For a certain value of 'c' `lim _(x to oo) [(x ^(5) +7x^(4)+2 )^(c ) -x]` is finite and non-zero. Then the vlaue oimit is :

Answer

To solve the limit problem, we need to find the value of \( c \) such that

\[ \lim_{x \to \infty} \left[ (x^5 + 7x^4 + 2)^c - x \right] \]

is finite and non-zero.

### Step-by-Step Solution:

1. **Identify the dominant term**: As \( x \) approaches infinity, the term \( x^5 \) in \( x^5 + 7x^4 + 2 \) dominates. Thus, we can factor out \( x^5 \):
\[ (x^5 + 7x^4 + 2)^c = x^{5c} \left(1 + \frac{7}{x} + \frac{2}{x^5}\right)^c \]

2. **Simplify the expression inside the limit**:
As \( x \to \infty \), the terms \( \frac{7}{x} \) and \( \frac{2}{x^5} \) approach 0. Therefore, we can approximate:
\[ \left(1 + \frac{7}{x} + \frac{2}{x^5}\right)^c \approx 1 + c\left(\frac{7}{x}\right) \text{ (using the binomial expansion)} \]

3. **Substituting back into the limit**:
Now substituting back into our limit:
\[ \lim_{x \to \infty} \left[ x^{5c} \left(1 + c\frac{7}{x}\right) - x \right] \]
This simplifies to:
\[ \lim_{x \to \infty} \left[ x^{5c} + 7c x^{5c - 1} - x \right] \]

4. **Setting the limit to be finite**:
For the limit to be finite and non-zero, the powers of \( x \) must balance. This means:
\[ 5c = 1 \quad \text{(to match the power of } x \text{ with } x^1\text{)} \]
Solving for \( c \):
\[ c = \frac{1}{5} \]

5. **Substituting \( c \) back into the limit**:
Now we substitute \( c = \frac{1}{5} \) back into our limit:
\[ \lim_{x \to \infty} \left[ x^{5 \cdot \frac{1}{5}} + 7 \cdot \frac{1}{5} x^{5 \cdot \frac{1}{5} - 1} - x \right] \]
This simplifies to:
\[ \lim_{x \to \infty} \left[ x + \frac{7}{5} - x \right] = \lim_{x \to \infty} \frac{7}{5} = \frac{7}{5} \]

### Final Answer:
Thus, the value of the limit is:
\[ \frac{7}{5} \]