If `f(x)={{:((5|x|+4tanx)/(x),xne0),(k,x=0):}`, then `f(x)` is continuous at x = 0 for

Options

A. k = 9
B. `k=-1`
C. no value of k
D. `k=2`

Answer

To determine the value of \( k \) for which the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit and the right-hand limit at \( x = 0 \) are equal to \( f(0) = k \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \begin{cases} \frac{5|x| + 4\tan x}{x} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

2. **Find \( f(0) \)**:
\[ f(0) = k \]

3. **Calculate the left-hand limit as \( x \) approaches 0**:
\[ f(0^-) = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{5(-x) + 4\tan x}{x} = \lim_{x \to 0^-} \frac{-5x + 4\tan x}{x} \]
Since \( |x| = -x \) for \( x < 0 \).

4. **Simplify the limit**:
\[ = \lim_{x \to 0^-} \left(-5 + \frac{4\tan x}{x}\right) \]
As \( x \to 0 \), \( \frac{\tan x}{x} \to 1 \), so:
\[ = -5 + 4 \cdot 1 = -5 + 4 = -1 \]
Thus,
\[ f(0^-) = -1 \]

5. **Calculate the right-hand limit as \( x \) approaches 0**:
\[ f(0^+) = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{5x + 4\tan x}{x} = \lim_{x \to 0^+} \left(5 + \frac{4\tan x}{x}\right) \]
Again, as \( x \to 0 \), \( \frac{\tan x}{x} \to 1 \), so:
\[ = 5 + 4 \cdot 1 = 5 + 4 = 9 \]
Thus,
\[ f(0^+) = 9 \]

6. **Set the limits equal to \( k \)**:
For \( f(x) \) to be continuous at \( x = 0 \):
\[ f(0^-) = f(0) = f(0^+) \]
This gives us:
\[ -1 = k = 9 \]

7. **Conclusion**:
Since \( -1 \neq 9 \), there is no value of \( k \) that makes \( f(x) \) continuous at \( x = 0 \).

### Final Answer:
There is no value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \).