The value of the function f at x=0 so that the function `f(x)=(2^(x)-2^(-x))/(x), x ne0` , is continuous at x=0 , is

Options

A. 0
B. log 2
C. log 4
D. `2^(4)`

Answer

To determine the value of the function \( f \) at \( x = 0 \) such that the function

\[ f(x) = \frac{2^x - 2^{-x}}{x} \quad \text{for } x \neq 0 \]

is continuous at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Evaluate the limit
We start by calculating the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{2^x - 2^{-x}}{x} \]

### Step 2: Apply L'Hôpital's Rule
Since both the numerator and denominator approach 0 as \( x \) approaches 0, we can apply L'Hôpital's Rule, which states that if the limit results in an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and denominator:

\[ \lim_{x \to 0} \frac{2^x - 2^{-x}}{x} = \lim_{x \to 0} \frac{(2^x \ln 2) - (-2^{-x} \ln 2)}{1} \]

### Step 3: Simplify the expression
This simplifies to:

\[ \lim_{x \to 0} \left( 2^x \ln 2 + 2^{-x} \ln 2 \right) \]

### Step 4: Substitute \( x = 0 \)
Now we can substitute \( x = 0 \):

\[ = 2^0 \ln 2 + 2^{0} \ln 2 = \ln 2 + \ln 2 = 2 \ln 2 \]

### Step 5: Rewrite using properties of logarithms
Using the property of logarithms, we can rewrite \( 2 \ln 2 \):

\[ 2 \ln 2 = \ln(2^2) = \ln 4 \]

### Step 6: Conclusion
Thus, for the function \( f(x) \) to be continuous at \( x = 0 \), we must have:

\[ f(0) = \lim_{x \to 0} f(x) = \ln 4 \]

Therefore, the value of the function \( f \) at \( x = 0 \) is:

\[ \boxed{\ln 4} \]

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