Math Doubts

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

Two trigonometric functions $\sin{\pi x}$ and $\sin{3\pi x}$ and algebraic function $(x-1)^3$ are formed by considering $x$ represents a variable mathematically. The three functions formed a rational expression.

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

We have to calculate the limit of this rational expression as $x$ approaches $1$.

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

Evaluate the limit by direct substitution

Firstly, let us evaluate the limit of the function by the direct substitution method as $x$ approaches $1$.

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

$=\,\,\,$ $\dfrac{3\sin{(\pi \times 1)}-\sin{(3\pi \times 1)}}{(0)^3}$

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

According to the trigonometry,

  1. $\sin{(\pi)}$ $\,=\,$ $0$
  2. $\sin{(3\pi)}$ $\,=\,$ $0$

Now, substitute the values of both sine functions in the expression.

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

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

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

It is evaluated that the limit of the rational expression, which contains both trigonometric and algebraic functions, is indeterminate. So, it is impossible to evaluate the limit of the given function by the direct substitution method initially. Hence, we have to think for another mathematical approach.

Simplify the Rational expression

In this step, let’s think about the possible way of simplifying the whole rational expression.

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

The rational expression can be simplified possibly in two ways.

  1. In the numerator, the term $\sin{3\pi x}$ is expanded by the sin triple angle identity.
  2. In the denominator, the term $(x-1)^3$ can also be expanded by the cube rule of difference of terms.

There are two terms in the numerator. Hence, let’s simplify the expression in the numerator. There is only one term in the denominator. Hence, no need to simplify the expression in the denominator. Therefore, expand the sin triple angle function in the numerator by the sin triple angle identity.

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

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

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

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

When $x \,\to\,1$, the $x-1 \,\to\,1-1$. Therefore, $x-1 \,\to\,1-1$. It is derived that $x-1$ approaches $0$ when $x$ tends to $1$.

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

Assume, $v = x-1$, then $x = v+1$. Now, convert the whole mathematical expression in terms of $v$.

$\implies$ $\displaystyle \large \lim_{x-1\,\to\,0}{\normalsize \dfrac{4\sin^3{(\pi x)}}{(x-1)^3}}$ $\,=\,$ $\displaystyle \large \lim_{v\,\to\,0}{\normalsize \dfrac{4\sin^3{\Big(\pi (v+1)\Big)}}{(v)^3}}$

Now, focus on simplifying the whole expression further.

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

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

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

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

The sine of sum of two angles can be function can be expanded by the sin of angle sum formula.

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

$=\,\,\,$ $\displaystyle \large \lim_{v\,\to\,0}{\normalsize \dfrac{4\Big(\sin{(\pi v)} \times \cos{(\pi)}+\cos{(\pi v)} \times \sin{(\pi)}\Big)^3}{v^3}}$

Now, come back to the trigonometry,

  1. $\sin{(\pi)}$ $\,=\,$ $0$
  2. $\cos{(\pi)}$ $\,=\,$ $-1$

Substitute the values of $\sin{(\pi)}$ and $\cos{(\pi)}$ in the rational expression.

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

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

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

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

$=\,\,\,$ $\displaystyle \large \lim_{v\,\to\,0}{\normalsize \dfrac{4\Big(-1 \times \sin{(\pi v)}\Big)^3}{v^3}}$

$=\,\,\,$ $\displaystyle \large \lim_{v\,\to\,0}{\normalsize \dfrac{4 \times \Big(-1 \times \sin{(\pi v)}\Big)^3}{v^3}}$

The second factor can be expanded by the power of product rule.

$=\,\,\,$ $\displaystyle \large \lim_{v\,\to\,0}{\normalsize \dfrac{4 \times (-1)^3 \times \Big(\sin{(\pi v)}\Big)^3}{v^3}}$

$=\,\,\,$ $\displaystyle \large \lim_{v\,\to\,0}{\normalsize \dfrac{4 \times (-1) \times \Big(\sin{(\pi v)}\Big)^3}{v^3}}$

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

$=\,\,\,$ $\displaystyle \large \lim_{v\,\to\,0}{\normalsize \Bigg(-4 \times \dfrac{\Big(\sin{(\pi v)}\Big)^3}{v^3}\Bigg)}$

It can be simplified further by the constant multiple rule of limits.

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

The quotient of cubes of functions can be simplified by the power of quotient rule.

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

Finally, use the power rule of limits before focusing on evaluating the limit of the function.

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

Evaluate the limit of the function

The mathematical expression inside the cube is similar to the trigonometric limit rule in sine function. If this expression is adjusted same as the limit rule, the limit of the simplified function can be evaluated.

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

The expression in the denominator is required $\pi$ in product form. Hence, we use an acceptable mathematical technique for including it in the expression.

$=\,\,\,$ $-4 \times \Bigg(\displaystyle \large \lim_{v\,\to\,0}{\normalsize \Big(1 \times \dfrac{\sin{(\pi v)}}{v}\Big)\Bigg)^3}$

$=\,\,\,$ $-4 \times \Bigg(\displaystyle \large \lim_{v\,\to\,0}{\normalsize \Big(\dfrac{\pi}{\pi} \times \dfrac{\sin{(\pi v)}}{v}\Big)\Bigg)^3}$

$=\,\,\,$ $-4 \times \Bigg(\displaystyle \large \lim_{v\,\to\,0}{\normalsize \Big(\pi \times \dfrac{\sin{(\pi v)}}{\pi \times v}\Big)\Bigg)^3}$

$=\,\,\,$ $-4 \times \Bigg(\displaystyle \large \lim_{v\,\to\,0}{\normalsize \Big(\pi \times \dfrac{\sin{(\pi v)}}{\pi v}\Big)\Bigg)^3}$

Here, $\pi$ is a constant and it can be separated from the limit operation by using the constant multiple rule of limits.

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

It can be simplified further by the power of a product rule.

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

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

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

The rational expression is almost similar to the trigonometric limit rule in sine function form. The input of the limit operation must be same as the angle in sine function or expression in the denominator.

If $v \,\to\, 0$, then $\pi \times v \,\to\, \pi \times 0$. Therefore, $\pi v \,\to\, 0$.

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

According to limit rule of sinx/x as x approaches 0, the limit of the function as $\pi v$ approaches zero should also be one.

$=\,\,\,$ $-4\pi^3 \times (1)^3$

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

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

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