Let `y=f(x) ={{:(e^((1)/(x^(2))), if x ne 0),( 0, if x=0):}` then which of the following can best represent the graph of `y=f(x)?`

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the function defined as:

\[ y = f(x) = \begin{cases} e^{\frac{1}{x^2}} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

### Step 1: Analyze the function for \( x \neq 0 \)

For \( x \neq 0 \), the function \( f(x) = e^{\frac{1}{x^2}} \) is defined. As \( x \) approaches 0 from either side (positive or negative), the term \( \frac{1}{x^2} \) approaches infinity. Therefore, \( e^{\frac{1}{x^2}} \) approaches infinity as well.

### Step 2: Analyze the function at \( x = 0 \)

At \( x = 0 \), the function is defined as \( f(0) = 0 \).

### Step 3: Check the limits as \( x \) approaches 0

- **Right-hand limit** as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{\frac{1}{x^2}} = +\infty \]

- **Left-hand limit** as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^{\frac{1}{x^2}} = +\infty \]

### Step 4: Check differentiability at \( x = 0 \)

To check differentiability at \( x = 0 \), we need to find the right-hand and left-hand derivatives.

- **Right-hand derivative**:
\[ f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{e^{\frac{1}{x^2}} - 0}{x} = \lim_{x \to 0^+} \frac{e^{\frac{1}{x^2}}}{x} \]
As \( x \to 0^+ \), \( e^{\frac{1}{x^2}} \to +\infty \) and \( x \to 0^+ \), thus this limit approaches \( +\infty \).

- **Left-hand derivative**:
\[ f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{e^{\frac{1}{x^2}} - 0}{x} = \lim_{x \to 0^-} \frac{e^{\frac{1}{x^2}}}{x} \]
As \( x \to 0^- \), \( e^{\frac{1}{x^2}} \to +\infty \) and \( x \to 0^- \), thus this limit approaches \( -\infty \).

Since the right-hand derivative approaches \( +\infty \) and the left-hand derivative approaches \( -\infty \), the function is not differentiable at \( x = 0 \).

### Step 5: Determine the behavior as \( x \) approaches \( +\infty \) and \( -\infty \)

- **Limit as \( x \to +\infty \)**:
\[ \lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} e^{\frac{1}{x^2}} = e^0 = 1 \]

- **Limit as \( x \to -\infty \)**:
\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} e^{\frac{1}{x^2}} = e^0 = 1 \]

### Step 6: Conclusion about the graph

The function \( f(x) \) approaches 1 as \( x \) goes to both \( +\infty \) and \( -\infty \), and it is not defined at \( x = 0 \) in a differentiable manner. Therefore, the graph of \( f(x) \) approaches the horizontal line \( y = 1 \) as \( x \) goes to both infinities, and it has a discontinuity at \( x = 0 \).

### Final Answer

The graph that best represents \( y = f(x) \) is the one that shows a discontinuity at \( x = 0 \) and approaches \( y = 1 \) as \( x \) approaches \( \pm \infty \).

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