Let `f(x)=g(x)(e^(1//x) -e^(-1//x))/(e^(1//x) + e^(-1//x))`, where g is a continuous function then `lim_(x to 0)` f(x) exist if

Options

A. g(x)=x+2
B. `g(x)=x^2+4`
C. `g(x) =xh(x)`, `h(x)` is a polynomial
D. g(x) is a constant function

Answer

To solve the problem, we need to analyze the function \( f(x) = g(x) \cdot \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \) as \( x \) approaches 0.

### Step 1: Simplify the expression for \( f(x) \)

The expression inside the limit can be simplified. We know that:

\[ \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} = \tanh\left(\frac{1}{x}\right) \]

Thus, we can rewrite \( f(x) \) as:

\[ f(x) = g(x) \cdot \tanh\left(\frac{1}{x}\right) \]

### Step 2: Analyze the behavior of \( \tanh\left(\frac{1}{x}\right) \)

As \( x \to 0^- \) (from the left), \( \frac{1}{x} \to -\infty \), and thus:

\[ \tanh\left(\frac{1}{x}\right) \to -1 \]

As \( x \to 0^+ \) (from the right), \( \frac{1}{x} \to +\infty \), and thus:

\[ \tanh\left(\frac{1}{x}\right) \to 1 \]

### Step 3: Find the left-hand limit and right-hand limit of \( f(x) \)

#### Left-hand limit as \( x \to 0^- \):

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

#### Right-hand limit as \( x \to 0^+ \):

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

### Step 4: Set the left-hand limit equal to the right-hand limit

For the limit \( \lim_{x \to 0} f(x) \) to exist, the left-hand limit must equal the right-hand limit:

\[ -g(0) = g(0) \]

This implies:

\[ 2g(0) = 0 \implies g(0) = 0 \]

### Conclusion

The limit \( \lim_{x \to 0} f(x) \) exists if \( g(0) = 0 \). Since \( g \) is a continuous function, this means that \( g(x) \) must approach 0 as \( x \) approaches 0.