Let f(x)=`{:{((3|x|+4tanx)/x, x ne 0),(k , x =0):}`
Then f(x) is continuous at x = 0 for ,

Options

A. k =7
B. k =1
C. No 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 analyze the function given by:

\[ f(x) = \begin{cases} \frac{3|x| + 4\tan x}{x}, & x \neq 0 \\ k, & x = 0 \end{cases} \]

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0 from the right (positive side).

For \( x > 0 \), we have \( |x| = x \). Thus, we can rewrite \( f(x) \) as:

\[ f(x) = \frac{3x + 4\tan x}{x} = 3 + \frac{4\tan x}{x} \]

Now we need to find:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left( 3 + \frac{4\tan x}{x} \right) \]

Using the known limit \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \):

\[ \lim_{x \to 0^+} f(x) = 3 + 4 \cdot 1 = 7 \]

### Step 2: Find the limit of \( f(x) \) as \( x \) approaches 0 from the left (negative side).

For \( x < 0 \), we have \( |x| = -x \). Thus, we can rewrite \( f(x) \) as:

\[ f(x) = \frac{3(-x) + 4\tan x}{x} = \frac{-3x + 4\tan x}{x} = -3 + \frac{4\tan x}{x} \]

Now we need to find:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left( -3 + \frac{4\tan x}{x} \right) \]

Again, using the limit \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \):

\[ \lim_{x \to 0^-} f(x) = -3 + 4 \cdot 1 = 1 \]

### Step 3: Check for continuity at \( x = 0 \).

For \( f(x) \) to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) = k \]

From our calculations:

- \( \lim_{x \to 0^+} f(x) = 7 \)
- \( \lim_{x \to 0^-} f(x) = 1 \)

Since \( 7 \neq 1 \), the limits from the left and right do not match. Therefore, there is no value of \( k \) that can make \( f(x) \) continuous at \( x = 0 \).

### Conclusion:

The function \( f(x) \) is not continuous at \( x = 0 \) for any value of \( k \).

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