# 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. According to the direct substitution method, it is already calculated that the limit of $x$ minus $\sin{x}$ divided by cube of $x$ is indeterminate as the value of $x$ is closer to zero.

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

The indeterminate form indicates that the direct substation method should not be used to find the limit of $x$ minus sine of angle $x$ divided by $x$ cubed as the value of $x$ tends to zero.

### Trick to minimize the complexity of the function

Observe the rational function carefully. In this function, there is a sine function in numerator and an algebraic function in cube form in denominator. This combination of them creates complications while finding the limit.

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

According to the triple angle trigonometric identity of sine function, the sine function can be expanded in terms of cube form if the angle inside the sine function is a triple angle. In this case, the sine function is not contained the triple angle. The problem can be overcome by replacing the angle with a triple angle acceptably.

So, assume $x = 3y$. Now, eliminate the variable $x$ in the function by its equivalent value.

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

The above equation expresses that the limit of function in $x$ can be calculated by finding the limit of the function in $y$.

### Simplify the whole expression for finding the Limit

Now, let us focus on simplifying the function in $y$ for preparing the function to find its limit.

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

Firstly, let’s concentrate on simplifying the mathematical expression in the numerator. The first term $3y$ is already in simplified form and the second term $\sin{3y}$ is also in simplified form but it can be expanded as per the sine triple angle identity for expressing it in cube form.

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3y-(3\sin{y}-4\sin^3{y})}{{(3y)}^3}}$

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3y-3\sin{y}+4\sin^3{y}}{{(3y)}^3}}$

Now, it is time to focus on the expression in the denominator. The cube of $3y$ is already in simplified form but it is essential to separate the variable $y$ from whole cube and it can be done as per the power of a product rule.

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3y-3\sin{y}+4\sin^3{y}}{3^3 \times y^3}}$

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3y-3\sin{y}+4\sin^3{y}}{27 \times y^3}}$

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3y-3\sin{y}+4\sin^3{y}}{27y^3}}$

The sine function in third term of numerator and the expression in denominator are in cube form. Separating them from the function is useful for finding the limit of that function as per trigonometric limit rule in sine function. So, let’s split the rational function as a sum of two fractional functions.

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg(\dfrac{3y-3\sin{y}}{27y^3}+\dfrac{4\sin^3{y}}{27y^3}\Bigg)}$

The number $3$ is a common factor in both terms of numerator in the first term, inside the limit operation. So, let us take out the common factor from the terms for simplifying the expression further.

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg(\dfrac{3(y-\sin{y})}{27y^3}+\dfrac{4\sin^3{y}}{27y^3}\Bigg)}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg(\dfrac{\cancel{3}(y-\sin{y})}{\cancel{27}y^3}+\dfrac{4\sin^3{y}}{27y^3}\Bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg(\dfrac{y-\sin{y}}{9y^3}+\dfrac{4\sin^3{y}}{27y^3}\Bigg)}$

$\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg(\dfrac{y-\sin{y}}{9y^3}+\dfrac{4\sin^3{y}}{27y^3}\Bigg)}$

### Evaluate the Limit of given function

According to sum rule of limits, the limit of sum of two functions is equal to sum of their limits.

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{9y^3}}$ $+$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{4\sin^3{y}}{27y^3}}$

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{9 \times y^3}}$ $+$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{4 \times \sin^3{y}}{27 \times y^3}}$

$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg( \dfrac{1}{9} \times \dfrac{y-\sin{y}}{y^3} \Bigg)}$ $+$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg(\dfrac{4}{27} \times \dfrac{\sin^3{y}}{y^3} \Bigg)}$

The constant from each term can be separated by the constant multiple rule of limits.

$=\,\,\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

The above expression is the expansion of the limit of given rational function in $x$ as per our assumption.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

Now, compare the expression in the left hand side of the equation and second factor of the first term in the right hand side expression of the equation.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

The comparison clears that the expressions $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ and $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ are almost similar but their input expressions in the limit operations are dissimilar. However, this dissimilarity can be erased mathematically.

If $3y\,\to\,0$ then $\dfrac{3y}{3}\,\to\,\dfrac{0}{3}$, which means $\dfrac{\cancel{3}y}{\cancel{3}}\,\to\,0$. Therefore, $y\,\to\,0$. It clears that when $3y$ approaches to $0$, the variable $y$ also approaches $0$.

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

Now, we can replace the expression $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ by its equivalent expression $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ in the above equation.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

Now, shift the first term of the right hand side expression to left hand side of the equation.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $-$ $\dfrac{1}{9} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $\,=\,$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

Now compare the first term $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ with the second factor of second term $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ in left hand side of equation. It clears us that the two mathematical expressions are similar. Hence, their limits should be equal.

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

The expression in the left hand side of the simplified equation can be expressed purely in $x$.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $-$ $\dfrac{1}{9} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

$\implies$ $1 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $-$ $\dfrac{1}{9} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

Now, we can factorize the left hand side expression of the equation by taking out the common factor in the expression.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\times$ $\Bigg(1-\dfrac{1}{9}\Bigg)$ $\,=\,$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

$\implies$ $\Bigg(1-\dfrac{1}{9}\Bigg) \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

$\implies$ $\Bigg(\dfrac{1}{1}-\dfrac{1}{9}\Bigg) \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

$\implies$ $\Bigg(\dfrac{1 \times 9-1\times 1}{9}\Bigg) \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

$\implies$ $\Bigg(\dfrac{9-1}{9}\Bigg) \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

$\implies$ $\dfrac{8}{9} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

The numerical factor in the left hand side of expression can be shifted to right hand side of the equation.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{9}{8} \times \dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{9 \times 4}{8 \times 27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{\cancel{9} \times \cancel{4}}{\cancel{8} \times \cancel{27}} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

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

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

We have already evaluated that $y$ also approaches $0$ when $3y$ approaches to $0$.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{(\sin{y})^3}{y^3}}$

The expression in the right hand side of the equation can be simplified as follows by the power of quotient rule.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{y}}{y}\Bigg)^3}$

It can be further simplified by the power rule of limits.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6} \times \Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{y}}{y}\Bigg)^3}$

According to the limit formula of sinx/x as x approaches 0, the limit of the $\dfrac{\sin{y}}{y}$ as $y$ approaches $0$ is equal to one.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6} \times (1)^3$

Now, simplify the expression in the right hand side of the equation for evaluating the limit of the rational expression.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6} \times 1^3$

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6} \times 1$

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6}$

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