The direct substitution method clears that the limit of $x$ minus sine of angle $x$ divided by cube of $x$ is indeterminate as $x$ approaches zero.

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

The indeterminate form hints us to use l’hospital’s rule to find the limit of $x$ minus $\sin{x}$ divided by $x$ cube as the variable $x$ tends to $0$ in this limit problem.

The $x$ minus $\sin{x}$ is an expression in the numerator and $x$ cube is an expression in the denominator of the given rational function. The two expressions are defined in terms of $x$. So, differentiate both expressions with respect to $x$ to use the l’hospital’s rule mathematically.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}(x-\sin{x})}{\dfrac{d}{dx}(x^3)}}$

It is time to focus on the mathematical expression on the right-hand side of the above equation. Firstly, let us concentrate on the numerator. The derivative of difference of the functions can be evaluated by the difference of their derivatives as per the difference rule of the derivatives.

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}{x}-\dfrac{d}{dx}\sin{x}}{\dfrac{d}{dx}(x^3)}}$

The derivative of $x$ with respect to $x$ can be calculated by the derivative rule of a variable and the derivative of sine of angle $x$ with respect to $x$ can also be calculated by the derivative rule of sine function. Similarly, the derivative of $x$ cubed with respect to $x$ can be evaluated by the power rule of the derivatives.

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-\cos{x}}{3x^2}}$

The differentiation is completed. Now, let’s try to find the limit by the direct substitution method.

$\,\,=\,\,$ $\dfrac{1-\cos{(0)}}{3(0)^2}$

$\,\,=\,\,$ $\dfrac{1-\cos{0}}{3 \times 0^2}$

According to the trigonometry, the cosine of angle zero radian is one and substitute its value in the numerator.

$\,\,=\,\,$ $\dfrac{1-1}{3 \times 0}$

It is time to evaluate the rational expression by simplification.

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

The limit is indeterminate even though the l’hôpital’s rule is used. Use the l’hospital’s rule one more time due to the indeterminate form.

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}(1-\cos{x})}{\dfrac{d}{dx}(3x^2)}}$

The derivative of the difference of the two functions can be calculated by the subtraction rule of the differentiation.

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}(1)-\dfrac{d}{dx}\cos{x}}{\dfrac{d}{dx}(3 \times x^2)}}$

Now, look at the denominator, there are two factors in the function and one of them is a constant. The constant factor can be excluded from the differentiation by the constant multiple rule of derivatives.

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}(1)-\dfrac{d}{dx}\cos{x}}{3 \times \dfrac{d}{dx}(x^2)}}$

According to the derivative rule of a constant, the derivative of one is zero, and the derivative of cosine of angle $x$ with respect to $x$ is negative of sine of angle $x$ as per the derivative rule of cos function. Similarly, the derivative of $x$ square can be evaluated with respect to $x$ by the power rule of differentiation.

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{0-(-\sin{x})}{3 \times 2x}}$

The process of differentiating the expressions in both numerator and denominator is completed. Now, it is time to simplify the whole rational function.

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{3 \times 2x}}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{6x}}$

Now, use the direct substitution method to find the limit of the rational function.

$\,\,=\,\,$ $\dfrac{\sin{(0)}}{6(0)}$

$\,\,=\,\,$ $\dfrac{\sin{0}}{6 \times 0}$

The sine of zero radian is zero as per the trigonometry. Now, replace the sine of angle zero radian by its value in the numerator of the rational function.

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

Now, continue to the process of using the l’hôpital’s rule due to indeterminate form.

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}\sin{x}}{\dfrac{d}{dx}(6x)}}$

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}\sin{x}}{\dfrac{d}{dx}(6 \times x)}}$

There is a constant factor involved in the differentiation in the denominator of the rational function. So, it can be taken out from the differentiation by using the constant multiple rule of differentiation.

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{d}{dx}\sin{x}}{6 \times \dfrac{d}{dx}(x)}}$

Now, evaluate the derivative of sine of angle $x$ with respect to $x$ by the differentiation rule of sine function and the derivative of variable $x$ with respect to $x$ can be evaluated by the derivative rule of a variable.

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\cos{x}}{6 \times 1}}$

$\,\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\cos{x}}{6}}$

Now, find the limit of the cosine of angle $x$ divided by $6$ by the direct substitution as the $x$ tends to zero.

$\,\,=\,\,$ $\dfrac{\cos{0}}{6}$

The cosine of angle zero radian is one as per the trigonometric mathematics.

$=\,\,$ $\dfrac{1}{6}$

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