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

Options

A. `7/5`
B. `1`
C. `2/5`
D. None of these

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 1: Analyze the expression inside the limit

As \( x \to \infty \), the term \( x^5 \) dominates the expression \( x^5 + 7x^4 + 2 \). Thus, we can approximate:

\[ x^5 + 7x^4 + 2 \approx x^5 \quad \text{(for large } x\text{)} \]

### Step 2: Rewrite the limit

Now we can rewrite the limit as:

\[ \lim_{x \to \infty} \left( (x^5(1 + \frac{7}{x} + \frac{2}{x^5}))^c - x \right) \]

### Step 3: Simplify the expression

Using the property of exponents, we get:

\[ = \lim_{x \to \infty} \left( x^{5c} \left( 1 + \frac{7}{x} + \frac{2}{x^5} \right)^c - x \right) \]

### Step 4: Expand the binomial expression

As \( x \to \infty \), \( \frac{7}{x} \) and \( \frac{2}{x^5} \) approach 0. We can use the binomial expansion:

\[ \left( 1 + \frac{7}{x} + \frac{2}{x^5} \right)^c \approx 1 + c\left(\frac{7}{x}\right) \quad \text{(for large } x\text{)} \]

Thus, we have:

\[ \lim_{x \to \infty} \left( x^{5c} \left( 1 + c\frac{7}{x} \right) - x \right) \]

### Step 5: Substitute and simplify

Substituting this back into the limit gives:

\[ = \lim_{x \to \infty} \left( x^{5c} + c \cdot 7x^{5c - 1} - x \right) \]

### Step 6: Set conditions for the limit to be finite and non-zero

For the limit to be finite and non-zero, the highest power of \( x \) must cancel out. This means we need:

\[ 5c = 1 \quad \Rightarrow \quad c = \frac{1}{5} \]

### Step 7: Substitute \( c \) back into the limit

Now, substituting \( c = \frac{1}{5} \):

\[ = \lim_{x \to \infty} \left( x + \frac{7}{5} \cdot x^{1 - 1} - x \right) \]

This simplifies to:

\[ = \lim_{x \to \infty} \left( \frac{7}{5} \right) = \frac{7}{5} \]

Thus, the value of the limit is:

\[ \boxed{\frac{7}{5}} \]