If `lim_(xrarr0)((cotx)(e^(x)-1)-cos^(2)x)/(sinx)=K,Kepsilonr` then find the value of `[K+11/7]`, where `[.]` represents the greatest integer function.

Explanation

To solve the limit problem given, we will proceed step by step.

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{(\cot x)(e^x - 1) - \cos^2 x}{\sin x} \]
We know that \(\cot x = \frac{\cos x}{\sin x}\). Therefore, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\frac{\cos x}{\sin x}(e^x - 1) - \cos^2 x}{\sin x} \]

### Step 2: Combine terms
Next, we can take a common denominator in the numerator:
\[ = \lim_{x \to 0} \frac{\cos x (e^x - 1) - \cos^2 x \sin x}{\sin^2 x} \]

### Step 3: Factor out \(\cos x\)
Now, we factor out \(\cos x\) from the numerator:
\[ = \lim_{x \to 0} \frac{\cos x \left(e^x - 1 - \cos x \sin x\right)}{\sin^2 x} \]

### Step 4: Expand \(e^x\) and \(\sin x\)
Using Taylor series expansions around \(x = 0\):
- \(e^x \approx 1 + x + \frac{x^2}{2} + O(x^3)\)
- \(\sin x \approx x - \frac{x^3}{6} + O(x^5)\)
- \(\cos x \approx 1 - \frac{x^2}{2} + O(x^4)\)

Substituting these approximations into the limit:
\[ e^x - 1 \approx x + \frac{x^2}{2} \]
\[ \cos x \sin x \approx (1 - \frac{x^2}{2})(x - \frac{x^3}{6}) \approx x - \frac{x^3}{6} - \frac{x^3}{2} \approx x - \frac{2x^3}{3} \]

### Step 5: Simplify the expression
Now substituting back into the limit:
\[ e^x - 1 - \cos x \sin x \approx \left(x + \frac{x^2}{2}\right) - \left(x - \frac{2x^3}{3}\right) = \frac{x^2}{2} + \frac{2x^3}{3} \]

Thus, we have:
\[ = \lim_{x \to 0} \frac{\cos x \left(\frac{x^2}{2} + \frac{2x^3}{3}\right)}{\sin^2 x} \]

### Step 6: Substitute \(\sin^2 x\)
Using \(\sin^2 x \approx x^2\):
\[ = \lim_{x \to 0} \frac{\cos x \left(\frac{x^2}{2} + \frac{2x^3}{3}\right)}{x^2} \]

### Step 7: Evaluate the limit
Now we can evaluate the limit:
\[ = \lim_{x \to 0} \left(\cos x \left(\frac{1}{2} + \frac{2x}{3}\right)\right) \]
As \(x \to 0\), \(\cos x \to 1\):
\[ = \frac{1}{2} + 0 = \frac{1}{2} \]

Thus, we have found that \(K = \frac{1}{2}\).

### Step 8: Find \(K + \frac{11}{7}\)
Now we need to calculate:
\[ K + \frac{11}{7} = \frac{1}{2} + \frac{11}{7} \]
Finding a common denominator (which is 14):
\[ = \frac{7}{14} + \frac{22}{14} = \frac{29}{14} \]

### Step 9: Apply the greatest integer function
Now we apply the greatest integer function:
\[ \left[\frac{29}{14}\right] = 2 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{2} \]