The graph of the function `y = f(x)` has a unique tangent at the point `(e^a,0)` through which the graph passes then `lim__(x->e^a) (log_e {1+7f (x)} - sin f(x))/(3f(x))`

Options

A. 1
B. 2
C. 7
D. None of these

Answer

To solve the limit problem given in the question, we will follow these steps:

### Step 1: Identify the limit expression
We need to evaluate the limit:
\[ \lim_{x \to e^a} \frac{\log_e(1 + 7f(x)) - \sin(f(x))}{3f(x)} \]

### Step 2: Substitute the point of tangency
Given that the graph passes through the point \((e^a, 0)\), we know:
\[ f(e^a) = 0 \]
Substituting \(x = e^a\):
\[ \log_e(1 + 7f(e^a)) - \sin(f(e^a)) = \log_e(1 + 7 \cdot 0) - \sin(0) = \log_e(1) - 0 = 0 \]
Thus, the numerator becomes \(0\).

### Step 3: Evaluate the denominator
Similarly, substituting \(x = e^a\) into the denominator:
\[ 3f(e^a) = 3 \cdot 0 = 0 \]
So, we have a \(0/0\) indeterminate form:
\[ \frac{0}{0} \]

### Step 4: Apply L'Hôpital's Rule
Since we have a \(0/0\) form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
if the limit on the right exists.

### Step 5: Differentiate the numerator and denominator
Differentiate the numerator:
\[ \frac{d}{dx} \left( \log_e(1 + 7f(x)) - \sin(f(x)) \right) = \frac{7f'(x)}{1 + 7f(x)} - \cos(f(x)) \cdot f'(x) \]
Differentiate the denominator:
\[ \frac{d}{dx}(3f(x)) = 3f'(x) \]

### Step 6: Rewrite the limit using derivatives
Now we can rewrite the limit:
\[ \lim_{x \to e^a} \frac{\frac{7f'(x)}{1 + 7f(x)} - \cos(f(x)) f'(x)}{3f'(x)} \]

### Step 7: Simplify the expression
We can factor out \(f'(x)\) from the numerator:
\[ = \lim_{x \to e^a} \frac{f'(x) \left( \frac{7}{1 + 7f(x)} - \cos(f(x)) \right)}{3f'(x)} \]
Assuming \(f'(x) \neq 0\) near \(x = e^a\), we can cancel \(f'(x)\):
\[ = \lim_{x \to e^a} \frac{7}{3(1 + 7f(x))} - \frac{\cos(f(x))}{3} \]

### Step 8: Substitute \(x = e^a\)
Now substituting \(x = e^a\):
\[ = \frac{7}{3(1 + 7 \cdot 0)} - \frac{\cos(0)}{3} = \frac{7}{3 \cdot 1} - \frac{1}{3} = \frac{7}{3} - \frac{1}{3} = \frac{6}{3} = 2 \]

### Final Answer
Thus, the limit is:
\[ \boxed{2} \]