If `f(x) = (2x+ tanx)/(x) , x!=0`, is continuous at x = 0, then f(0) equals

Options

A. 0
B. 1
C. 2
D. 3

Answer

To determine the value of \( f(0) \) such that the function \( f(x) = \frac{2x + \tan x}{x} \) is continuous at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0. The function is defined for \( x \neq 0 \), so we will evaluate the limit as follows:

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

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{2x + \tan x}{x} \]

### Step 2: Simplify the expression.

We can separate the limit into two parts:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{2x}{x} + \frac{\tan x}{x} \right) \]

This simplifies to:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} (2 + \frac{\tan x}{x}) \]

### Step 3: Evaluate \( \lim_{x \to 0} \frac{\tan x}{x} \).

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

\[ \lim_{x \to 0} f(x) = 2 + 1 = 3 \]

### Step 4: Define \( f(0) \).

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

\[ f(0) = \lim_{x \to 0} f(x) = 3 \]

### Conclusion:

Thus, \( f(0) = 3 \).

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