Math Doubts

Simplify $\displaystyle \lim_{x \to 0} \dfrac{\cos 3x -\cos 4x}{x \sin 2x}$

$\displaystyle \lim_{x \to 0} \dfrac{\cos 3x -\cos 4x}{x \sin 2x}$

The trigonometric function cosine formed two functions and sine formed a function in multiple angle form. The ratio of subtraction of $\cos 4x$ from $\cos 3x$ to product of $x$ and $\sin 2x$ function has to be calculated when the value of limit $x$ approaches zero.

Step: 1

The numerator contains two cos terms with two different angles in subtraction form and it can be expressed in product form by the following trigonometric formula.

$\cos \alpha -\cos \beta$ $=$ $-2 \sin \Bigg[\dfrac{\alpha + \beta}{2} \Bigg] \sin \Bigg[\dfrac{\alpha -\beta}{2} \Bigg]$

$$\implies \lim_{x \to 0} \dfrac{-2\sin \Bigg[\dfrac{3x+4x}{2}\Bigg]\sin \Bigg[\dfrac{3x-4x}{2}\Bigg]}{x \sin 2x}$$

$$\implies \lim_{x \to 0} \dfrac{-2\sin \Bigg[\dfrac{7x}{2}\Bigg]\sin \Bigg[\dfrac{-x}{2}\Bigg]}{x \sin 2x}$$

$$\implies \lim_{x \to 0} \dfrac{-2\sin \Bigg[\dfrac{7x}{2}\Bigg]\Bigg(-\sin \Bigg[\dfrac{x}{2}\Bigg]\Bigg)}{x \sin 2x}$$

$$\implies \lim_{x \to 0} \dfrac{2\sin \Bigg[\dfrac{7x}{2}\Bigg]\sin \Bigg[\dfrac{x}{2}\Bigg]}{x \sin 2x}$$

Step: 2

Split the entire function into product of three sine functions.

$$\implies \lim_{x \to 0} \dfrac{2\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x} \times \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\sin 2x}$$

$$\implies \lim_{x \to 0} \dfrac{2\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x} \times \sin \Bigg[\dfrac{x}{2}\Bigg] \times \dfrac{1}{\sin 2x}$$

Step: 3

There is a straight formula in limits for sine function. In other words, the ratio of sine of an angle to angle is one when limit of angle tends to zero. So, adjust each sine function to be in the form of the following identity.

$\displaystyle \lim_{\theta \to 0} \dfrac{\sin \theta}{\theta} = 1$

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x} \times \sin \Bigg[\dfrac{x}{2}\Bigg] \times \dfrac{1}{\sin 2x}$

It is required an $x$ term under the terms $\sin \Bigg[\dfrac{x}{2}\Bigg]$ and $\sin 2x$.

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x}$ $\times$ $\sin \Bigg[\dfrac{x}{2}\Bigg]$ $\times$ $\dfrac{1}{\sin 2x} \times 1$

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x}$ $\times$ $\sin \Bigg[\dfrac{x}{2}\Bigg]$ $\times$ $\dfrac{1}{\sin 2x} \times \dfrac{x}{x}$

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x}$ $\times$ $\dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{x}$ $\times$ $\dfrac{x}{\sin 2x}$

The limit belongs to all three multiplying functions. So, apply limit to each function.

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{x}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{x}{\sin 2x}$

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{x}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{1}{\dfrac{\sin 2x}{x}}$

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{x}$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{x}}$

The first, second and third sine functions contain $\dfrac{7x}{2}$, $\dfrac{x}{2}$ and $2x$ as angles. So, adjust the denominator of the each function to have same angle.

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x \times 1}$ $\times$ $\lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{x \times 1}$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{x \times 1}}$

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{x \times \dfrac{7}{7}}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{x \times \dfrac{2}{2}}$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{x \times \dfrac{2}{2}}}$

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\dfrac{7x}{7}}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{2 \times \dfrac{x}{2}}$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{\dfrac{2x}{2}}}$

$\implies 2 \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\dfrac{7x}{7} \times 1}$ $\times$ $\Bigg(\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]} \times \dfrac{1}{2}\Bigg)$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{2 \sin 2x}{2x}}$

$\implies 2 \times \dfrac{1}{2} \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\dfrac{7x}{7} \times \dfrac{2}{2}}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$ $\times$ $\Bigg(\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{2x}} \times \dfrac{1}{2}\Bigg)$

$\implies \dfrac{2}{2} \times \dfrac{1}{2} \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\dfrac{7x}{2} \times \dfrac{2}{7}}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{2x}}$

$\implies \require{cancel} \dfrac{\cancel{2}}{\cancel{2}} \times \dfrac{1}{2} \displaystyle \Bigg(\lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\dfrac{7x}{2}} \times \dfrac{1}{\dfrac{2}{7}}\Bigg)$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{2x}}$

$\implies \dfrac{1}{2} \displaystyle \Bigg(\lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\dfrac{7x}{2}} \times \dfrac{7}{2}\Bigg)$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{2x}}$

$\implies \dfrac{1}{2} \times \dfrac{7}{2} \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\dfrac{7x}{2}}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{2x}}$

$\implies \dfrac{7}{4} \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\Bigg[\dfrac{7x}{2}\Bigg]}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{2x}}$

Step: 4

Each multiplying function is successfully converted as a ratio of sine of angle to angle. Now, adjust the limit condition and it should be same as the angle of the respective sine function.

Set Limit of first multiplying function

We know that $x \to 0 \implies 7x \to 7 \times 0$

$\implies 7x \to 0 \implies \dfrac{7x}{2} \to \dfrac{0}{2}$

$\therefore \,\,\,\,\,\, x \to 0 \implies \dfrac{7x}{2} \to 0$

Therefore, when $x$ tends to zero, the value of $\dfrac{7x}{2}$ also tends to zero.

$\therefore \,\,\,\,\,\, \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\Bigg[\dfrac{7x}{2}\Bigg]}$ $=$ $\displaystyle \lim_{\dfrac{7x}{2} \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\Bigg[\dfrac{7x}{2}\Bigg]}$

Set Limit of second multiplying function

We know that $x \to 0 \implies \dfrac{x}{2} \to \dfrac{0}{2}$

$\therefore \,\,\,\,\,\, x \to 0 \implies \dfrac{x}{2} \to 0$

Therefore, the value of $\dfrac{x}{2}$ tends to zero when the value of $x$ tends to zero.

$\therefore \,\,\,\,\,\, \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$ $=$ $\displaystyle \lim_{\dfrac{x}{2} \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$

Set Limit of Third multiplying function

We know that $x \to 0 \implies 2x \to 2 \times 0$

$\therefore \,\,\,\,\,\, x \to 0 \implies 2x \to 0$

It is proved that the value of $2x$ tends to zero if $x$ tends to zero.

$\therefore \,\,\,\,\,\, \displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{2x} = \lim_{2x \to 0} \dfrac{\sin 2x}{2x}$

Step: 5

Transform the entire multiplying function by replacing the values of limits.

$\implies \dfrac{7}{4} \displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\Bigg[\dfrac{7x}{2}\Bigg]}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$ $\times$ $\dfrac{1}{\displaystyle \lim_{x \to 0} \dfrac{\sin 2x}{2x}}$ $=$ $\dfrac{7}{4} \displaystyle \lim_{\dfrac{7x}{2} \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\Bigg[\dfrac{7x}{2}\Bigg]}$ $\times$ $\displaystyle \lim_{\dfrac{x}{2} \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$ $\times$ $\dfrac{1}{\displaystyle \lim_{2x \to 0} \dfrac{\sin 2x}{2x}}$

Step: 6

Simplify the each function by applying ratio of sine of angle to angle rule when angle tends to zero.

$=$ $\dfrac{7}{4} \displaystyle \lim_{\dfrac{7x}{2} \to 0} \dfrac{\sin \Bigg[\dfrac{7x}{2}\Bigg]}{\Bigg[\dfrac{7x}{2}\Bigg]}$ $\times$ $\displaystyle \lim_{\dfrac{x}{2} \to 0} \dfrac{\sin \Bigg[\dfrac{x}{2}\Bigg]}{\Bigg[\dfrac{x}{2}\Bigg]}$ $\times$ $\dfrac{1}{\displaystyle \lim_{2x \to 0} \dfrac{\sin 2x}{2x}}$

According to this limit property, the value of ratio of sine of an angle to angle is one when the angle tends to zero.

$=$ $\dfrac{7}{4} \times 1 \times 1 \times 1$

$\therefore \,\,\,\,\,\, \displaystyle \lim_{x \to 0} \dfrac{\cos 3x -\cos 4x}{x \sin 2x} = \dfrac{7}{4}$

It is the required solution for this limits problem in according to calculus.



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