Suppose `f (x)` is differentiable at `x=1 and lim _(h to0) (f(1+h))/(h) =5.` Then `f '(1)` is equal to-

Options

A. 6
B. 5
C. 4
D. 3

Answer

To solve the problem, we need to find the value of \( f'(1) \) given that \( \lim_{h \to 0} \frac{f(1+h)}{h} = 5 \) and that \( f(x) \) is differentiable at \( x = 1 \).

### Step-by-step Solution:

1. **Understanding the Limit**:
We are given that:
\[ \lim_{h \to 0} \frac{f(1+h)}{h} = 5 \]
This limit can be interpreted as the behavior of the function \( f \) around \( x = 1 \).

2. **Using the Definition of the Derivative**:
The derivative of \( f \) at \( x = 1 \) is defined as:
\[ f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \]
Since \( f \) is differentiable at \( x = 1 \), this limit exists.

3. **Rearranging the Given Limit**:
We can rewrite the given limit:
\[ \lim_{h \to 0} \frac{f(1+h)}{h} = \lim_{h \to 0} \left( \frac{f(1+h) - f(1) + f(1)}{h} \right) \]
This can be split into two parts:
\[ \lim_{h \to 0} \left( \frac{f(1+h) - f(1)}{h} + \frac{f(1)}{h} \right) \]

4. **Identifying the Limits**:
The first part of the limit is \( f'(1) \):
\[ f'(1) + \lim_{h \to 0} \frac{f(1)}{h} = 5 \]
The second part, \( \lim_{h \to 0} \frac{f(1)}{h} \), will approach infinity unless \( f(1) = 0 \).

5. **Setting \( f(1) = 0 \)**:
To avoid the second term going to infinity, we must have:
\[ f(1) = 0 \]

6. **Substituting Back**:
Now substituting \( f(1) = 0 \) into our limit:
\[ f'(1) + \lim_{h \to 0} \frac{0}{h} = 5 \]
The limit \( \lim_{h \to 0} \frac{0}{h} = 0 \).

7. **Final Calculation**:
Therefore, we have:
\[ f'(1) + 0 = 5 \implies f'(1) = 5 \]

### Conclusion:
Thus, the value of \( f'(1) \) is \( 5 \).