`"Let "g(x)=|{:(f(x+c),f(x+2c),f(x+3c)),(f(c),f(2c),f(3c)),(f'(c),f'(2c),f'(3c)):}|,`
where c is constant, then find `lim_(xrarr0) (g(x))/(x).`

Explanation

To solve the problem, we need to find the limit:

\[ \lim_{x \to 0} \frac{g(x)}{x} \]

where

\[ g(x) = \begin{vmatrix} f(x+c) & f(x+2c) & f(x+3c) \\ f(c) & f(2c) & f(3c) \\ f'(c) & f'(2c) & f'(3c) \end{vmatrix} \]

### Step-by-Step Solution:

**Step 1: Evaluate \( g(0) \)**

First, we need to evaluate \( g(0) \):

\[ g(0) = \begin{vmatrix} f(c) & f(2c) & f(3c) \\ f(c) & f(2c) & f(3c) \\ f'(c) & f'(2c) & f'(3c) \end{vmatrix} \]

Notice that the first two rows are identical, which means the determinant is zero:

\[ g(0) = 0 \]

**Step 2: Check the limit form**

Since \( g(0) = 0 \), we have:

\[ \lim_{x \to 0} \frac{g(x)}{x} = \frac{0}{0} \]

This is an indeterminate form, so we can apply L'Hôpital's rule.

**Step 3: Apply L'Hôpital's Rule**

We differentiate \( g(x) \) with respect to \( x \):

Using properties of determinants, we differentiate the first row of the determinant:

\[ g'(x) = \begin{vmatrix} f'(x+c) & f'(x+2c) & f'(x+3c) \\ f(c) & f(2c) & f(3c) \\ f'(c) & f'(2c) & f'(3c) \end{vmatrix} \]

**Step 4: Evaluate \( g'(0) \)**

Now we evaluate \( g'(0) \):

\[ g'(0) = \begin{vmatrix} f'(c) & f'(2c) & f'(3c) \\ f(c) & f(2c) & f(3c) \\ f'(c) & f'(2c) & f'(3c) \end{vmatrix} \]

Again, the first and third rows are identical, so the determinant is zero:

\[ g'(0) = 0 \]

**Step 5: Apply L'Hôpital's Rule Again**

Since we still have \( \frac{0}{0} \), we apply L'Hôpital's rule again:

\[ \lim_{x \to 0} \frac{g'(x)}{1} \]

Differentiate \( g'(x) \) again:

\[ g''(x) = \begin{vmatrix} f''(x+c) & f''(x+2c) & f''(x+3c) \\ f(c) & f(2c) & f(3c) \\ f'(c) & f'(2c) & f'(3c) \end{vmatrix} \]

**Step 6: Evaluate \( g''(0) \)**

Now we evaluate \( g''(0) \):

\[ g''(0) = \begin{vmatrix} f''(c) & f''(2c) & f''(3c) \\ f(c) & f(2c) & f(3c) \\ f'(c) & f'(2c) & f'(3c) \end{vmatrix} \]

This determinant is not guaranteed to be zero, so we can compute it directly.

### Final Result

Thus, the limit we are looking for is:

\[ \lim_{x \to 0} \frac{g(x)}{x} = \frac{g''(0)}{1} = g''(0) \]