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.

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


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.


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.

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

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

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

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


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 $x$.

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

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

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


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.

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

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

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

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

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


Final adjustments

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.

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


Obtaining the Result

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

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

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

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

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

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