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Solve $$\lim_{\theta \to 0} \dfrac{\sin 5\theta -\sin 3\theta}{\theta}$$

The limits problem contains trigonometric function sine with two multiple angles $5 \theta$ and $3 \theta$ and this problem can be solved in two different methods in calculus.

$$\lim_{\theta \to 0} \dfrac{\sin 5\theta -\sin 3\theta}{\theta}$$

Step: 1

The angle theta in denominator belongs to both terms of the numerator.

$$= \lim_{\theta \to 0} \Bigg[\dfrac{\sin 5\theta}{\theta} -\dfrac{\sin 3\theta}{\theta}\Bigg]$$

Step: 2

The limit belongs to both terms.

$$= \lim_{\theta \to 0} \dfrac{\sin 5\theta}{\theta} -\lim_{\theta \to 0}\dfrac{\sin 3\theta}{\theta}$$

Step: 3

Adjust the denominator of each term to make denominator of each term to contain same angle of the respective sine function. For this, the term is expressed as the product of the term and one.

$$= \lim_{\theta \to 0} \dfrac{\sin 5\theta}{\theta} \times 1 -\lim_{\theta \to 0}\dfrac{\sin 3\theta}{\theta} \times 1$$

$$= \lim_{\theta \to 0} \dfrac{\sin 5\theta}{\theta} \times \dfrac{5}{5} -\lim_{\theta \to 0}\dfrac{\sin 3\theta}{\theta} \times \dfrac{3}{3}$$

$$= \lim_{\theta \to 0} \dfrac{5 \sin 5\theta}{5\theta} -\lim_{\theta \to 0}\dfrac{3 \sin 3\theta}{3 \theta}$$

$$= 5\lim_{\theta \to 0} \dfrac{\sin 5\theta}{5\theta} -3\lim_{\theta \to 0}\dfrac{\sin 3\theta}{3 \theta}$$

Step: 4

If theta tends to zero ($\theta \to 0$), then $5$ theta also tends to zero ($5\theta \to 0$). Similarly, if theta tends to zero, then $3$ theta also tends to zero ($5\theta \to 0$).

$$= 5\lim_{5\theta \to 0} \dfrac{\sin 5\theta}{5\theta} -3\lim_{3\theta \to 0}\dfrac{\sin 3\theta}{3 \theta}$$

According to limits law, the ratio sin of an angle to angle is one when limit angle tends to zero.

$= 5 \times 1 -3 \times 1$

$= 5 -3$

$$\therefore \,\,\,\,\, \lim_{\theta \to 0} \dfrac{\sin 5\theta -\sin 3\theta}{\theta} = 2$$

Method: 2

The problem can also be solved in another method by using a trigonometry identity.

$$\lim_{\theta \to 0} \dfrac{\sin 5\theta -\sin 3\theta}{\theta}$$

Step: 1

Use $\sin x -\sin y$ $=$ $2\cos \Bigg(\dfrac{x+y}{2}\Bigg)\sin \Bigg(\dfrac{x-y}{2}\Bigg)$ formula to express the numerator in simplified form.

$$= \lim_{\theta \to 0} \dfrac{2\cos \Bigg(\dfrac{5\theta + 3\theta}{2}\Bigg)\sin \Bigg(\dfrac{5\theta -3\theta}{2}\Bigg) }{\theta}$$

$$= \lim_{\theta \to 0} \dfrac{2\cos \Bigg(\dfrac{8\theta}{2}\Bigg)\sin \Bigg(\dfrac{2\theta}{2}\Bigg) }{\theta}$$

$$= \lim_{\theta \to 0} \dfrac{2\cos 4\theta \sin \theta}{\theta}$$

Step: 2

Express the fraction as the product of two multiplying factors.

$$= \lim_{\theta \to 0} 2\cos 4\theta \times \dfrac{\sin \theta}{\theta}$$

$$= 2\lim_{\theta \to 0} \cos 4\theta \times \dfrac{\sin \theta}{\theta}$$

Step: 3

Apply limit to both multiplying factors.

$$= 2\lim_{\theta \to 0} \cos 4\theta \times \lim_{\theta \to 0} \dfrac{\sin \theta}{\theta}$$

The value of ratio of sin of angle theta to theta is one if limit theta approaches theta.

$= 2\cos (4 \times 0) \times 1$

$= 2\cos 0 \times 1$

$= 2 \times 1 \times 1$

$$\therefore \,\,\,\,\, \lim_{\theta \to 0} \dfrac{\sin 5\theta -\sin 3\theta}{\theta} = 2$$



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