# Solve $\displaystyle \Large \lim_{x \,\to\, 1} \normalsize \dfrac{3\sin{\pi x}-\sin{3\pi x}}{{(x-1)}^3}$

$x$ is a literal number and is used to form an algebraic trigonometric function.

$\dfrac{3\sin{\pi x}-\sin{3\pi x}}{{(x-1)}^3}$

As $x$ tends to one, the value of this function should be calculated mathematically and it is symbolically expressed as follows.

$\displaystyle \Large \lim_{x \,\to\, 1} \normalsize \dfrac{3\sin{\pi x}-\sin{3\pi x}}{{(x-1)}^3}$

01

### Evaluate function as x approaches 1

Firstly, find the value of the function when the value of $x$ tends to $1$. Substitute, $x = 1$ and evaluate the function.

$=\,\,\,$ $\dfrac{3\sin{(\pi (1))}-\sin{(3\pi (1))}}{{(1-1)}^3}$

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

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

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

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

As the value of $x$ approaches $1$, the value of the function is indeterminate. Therefore, an alternative method should be used to solve this function in calculus mathematics.

02

### Applying Triple angle identity

The second term in the numerator is a triple angle trigonometric function. It can be expanded by using the sine triple angle formula.

According to sine triple angle rule.

$\sin{3\pi x} \,=\, 3\sin{\pi x}-4\sin^3{\pi x}$

Replace the term $\sin{3\pi x}$ by its expansion in the numerator.

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 1} \normalsize \dfrac{3\sin{\pi x}-[3\sin{\pi x}-4\sin^3{\pi x}]}{{(x-1)}^3}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 1} \normalsize \dfrac{3\sin{\pi x}-3\sin{\pi x}+4\sin^3{\pi x}}{{(x-1)}^3}$

$=\,\,\,$ $\require{cancel} \displaystyle \Large \lim_{x \,\to\, 1} \normalsize \dfrac{\cancel{3\sin{\pi x}}-\cancel{3\sin{\pi x}}+4\sin^3{\pi x}}{{(x-1)}^3}$

$=\,\,\,$ $\displaystyle \Large \lim_{x \,\to\, 1} \normalsize \dfrac{4\sin^3{\pi x}}{{(x-1)}^3}$

03

### Transformation of limit of the function

When $x \to 1$, then $x-1 \to 0$.

$=\,\,\,$ $\displaystyle \Large \lim_{x-1 \,\to\, 0} \normalsize \dfrac{4\sin^3{\pi x}}{{(x-1)}^3}$

Take $v = x-1$, then $x = v+1$. Convert the entire expression purely in terms of $v$ from $x$.

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize \dfrac{4\sin^3{\pi (v+1)}}{v^3}$

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize \dfrac{4 {[\sin{\pi (v+1)}]}^3}{v^3}$

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize \dfrac{4 {[\sin{(\pi v+\pi )}]}^3}{v^3}$

04

### Simplification of the numerator

There is a sine function with compound angle in numerator and it can be expanded by using the sine of sum of angles rule.

According to sine of sum of angles rule,

$\sin{(\alpha + \beta)} \,=\, \sin{\alpha} \cos{\beta} + \cos{\alpha} \sin{\beta}$

Try this rule to expand the trigonometric function in the numerator.

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize \dfrac{4 {[ \sin{\pi v} \cos{\pi} + \cos{\pi v} \sin{\pi}]}^3}{v^3}$

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize \dfrac{4 {[ \sin{\pi v} \times (-1) + \cos{\pi v} \times 0]}^3}{v^3}$

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize \dfrac{4 {[-\sin{\pi v} + 0]}^3}{v^3}$

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize \dfrac{4 {[-\sin{\pi v}]}^3}{v^3}$

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize \dfrac{-4{[\sin{\pi v}]}^3}{v^3}$

Use quotient rule of exponents, to write both numerator and denominator as a term.

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize -4 \times {\Bigg[\dfrac{\sin{\pi v}}{v}\Bigg]}^3$

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize -4 \times {\Bigg[\dfrac{\pi \times \sin{\pi v}}{\pi \times v}\Bigg]}^3$

$=\,\,\,$ $\displaystyle \Large \lim_{v \,\to\, 0} \normalsize -4{\pi}^3 \times {\Bigg[\dfrac{\sin{\pi v}}{\pi v}\Bigg]}^3$

$=\,\,\,$ $\displaystyle -4{\pi}^3 \Large \lim_{v \,\to\, 0} \normalsize {\Bigg[\dfrac{\sin{\pi v}}{\pi v}\Bigg]}^3$

$=\,\,\,$ $\displaystyle -4{\pi}^3 \normalsize {\Bigg[ \Large \lim_{v \,\to\, 0} \normalsize \dfrac{\sin{\pi v}}{\pi v}\Bigg]}^3$

05

If $v \to 0$, then $\pi \times v \to \pi \times 0$. Therefore, $\pi v \to 0$. It is proved that, as the value of $v$ tends to zero, the value of $\pi v$ also approaches zero. Hence, the the value of limit can be changed mathematically.

$=\,\,\,$ $\displaystyle -4{\pi}^3 \normalsize {\Bigg[ \Large \lim_{\pi v \,\to\, 0} \normalsize \dfrac{\sin{\pi v}}{\pi v}\Bigg]}^3$

06

### Obtaining the Result

As per limit x tends to 0 sinx/x rule, the value of the function is one.

$=\,\,\,$ $-4{\pi}^3 {}^3$

$=\,\,\,$ $-4{\pi}^3 \times 1$

$=\,\,\,$ $-4{\pi}^3$

Therefore, $-4{\pi}^3$ is the required solution for this limit math problem in calculus mathematics.

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