# Simplify $\displaystyle \lim_{x \to 0} \dfrac{1-\cos x \sqrt{\cos 2x}}{x^2}$

$\displaystyle \lim_{x \to 0} \dfrac{1-\cos x \sqrt{\cos 2x}}{x^2}$

$x$ is a literal but represents an angle in radian. In this limits problem, the trigonometric functions $\cos x$ and $\cos 2x$ are involved. The product of $\cos x$ and square root of $\cos 2x$ is subtracted from one. The subtraction of them is compared with an algebraic term $x^2$ to find the value of the ratio between them when the value of the $x$ approaches zero.

$= \displaystyle \lim_{x \to 0} \dfrac{1-\cos x \sqrt{\cos 2x}}{x^2} \times 1$

$= \displaystyle \lim_{x \to 0} \dfrac{1-\cos x \sqrt{\cos 2x}}{x^2}$ $\times$ $\dfrac{1+\cos x \sqrt{\cos 2x}}{1+\cos x \sqrt{\cos 2x}}$

$= \displaystyle \lim_{x \to 0} \dfrac{(1-\cos x \sqrt{\cos 2x})(1+\cos x \sqrt{\cos 2x})}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= \displaystyle \lim_{x \to 0} \dfrac{{(1)}^2-{(\cos x \sqrt{\cos 2x})}^2}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= \displaystyle \lim_{x \to 0} \dfrac{1-\cos^2 x \cos 2x}{x^2 (1+\cos x \sqrt{\cos 2x})}$

The limits problem can be simplified in two methods from this step.

### Method: 1

In this method, each cosine term in the numerator is expressed in terms of sine function.

$= \displaystyle \lim_{x \to 0} \dfrac{1-\cos^2 x \cos 2x}{x^2 (1+\cos x \sqrt{\cos 2x})}$

#### Step: 1

Express $\cos^2 x$ in terms of $sin x$ and expand $\cos 2x$ in terms of $\sin x$.

According to Pythagorean identity of sine and cosine, $\cos^2 x = 1-\sin^2 x$ and $\cos 2x = 1-2\sin^2 x$ as per the expansion of the $\cos 2x$ formula in terms of sine.

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

$= \displaystyle \lim_{x \to 0} \dfrac{1-(1-2\sin^2 x -\sin^2 x +2\sin^4 x)}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= \displaystyle \lim_{x \to 0} \dfrac{1-(1-3\sin^2 x + 2\sin^4 x)}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= \displaystyle \lim_{x \to 0} \dfrac{1-1+3\sin^2 x -2\sin^4 x}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$\require{cancel} = \displaystyle \lim_{x \to 0} \dfrac{\cancel{1}-\cancel{1}+3\sin^2 x -2\sin^4 x}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= \displaystyle \lim_{x \to 0} \dfrac{3\sin^2 x -2\sin^4 x}{x^2 (1+\cos x \sqrt{\cos 2x})}$

#### Step: 2

Dissect the ratio into two multiplying factors.

$= \displaystyle \lim_{x \to 0} \dfrac{3\sin^2 x -2\sin^4 x}{x^2}$ $\times$ $\dfrac{1}{(1+\cos x \sqrt{\cos 2x})}$

#### Step: 3

Apply limit to both multiplying factors because it belongs to both of them and then continue the procedure of the simplification.

$= \displaystyle \lim_{x \to 0} \dfrac{3\sin^2 x -2\sin^4 x}{x^2}$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{1}{(1+\cos x \sqrt{\cos 2x})}$

$= \displaystyle \lim_{x \to 0} \Bigg[\dfrac{3\sin^2 x}{x^2} -\dfrac{2\sin^4 x}{x^2}\Bigg]$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{1}{(1+\cos x \sqrt{\cos 2x})}$

$= \Bigg[\displaystyle \lim_{x \to 0} \dfrac{3\sin^2 x}{x^2}$ $-$ $\displaystyle \lim_{x \to 0} \dfrac{2\sin^4 x}{x^2}\Bigg]$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{1}{(1+\cos x \sqrt{\cos 2x})}$

$= \Bigg[\displaystyle 3\lim_{x \to 0} \dfrac{\sin^2 x}{x^2}$ $-$ $\displaystyle 2\lim_{x \to 0} \dfrac{\sin^4 x}{x^2}\Bigg]$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{1}{(1+\cos x \sqrt{\cos 2x})}$

$= \Bigg[3 \displaystyle \lim_{x \to 0} {\Bigg(\dfrac{\sin x}{x}\Bigg)}^2$ $-$ $\displaystyle 2\lim_{x \to 0} \dfrac{\sin^2 x \times \sin^2 x}{x^2}\Bigg]$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{1}{(1+\cos x \sqrt{\cos 2x})}$

$= \Bigg[ 3 {\Bigg(\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}\Bigg)}^2$ $-$ $\displaystyle 2\lim_{x \to 0} \sin^2 x \times \dfrac{\sin^2 x}{x^2}\Bigg]$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{1}{(1+\cos x \sqrt{\cos 2x})}$

$= \Bigg[ 3 {\Bigg(\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}\Bigg)}^2$ $-$ $\displaystyle 2\lim_{x \to 0} \sin^2 x \times {\Bigg(\dfrac{\sin x}{x}\Bigg)}^2 \Bigg]$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{1}{(1+\cos x \sqrt{\cos 2x})}$

$= \Bigg[ 3 {\Bigg(\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}\Bigg)}^2$ $-$ $2 \Bigg[ \displaystyle \lim_{x \to 0} \sin^2 x \times \displaystyle \lim_{x \to 0} {\Bigg(\dfrac{\sin x}{x}\Bigg)}^2 \Bigg] \Bigg]$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{1}{(1+\cos x \sqrt{\cos 2x})}$

$= \Bigg[ 3 {\Bigg(\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}\Bigg)}^2$ $-$ $2 \Bigg[ \displaystyle \lim_{x \to 0} \sin^2 x \times {\Bigg(\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}\Bigg)}^2 \Bigg] \Bigg]$ $\times$ $\displaystyle \lim_{x \to 0} \dfrac{1}{(1+\cos x \sqrt{\cos 2x})}$

#### Step: 4

Substitute $x = 0$ and evaluate its value.

$= \Big[3{(1)}^2$ $-$ $2 \Big[\sin^2 (0) \times {(1)}^2 \Big] \Big]$ $\times$ $\dfrac{1}{(1+\cos (0) \sqrt{\cos 2(0)})}$

$= \Big[3 \times 1$ $-$ $2 [0 \times 1 ] \Big]$ $\times$ $\dfrac{1}{(1+1 \times \sqrt{\cos (0)})}$

$= \Big[3-2 \times [0] \Big]$ $\times$ $\dfrac{1}{(1+1 \times \sqrt{1})}$

$= \Big[3-0 \Big]$ $\times$ $\dfrac{1}{(1+1 \times 1)}$

$= \Big[3\Big]$ $\times$ $\dfrac{1}{(1+1)}$

$= \Big[3\Big]$ $\times$ $\dfrac{1}{(2)}$

$= 3 \times \dfrac{1}{2}$

$\therefore \,\,\,\,\,\, \displaystyle \lim_{x \to 0} \dfrac{1-\cos x \sqrt{\cos 2x}}{x^2} = \dfrac{3}{2}$

It is the required answer for this calculus problem.

### Method: 2

The problem can be solved in another method but Keep cosine terms in numerator but expand the $\cos 2x$ in terms of cosine function.

#### Step: 1

As per the expansion of the cosine of double angle, $\cos 2x$ is written in terms of cosine function as $\cos 2x = 2\cos^2 x -1$

$= \displaystyle \lim_{x \to 0} \dfrac{1-\cos^2 x (2\cos^2 x -1)}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= \displaystyle \lim_{x \to 0} \dfrac{1-2\cos^4 x + \cos^2 x}{x^2 (1+\cos x \sqrt{\cos 2x})}$

#### Step: 2

The numerator can be expressed as two multiplying factors to simplify the numerator.

$= \displaystyle \lim_{x \to 0} \dfrac{-(2\cos^4 x-\cos^2 x-1)}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= \displaystyle \lim_{x \to 0} \dfrac{-(2\cos^4 x-2\cos^2 x +\cos^2 x -1)}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= -\displaystyle \lim_{x \to 0} \dfrac{2\cos^4 x-2\cos^2 x +\cos^2 x -1}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= -\displaystyle \lim_{x \to 0} \dfrac{2\cos^2 x (\cos^2 x -1) + (\cos^2 x -1)}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= -\displaystyle \lim_{x \to 0} \dfrac{(2\cos^2 x +1)(\cos^2 x -1)}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= \displaystyle \lim_{x \to 0} \dfrac{(2\cos^2 x +1)(1-\cos^2 x)}{x^2 (1+\cos x \sqrt{\cos 2x})}$

$= \displaystyle \lim_{x \to 0} \dfrac{(2\cos^2 x +1) \times \sin^2 x}{(1+\cos x \sqrt{\cos 2x})x^2}$

$= \displaystyle \lim_{x \to 0} \dfrac{2\cos^2 x +1}{(1+\cos x \sqrt{\cos 2x})} \times \dfrac{\sin^2 x}{x^2}$

#### Step: 3

Apply limit to both multiplying factors and proceed simplification.

$= \displaystyle \lim_{x \to 0} \dfrac{2\cos^2 x +1}{(1+\cos x \sqrt{\cos 2x})} \times \displaystyle \lim_{x \to 0} \dfrac{\sin^2 x}{x^2}$

$= \displaystyle \lim_{x \to 0} \dfrac{2\cos^2 x +1}{(1+\cos x \sqrt{\cos 2x})} \times \displaystyle \lim_{x \to 0} {\Bigg(\dfrac{\sin x}{x}\Bigg)}^2$

$= \displaystyle \lim_{x \to 0} \dfrac{2\cos^2 x +1}{(1+\cos x \sqrt{\cos 2x})} \times {\Bigg(\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x}\Bigg)}^2$

#### Step: 4

Substitute $x = 0$ to obtain the required solution.

$= \dfrac{2\cos^2 (0) +1}{(1+\cos (0) \sqrt{\cos 2(0)})} \times {(1)}^2$

$= \dfrac{2 \times {(1)}^2 +1}{(1+ 1 \times \sqrt{\cos (0)})} \times 1$

$= \dfrac{2 \times 1 +1}{(1+ 1 \times \sqrt{1})} \times 1$

$= \dfrac{2+1}{(1+ 1 \times 1)} \times 1$

$= \dfrac{3}{(1+1)}$

$\therefore \,\,\,\,\,\, \displaystyle \lim_{x \to 0} \dfrac{1-\cos x \sqrt{\cos 2x}}{x^2} = \dfrac{3}{2}$

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