The value of `lim_(xrarr0)(((1-cos4x)(5+cosx))/(xtan5x))` is equal to

Answer

To find the limit \( \lim_{x \to 0} \frac{(1 - \cos 4x)(5 + \cos x)}{x \tan 5x} \), we can break it down step by step.

### Step 1: Rewrite the limit
We start with the limit:
\[ \lim_{x \to 0} \frac{(1 - \cos 4x)(5 + \cos x)}{x \tan 5x} \]

### Step 2: Use the limit identity for \(1 - \cos x\)
We know that:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]
Thus, we can express \(1 - \cos 4x\) in terms of \(x^2\):
\[ 1 - \cos 4x = \frac{1 - \cos 4x}{(4x)^2} \cdot (4x)^2 \]
So,
\[ \frac{1 - \cos 4x}{x^2} = \frac{1}{2} \cdot \frac{(4x)^2}{x^2} = \frac{16x^2}{x^2} = 16 \]

### Step 3: Rewrite the limit with \(1 - \cos 4x\)
Substituting this into our limit, we have:
\[ \lim_{x \to 0} \frac{(1 - \cos 4x)(5 + \cos x)}{x \tan 5x} = \lim_{x \to 0} \frac{(16)(5 + \cos x)(x^2)}{x \tan 5x} \]

### Step 4: Simplify the limit
This simplifies to:
\[ \lim_{x \to 0} \frac{16(5 + \cos x)x}{\tan 5x} \]

### Step 5: Use the limit identity for \(\tan x\)
We know that:
\[ \lim_{x \to 0} \frac{\tan x}{x} = 1 \]
Thus, we can express \(\tan 5x\) as:
\[ \tan 5x \approx 5x \text{ as } x \to 0 \]
So,
\[ \lim_{x \to 0} \frac{x}{\tan 5x} = \lim_{x \to 0} \frac{x}{5x} = \frac{1}{5} \]

### Step 6: Substitute back into the limit
Now we can substitute this back into our limit:
\[ \lim_{x \to 0} \frac{16(5 + \cos x)x}{\tan 5x} = \lim_{x \to 0} \frac{16(5 + \cos x)x}{5x} = \lim_{x \to 0} \frac{16(5 + \cos x)}{5} \]

### Step 7: Evaluate the limit as \(x\) approaches 0
As \(x\) approaches 0, \(\cos x\) approaches 1:
\[ \lim_{x \to 0} \frac{16(5 + \cos x)}{5} = \frac{16(5 + 1)}{5} = \frac{16 \cdot 6}{5} = \frac{96}{5} \]

### Final Result
Thus, the value of the limit is:
\[ \frac{96}{5} \]