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Evaluate $\displaystyle \lim_{\displaystyle x \to 0} \dfrac{2\sin x -\sin 2x}{x^3}$

$x$ is a literal number but represents angle of right angled triangle too. The quotient of subtraction of $\sin 2x$ from $2\sin x$ by $x$ cubed is required to find in this calculus problem when the limit x tends to zero.

$\displaystyle \lim_{\displaystyle x \to 0} \dfrac{2\sin x -\sin 2x}{x^3}$


Expand the sine of double angle

$2\sin x$ is a term in the numerator and the same term can be obtained by the expansion of the sine of double angle. Hence, expand the function $\sin 2x$.

$=\,\,$ $\displaystyle \lim_{\displaystyle x \to 0} \dfrac{2\sin x -2\sin x \cos x}{x^3}$

Take $2\sin x$ term common from them.

$=\,\,$ $\displaystyle \lim_{\displaystyle x \to 0} \dfrac{2\sin x(1-\cos x)}{x^3}$


Transform the 1 minus cos of angle

As per 1 minus cosine of angle transformation identity, the function $1-\cos x$ can be expressed as the square of the sine of half angle.

$=\,\,$ $\displaystyle \lim_{\displaystyle x \to 0} \dfrac{2\sin x \times 2\sin^2 \Big( \dfrac{x}{2} \Big)}{x^3}$

$=\,\,$ $\displaystyle \lim_{\displaystyle x \to 0} \dfrac{4\sin x \times \sin^2 \Big( \dfrac{x}{2} \Big)}{x^3}$

$=\,\,$ $\displaystyle 4 \times \lim_{\displaystyle x \to 0} \dfrac{\sin x \times \sin^2 \Big( \dfrac{x}{2} \Big)}{x^3}$


Split the function as two multiplying factors

The sum of the powers of both sine terms in the numerator is $3$ and the exponent of the $x$ in denominator is also $3$. There is a chance to simplify the limit problem by using the limit x tends to 0, sin x by x rule. Hence, split the function as two multiplying functions but the exponents of the terms in both numerator and denominator of each multiplying function should be same.

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

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


Adjust the denominator of the second function

There is a problem with the second multiplying function. The angle of the sine function is $\dfrac{x}{2}$ but it is not in the denominator. So, the limit rule cannot be applied directly. However, it can be adjusted to make the angle should be same in both numerator and denominator for the second multiplying function.

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

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

$=\,\,$ $\displaystyle 4 \lim_{\displaystyle x \to 0} \dfrac{\sin x}{x} \times \dfrac{1}{4} \times \dfrac{\sin^2 \Big(\dfrac{x}{2} \Big)}{{\Big(\dfrac{x}{2}\Big)}^2}$

$=\,\,$ $\displaystyle 4 \times \dfrac{1}{4} \lim_{\displaystyle x \to 0} \dfrac{\sin x}{x} \times \dfrac{\sin^2 \Big(\dfrac{x}{2} \Big)}{{\Big(\dfrac{x}{2}\Big)}^2}$

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

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

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


Obtain the result

The limit $x$ tends to zero rule belongs to both multiplying functions. So, it can be written to each function separately.

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

It is time to adjust limit of second function but don’t change anything to first function.

If $x \,\to\, 0$, then $\dfrac{x}{2}$ to $\dfrac{0}{2}$, therefore, $\dfrac{x}{2} \,\to\, 0$. Therefore, $x$ tends to $0$ can be replaced by $\dfrac{x}{2}$ tends to $0$ to make the second eligible to use our limit formula.

$=\,\,$ $\displaystyle \lim_{\displaystyle x \to 0} \dfrac{\sin x}{x} \times \lim_{\displaystyle \frac{x}{2} \to 0} {\Bigg[\dfrac{\sin \Big(\dfrac{x}{2} \Big)}{\Big(\dfrac{x}{2}\Big)}\Bigg]}^2$

$=\,\,$ $\displaystyle \lim_{\displaystyle x \to 0} \dfrac{\sin x}{x} \times {\lim_{\displaystyle \frac{x}{2} \to 0} \Bigg[\dfrac{\sin \Big(\dfrac{x}{2} \Big)}{\Big(\dfrac{x}{2}\Big)}\Bigg]}^2$

Now, use the limit x tends to 0, sin x by x rule and its value is 1 according to the rule of the calculus.

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

$=\,\,$ $1 \times 1$

$=\,\,$ $1$

Therefore, it is the requisite solution for this limit problem.

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