If `lim_(x to oo) x sin ((1)/(x)) =A and lim_(x to 0) x sin ((1)/(x)) =B`, then which one of the following is correct?

Options

A. A =1 and B = 0
B. A =0 and B = 1
C. A =0 and B = 0
D. A =1 and B = 1

Answer

To solve the problem, we need to evaluate the two limits:

1. \( A = \lim_{x \to \infty} x \sin\left(\frac{1}{x}\right) \)
2. \( B = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \)

### Step 1: Evaluate \( A \)

We start with the limit as \( x \) approaches infinity:

\[ A = \lim_{x \to \infty} x \sin\left(\frac{1}{x}\right) \]

To simplify this expression, we can substitute \( t = \frac{1}{x} \). As \( x \to \infty \), \( t \to 0 \). Therefore, we can rewrite the limit as:

\[ A = \lim_{t \to 0} \frac{\sin(t)}{t} \]

Using the well-known limit:

\[ \lim_{t \to 0} \frac{\sin(t)}{t} = 1 \]

Thus, we find:

\[ A = 1 \]

### Step 2: Evaluate \( B \)

Next, we evaluate the limit as \( x \) approaches 0:

\[ B = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \]

As \( x \) approaches 0, \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1. Therefore, we can bound \( x \sin\left(\frac{1}{x}\right) \):

\[ -x \leq x \sin\left(\frac{1}{x}\right) \leq x \]

As \( x \to 0 \), both bounds \( -x \) and \( x \) approach 0. By the Squeeze Theorem, we conclude:

\[ B = 0 \]

### Conclusion

Now we have:

- \( A = 1 \)
- \( B = 0 \)

Thus, the relationship between \( A \) and \( B \) is:

\[ A > B \]