If `f(x)` is derivable at `x=2` such that `f(2)=2` and `f'(2)=4 `, then the value of `lim_(h rarr0)(1)/(h^(2))(ln f(2+h^(2))-ln f(2-h^(2)))` is equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the limit

\[ \lim_{h \to 0} \frac{1}{h^2} \left( \ln f(2+h^2) - \ln f(2-h^2) \right), \]

we can start by using the properties of logarithms.

### Step 1: Rewrite the limit using properties of logarithms
Using the property of logarithms that states \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\), we can rewrite the limit as:

\[ \lim_{h \to 0} \frac{1}{h^2} \ln \left( \frac{f(2+h^2)}{f(2-h^2)} \right). \]

### Step 2: Apply L'Hôpital's Rule
As \(h\) approaches \(0\), both \(f(2+h^2)\) and \(f(2-h^2)\) approach \(f(2)\), which is \(2\). Therefore, the limit takes the indeterminate form \(\frac{0}{0}\). We can apply L'Hôpital's Rule, which states that if the limit results in \(\frac{0}{0}\), we can differentiate the numerator and the denominator.

### Step 3: Differentiate the numerator and the denominator
We differentiate the numerator and the denominator with respect to \(h\):

- The derivative of the numerator \(\ln \left( \frac{f(2+h^2)}{f(2-h^2)} \right)\) using the chain rule is:

\[ \frac{d}{dh} \left( \ln \left( \frac{f(2+h^2)}{f(2-h^2)} \right) \right) = \frac{1}{\frac{f(2+h^2)}{f(2-h^2)}} \cdot \left( \frac{f'(2+h^2) \cdot 2h}{f(2-h^2)} + \frac{-f'(2-h^2) \cdot 2h}{f(2+h^2)} \right). \]

- The derivative of the denominator \(h^2\) is \(2h\).

### Step 4: Substitute back into the limit
Now we substitute back into the limit:

\[ \lim_{h \to 0} \frac{\frac{1}{\frac{f(2+h^2)}{f(2-h^2)}} \cdot \left( f'(2+h^2) \cdot 2h - f'(2-h^2) \cdot 2h \right)}{2h}. \]

### Step 5: Simplify the expression
This simplifies to:

\[ \lim_{h \to 0} \frac{1}{2} \cdot \frac{f'(2+h^2) - f'(2-h^2)}{f(2+h^2) - f(2-h^2)}. \]

### Step 6: Evaluate the limit
As \(h\) approaches \(0\), \(f'(2+h^2)\) approaches \(f'(2)\) and \(f(2+h^2)\) approaches \(f(2)\). Thus, we can substitute \(f'(2) = 4\) and \(f(2) = 2\):

\[ \lim_{h \to 0} \frac{f'(2+h^2) - f'(2-h^2)}{f(2+h^2) - f(2-h^2)} = \frac{4 - 4}{2 - 2} = \frac{0}{0}. \]

### Step 7: Apply L'Hôpital's Rule again
We apply L'Hôpital's Rule again to differentiate the numerator and denominator again.

After simplification, we find that:

\[ \frac{f''(2)}{f'(2)}. \]

### Conclusion
Finally, substituting the known values, we find that the limit evaluates to:

\[ \frac{4}{2} = 4. \]

Thus, the value of the limit is

\[ \boxed{4}. \]