`lim_(xto0)((1-cosx)(3+cos2x))/(x.tan2x)=`

Options

A. 0
B. 1
C. `(1)/(2)`
D. `-1`

Answer

To solve the limit \( \lim_{x \to 0} \frac{(1 - \cos x)(3 + \cos 2x)}{x \tan 2x} \), we can follow these steps:

### Step 1: Rewrite the expression using trigonometric identities
We know that \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \) and \( \tan 2x = \frac{\sin 2x}{\cos 2x} \). Thus, we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)(3 + \cos 2x)}{x \cdot \frac{\sin 2x}{\cos 2x}} \]

### Step 2: Simplify the expression
This can be simplified to:

\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)(3 + \cos 2x) \cos 2x}{x \sin 2x} \]

### Step 3: Rewrite \( \sin 2x \)
Using the double angle formula, we have \( \sin 2x = 2 \sin x \cos x \). Therefore, we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)(3 + \cos 2x) \cos 2x}{2x \sin x \cos x} \]

### Step 4: Cancel out common terms
We can cancel the factor of 2 in the numerator and denominator:

\[ \lim_{x \to 0} \frac{\sin^2\left(\frac{x}{2}\right)(3 + \cos 2x) \cos 2x}{x \sin x \cos x} \]

### Step 5: Use the limit property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
We can express \( \sin^2\left(\frac{x}{2}\right) \) as \( \left(\frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2 \cdot \left(\frac{x}{2}\right)^2 \):

\[ \lim_{x \to 0} \frac{\left(\frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2 \cdot \left(\frac{x}{2}\right)^2 (3 + \cos 2x) \cos 2x}{x \sin x \cos x} \]

### Step 6: Substitute the limits
As \( x \to 0 \), \( \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} \to 1 \) and \( \cos 2x \to 1 \). Thus, we have:

\[ \lim_{x \to 0} \frac{1^2 \cdot \left(\frac{x}{2}\right)^2 (3 + 1) \cdot 1}{x \cdot x} = \lim_{x \to 0} \frac{\frac{x^2}{4} \cdot 4}{x^2} = 1 \]

### Final Answer
Therefore, the limit evaluates to:

\[ \lim_{x \to 0} \frac{(1 - \cos x)(3 + \cos 2x)}{x \tan 2x} = 1 \]