If `f(x) (10^(x)+7^(x)-14^(x)-5^(x))/(1-cos x), x != 0 ` is continuous at x = 0 , then f(0) = …

Options

A. 2 log 7
B. `2 log (10/7)`
C. `log(4)log (5/7)`
D. log 70

Answer

To determine the value of \( f(0) \) for the function

\[ f(x) = \frac{10^x + 7^x - 14^x - 5^x}{1 - \cos x}, \quad x \neq 0 \]

and ensure that it is continuous at \( x = 0 \), we will follow these steps:

### Step 1: Evaluate the limit as \( x \) approaches 0
To find \( f(0) \), we need to compute the limit:

\[ f(0) = \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{10^x + 7^x - 14^x - 5^x}{1 - \cos x} \]

### Step 2: Substitute \( x = 0 \)
Substituting \( x = 0 \) directly into the function gives:

\[ f(0) = \frac{10^0 + 7^0 - 14^0 - 5^0}{1 - \cos(0)} = \frac{1 + 1 - 1 - 1}{1 - 1} = \frac{0}{0} \]

This results in an indeterminate form \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

We differentiate the numerator and the denominator:

- **Numerator**:
\[ \frac{d}{dx}(10^x + 7^x - 14^x - 5^x) = 10^x \ln(10) + 7^x \ln(7) - 14^x \ln(14) - 5^x \ln(5) \]

- **Denominator**:
\[ \frac{d}{dx}(1 - \cos x) = \sin x \]

### Step 4: Rewrite the limit
Now we rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{10^x \ln(10) + 7^x \ln(7) - 14^x \ln(14) - 5^x \ln(5)}{\sin x} \]

### Step 5: Substitute \( x = 0 \) again
Substituting \( x = 0 \):

\[ = \frac{10^0 \ln(10) + 7^0 \ln(7) - 14^0 \ln(14) - 5^0 \ln(5)}{\sin(0)} = \frac{\ln(10) + \ln(7) - \ln(14) - \ln(5)}{0} \]

This still results in \( \frac{0}{0} \), so we apply L'Hôpital's Rule again.

### Step 6: Differentiate again
Differentiating again:

- **Numerator**:
\[ \frac{d}{dx}(10^x \ln(10) + 7^x \ln(7) - 14^x \ln(14) - 5^x \ln(5)) = 10^x (\ln(10))^2 + 7^x (\ln(7))^2 - 14^x (\ln(14))^2 - 5^x (\ln(5))^2 \]

- **Denominator**:
\[ \frac{d}{dx}(\sin x) = \cos x \]

### Step 7: Rewrite the limit again
Now we have:

\[ \lim_{x \to 0} \frac{10^x (\ln(10))^2 + 7^x (\ln(7))^2 - 14^x (\ln(14))^2 - 5^x (\ln(5))^2}{\cos x} \]

### Step 8: Substitute \( x = 0 \) one last time
Substituting \( x = 0 \):

\[ = \frac{1 \cdot (\ln(10))^2 + 1 \cdot (\ln(7))^2 - 1 \cdot (\ln(14))^2 - 1 \cdot (\ln(5))^2}{1} \]

### Step 9: Simplify the expression
Thus, we have:

\[ f(0) = (\ln(10))^2 + (\ln(7))^2 - (\ln(14))^2 - (\ln(5))^2 \]

Using the identity \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \):

\[ = \ln(10) + \ln(7) - \ln(14) - \ln(5) = \ln\left(\frac{10 \cdot 7}{14 \cdot 5}\right) = \ln\left(\frac{70}{70}\right) = \ln(1) = 0 \]

### Conclusion
Thus, the value of \( f(0) \) is:

\[ \boxed{0} \]