Math Doubts

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

It is most challenging problem in limits but the limit of the algebraic trigonometric function can be evaluated mathematically in calculus by using a mathematical technique.

Balance Numerator and Denominator

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

The denominator is a cube function but there is no cube function in the numerator. If sine function has triple angle then the sine function can be expanded in cube form. It helps us to evaluate the function by the limit rule of trigonometric functions.

Take $x = 3y$ and then transform the entire function in terms of $y$.

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

Simplify the limit function

Now, simplify the functions in both numerator and denominator for simplifying the whole function.

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

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

Expand $\sin{3y}$ term by the sine triple angle identity.

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

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

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

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

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

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

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

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

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

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

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

Find the limit of each function

Now, concentrate on the second term firstly, the limit of the trigonometric function is same as the limit of $\sin{x}/{x}$ as $x$ approaches $0$ formula. So, let’s evaluate this function.

If $3y \to 0$, then $y \to \dfrac{0}{3}$. Therefore, $y \to 0$. It states that if $3y$ approaches $0$, then $y$ also approaches $0$.

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

According to power rule of limits, the second term can be written as follows.

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

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

According to limit of sinx/x as x approaches 0 formula, the limit of the trigonometric function in the second term is equal to $1$.

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

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

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

Therefore, the limit of the algebraic trigonometric function is successfully evaluate and we are in one step ahead in evaluating the limit of the given algebraic trigonometric function.

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

Now, observe the mathematical equation. The limit of given algebraic trigonometric function and part of the first term are same but they are in $x$ and $y$ terms.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ and $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$. There is no difference between them but they are in $x$ and $y$ terms. However, their values should be equal mathematically. Therefore, $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$. Therefore, $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ can be written as $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ in this equation.

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

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

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

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

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

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

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

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

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

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

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

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