to find the limit of $x$ minus sine of angle $x$ divided by $x$ cube as $x$ approaches zero.

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