`lim_(x to 0) (sin nx[(a-n)nx - tanx])/x^(2)=0`, where n is non-zero positive integer, then a is equal to:

Options

A. `(n+1)//n`
B. `n^(2)`
C. `1/n`
D. `n + 1/n`

Answer

To solve the limit problem, we need to evaluate the expression:

\[ \lim_{x \to 0} \frac{\sin(nx) \left[(a-n)nx - \tan(x)\right]}{x^2} = 0 \]

where \( n \) is a non-zero positive integer. Let's go through the steps to find the value of \( a \).

### Step 1: Rewrite the limit
We can separate the limit into two parts:

\[ \lim_{x \to 0} \left( \frac{\sin(nx)}{x^2} \cdot \left[(a-n)nx - \tan(x)\right] \right) \]

### Step 2: Use the small angle approximation
As \( x \to 0 \), we can use the approximation \( \tan(x) \approx x \). Thus, we can rewrite the expression:

\[ \tan(x) \approx x \implies \lim_{x \to 0} \left[(a-n)nx - \tan(x)\right] = \lim_{x \to 0} \left[(a-n)nx - x\right] = \lim_{x \to 0} \left[(a-n)n - 1\right]x \]

### Step 3: Substitute back into the limit
Now we substitute this back into our limit:

\[ \lim_{x \to 0} \frac{\sin(nx)}{x^2} \cdot \left[(a-n)n - 1\right]x \]

This simplifies to:

\[ \lim_{x \to 0} \frac{\sin(nx)}{x} \cdot \frac{(a-n)n - 1}{x} \]

### Step 4: Evaluate the limit of \(\frac{\sin(nx)}{x}\)
Using the standard limit \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \) for any constant \( k \):

\[ \lim_{x \to 0} \frac{\sin(nx)}{x} = n \]

### Step 5: Combine the limits
Now we have:

\[ n \cdot \lim_{x \to 0} \frac{(a-n)n - 1}{x} \]

For the limit to equal zero, the term \((a-n)n - 1\) must equal zero:

\[ (a-n)n - 1 = 0 \]

### Step 6: Solve for \( a \)
Now we can solve for \( a \):

\[ (a-n)n = 1 \implies a-n = \frac{1}{n} \implies a = n + \frac{1}{n} \]

### Conclusion
Thus, the value of \( a \) is:

\[ \boxed{n + \frac{1}{n}} \]