Math Doubts

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

$x$ is a variable and $\sin{x}$ is a trigonometric function. They both formed a special function to find the value of the function as $x$ approaches $0$.

Check the value of the function as x tends to 0

Substitute $x = 0$ in the algebraic trigonometric function to find the value of the function.

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

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

$= \dfrac{0}{0}$

The value of this function is indeterminate as $x$ approaches zero. So, this limit problem should be solved in another mathematical approach.

Try Logical Approach

The numerator of the fraction contains sine function and denominator contains a variable in cube form. It is one of the most difficult problems in limit calculus but it can be solved by thinking logically.

Take $x = 3y$. It makes the sin function in numerator to express in cube form and then it could be solved by using lim sinx/x as x approaches 0 rule.

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

Expand sin of angle $3y$ by sin triple angle formula.

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

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

Separate expression as sum of two terms

Write whole expression as sum of two terms. It helps us to simplify this limit problem easily.

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

Use Limit Addition Rule to apply limit value to both terms.

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

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

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

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

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

Change the Limit value

Each expression is expressed in terms of $y$. So, change the limit value to $y$.

If $3y \to 0$, then $y \to \dfrac{0}{3}$. Therefore, $y \to 0$.

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

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

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

Obtaining Required Solution

According to limit of sinx/x as x approaches 0 rule, the value of limit of siny/y as y approaches 0 is equal to 1.

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

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

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

This expression represents the solution of $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{x-\sin{x}}{x^3}$

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

$\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{x-\sin{x}}{x^3}$ is limit of a function in terms of $x$ and $\displaystyle \large \lim_{y \,\to\, 0} \normalsize \dfrac{y-\sin{y}}{y^3}$ is limit of a function in terms of $y$. They both are in same form. So, the value of them should be equal though they are expressed in different variables.

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

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

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

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

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

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

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

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

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

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

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

It is the required solution for this advanced limit problem.



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