The limit of `x sin (e^(-1//x))` as `x rarr0`

Options

A. is equal to 0
B. is equal to 1/2
C. is equal to e/2
D. does not exist

Answer

To find the limit of \( x \sin(e^{-1/x}) \) as \( x \to 0 \), we will follow these steps:

### Step 1: Change of Variable
Let \( y = \frac{1}{x} \). As \( x \to 0 \), \( y \to \infty \). Thus, we can rewrite the limit in terms of \( y \):
\[ \lim_{x \to 0} x \sin(e^{-1/x}) = \lim_{y \to \infty} \frac{1}{y} \sin(e^{-y}). \]

### Step 2: Analyze the Sine Function
Next, we need to analyze \( \sin(e^{-y}) \) as \( y \to \infty \). As \( y \to \infty \), \( e^{-y} \to 0 \). Therefore, we can use the fact that \( \sin(z) \approx z \) when \( z \) is close to 0:
\[ \sin(e^{-y}) \approx e^{-y}. \]

### Step 3: Substitute Back into the Limit
Now, substituting this approximation back into our limit gives:
\[ \lim_{y \to \infty} \frac{1}{y} \sin(e^{-y}) \approx \lim_{y \to \infty} \frac{1}{y} e^{-y}. \]

### Step 4: Evaluate the Limit
Now we need to evaluate \( \lim_{y \to \infty} \frac{e^{-y}}{y} \). As \( y \to \infty \), \( e^{-y} \) approaches 0 much faster than \( y \) approaches infinity. Therefore, we can conclude that:
\[ \lim_{y \to \infty} \frac{e^{-y}}{y} = 0. \]

### Conclusion
Thus, the limit we are looking for is:
\[ \lim_{x \to 0} x \sin(e^{-1/x}) = 0. \]

### Final Answer
The limit is \( 0 \).

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