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

The limit of the quotient of trigonometric expression $\tan{x}-\sin{x}$ by the algebraic expression $x$ cubed has to calculate as the input $x$ approaches zero in this limit problem. First of all, let’s try to evaluate the given algebraic trigonometric function by the direct substitution method.

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

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

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

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

It is not possible to find the limit of the given algebraic trigonometric function by direct substitution method due to indeterminate form.

### Simplify the trigonometric expression

Let’s simplify the trigonometric expression $\tan{x}-\sin{x}$ firstly. According to quotient trigonometric rule of sine and cosine functions, the quotient of $\sin{x}$ by $\cos{x}$ is equal to $\tan{x}$.

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

The function $\sin{x}$ is a common factor in the both terms of the trigonometric expression in the numerator of the function.

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

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

Now, take the common factor $\sin{x}$ out from the terms for simplify the expression further.

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

Continue the process for simplifying the trigonometric expression in the numerator.

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

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

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

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

### Factorize the function as two factors

There is a $\tan{x}$ function in the numerator and also $x$ in the denominator. So, the ratio can be evaluated by using limit rules of trigonometric functions. Therefore, separate them from the function by factorisation. It can be achieved by splitting the algebraic function in the denominator as per product rule of exponents with same base.

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

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

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

Now, split the function as the product of following two functions by factorization.

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

Use product rule of limits to find the limit of the function by the product of their limits.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\tan{x}}{x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-\cos{x}}{x^2}}$

### Evaluate the limits of the functions

It is proved that the limit of tanx/x as x approaches 0 is equal to one.

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

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

As per the power reduction trigonometric identity, $2\sin^2{\Big(\dfrac{x}{2}\Big)} = 1-\cos{x}$. Therefore, replace the trigonometric expression $1-\cos{x}$ by its equivalent value $2\sin^2{\Big(\dfrac{x}{2}\Big)}$ in the numerator of the function.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{x}{2}\Big)}}{x^2}}$

The angle in sin function is $\dfrac{x}{2}$ and try to adjust the denominator same for applying a limit rule of trigonometric functions.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{x}{2}\Big)}}{1 \times x^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{x}{2}\Big)}}{\dfrac{4}{4} \times x^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{x}{2}\Big)}}{4 \times \dfrac{x^2}{4}}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{x}{2}\Big)}}{4 \times \dfrac{x^2}{2^2}}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{x}{2}\Big)}}{4 \times \Big(\dfrac{x}{2}\Big)^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2 \times \sin^2{\Big(\dfrac{x}{2}\Big)}}{4 \times \Big(\dfrac{x}{2}\Big)^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{2}{4} \times \dfrac{\sin^2{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)^2}\Bigg)}$

$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\cancel{2}}{\cancel{4}} \times \dfrac{\sin^2{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)^2}\Bigg)}$

Use the quotient rule of exponents with same base rule to simplify the expression in the denominator further.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{1}{2} \times \dfrac{\sin^2{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)^2}\Bigg)}$

According to the constant multiple rule of limits.

$= \,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)^2}}$

$= \,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\Bigg(\sin{\Big(\dfrac{x}{2}\Big)}\Bigg)^2}{\Big(\dfrac{x}{2}\Big)^2}}$

$= \,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)}\Bigg)^2}$

Now, use power rule of limits.

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)}\Bigg)^2}$

If $x$ approaches $0$, then $\dfrac{x}{2}$ approaches $\dfrac{0}{2}$. Therefore, $\dfrac{x}{2}$ also approaches zero.

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{\Large \frac{x}{2} \large \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)}\Bigg)^2}$

Now, take $y = \dfrac{x}{2}$ and express the whole function in terms of $y$.

$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{y}}{y}\Bigg)^2}$

According to the limit of sinx/x as approaches 0 rule, the limit of quotient of $\sin{y}$ by $y$ as $y$ tends to $0$ is equal to $1$.

$= \,\,\,$ $\dfrac{1}{2} \times \Big(1\Big)^2$

$= \,\,\,$ $\dfrac{1}{2} \times 1^2$

$= \,\,\,$ $\dfrac{1}{2} \times 1$

$= \,\,\,$ $\dfrac{1}{2}$

Therefore, it is evaluated that the limit of algebraic trigonometric function $\dfrac{\tan{x}-\sin{x}}{x^3}$ as $x$ approaches zero is equal to $\dfrac{1}{2}$ or $0.5$.

Latest Math Topics

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.