Evaluate the following :
`lim_(x to 0)(xtan4x)/(1-cos4x)`

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{x \tan(4x)}{1 - \cos(4x)} \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the given limit:
\[ \lim_{x \to 0} \frac{x \tan(4x)}{1 - \cos(4x)} \]

### Step 2: Substitute the definition of \(\tan\)
Recall that \( \tan(4x) = \frac{\sin(4x)}{\cos(4x)} \). Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{x \cdot \frac{\sin(4x)}{\cos(4x)}}{1 - \cos(4x)} = \lim_{x \to 0} \frac{x \sin(4x)}{\cos(4x) (1 - \cos(4x))} \]

### Step 3: Use the identity for \(1 - \cos(4x)\)
Using the identity \(1 - \cos(4x) = 2 \sin^2(2x)\), we can rewrite the limit:
\[ \lim_{x \to 0} \frac{x \sin(4x)}{\cos(4x) \cdot 2 \sin^2(2x)} \]

### Step 4: Simplify the expression
Now, we can simplify the limit:
\[ \lim_{x \to 0} \frac{x \sin(4x)}{2 \sin^2(2x) \cos(4x)} \]

### Step 5: Apply the small angle approximation
As \(x\) approaches \(0\), we can use the small angle approximation:
- \(\sin(4x) \approx 4x\)
- \(\sin(2x) \approx 2x\)

Thus, we have:
\[ \sin(4x) \approx 4x \quad \text{and} \quad \sin(2x) \approx 2x \]

Substituting these approximations into the limit gives:
\[ \lim_{x \to 0} \frac{x \cdot 4x}{2(2x)^2 \cdot \cos(4x)} = \lim_{x \to 0} \frac{4x^2}{8x^2 \cos(4x)} = \lim_{x \to 0} \frac{4}{8 \cos(4x)} \]

### Step 6: Evaluate the limit
As \(x\) approaches \(0\), \(\cos(4x)\) approaches \(1\):
\[ \lim_{x \to 0} \frac{4}{8 \cdot 1} = \frac{4}{8} = \frac{1}{2} \]

### Final Answer
Thus, the limit evaluates to:
\[ \lim_{x \to 0} \frac{x \tan(4x)}{1 - \cos(4x)} = \frac{1}{2} \]