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

When $x$ is used to represent a variable, a rational expression is formed by a trigonometric function $\sin{x}$ and the algebraic functions $x$ and $x^3$.

$\dfrac{x-\sin{x}}{x^3}$

Now, we have to evaluate the limit of this rational expression as $x$ approaches $0$ in this limit problem.

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

### Evaluate the Limit by Direct substitution

Let’s evaluate the limit of the quotient of $x-\sin{x}$ by cube of $x$ as $x$ tends to $0$ by the direct substitution method.

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

As per trigonometry, the value of sin of angle zero degrees is zero.

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

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

The limit of the function is indeterminate and it is actually evaluated by the direct substitution. The limit of the function should not be equal to indeterminate, which means that the direct substitution method is not recommendable method at this moment. Therefore, we have to explore for an alternative mathematical approach.

### Balancing the expressions in the function

It is a very special problem in the calculus and we cannot evaluate the limit by the standard methods, used in the limits. So, we have to think in advanced level for evaluating the limit of the given rational expression.

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

In the given limit problem, $x-\sin{x}$ and $x^3$ are expressions in the numerator and denominator. In limits, there is a limit rule in terms of $\sin{x}$ and $x$ but we have $\sin{x}$ in numerator but $x$ cubed in denominator. Hence, the trigonometric limit rule cannot be used in this problem. If the sine function is adjusted to cube form, then it clears route to us for balancing the expressions.

Actually, we cannot include anything in the given function for expressing the sine function in cube form. Fortunately, we can express the sine function in cube form by the expansion of sine triple angle identity.

Hence, assume $x = 3y$ and then transform the entire function in terms of $y$.

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

The equation expresses that the limit of the given rational expression in $x$ can be evaluated by evaluating the limit of the rational expression in $y$.

### Simplify the expression for evaluating limit

Now, we have to concentrate on simplifying the right hand side mathematical expression.

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

The expression in the denominator can be expressed in product form by the power rule of product.

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

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

Now, the sine triple angle identity can be used to expand the $\sin{3y}$ function in the numerator.

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

Now, focus on simplifying the rational expression for evaluating the limit of the function.

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

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

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

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

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

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.

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

### Evaluate the Limit of the function

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.

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

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

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

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

###### Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

###### Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

###### Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.