If `f(x)={(xsin,((1)/(x)),xne0),(0,,x=0):}` Then, `lim_(xrarr0) f(x)`

Options

A. is equal to 1
B. is equal to`-1`
C. is equal to 0
D. does not exist

Answer

To find the limit of the function \( f(x) \) as \( x \) approaches 0, we start with the definition of the function:

\[ f(x) = \begin{cases} x \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

We need to evaluate:

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

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

Since we are interested in the limit as \( x \) approaches 0, we will focus on the case when \( x \neq 0 \):

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

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

The function \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1 as \( x \) approaches 0. Therefore, we can say:

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

### Step 3: Multiply by \( x \)

Now, if we multiply the entire inequality by \( x \) (keeping in mind that \( x \) approaches 0), we have:

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

### Step 4: Apply the Squeeze Theorem

As \( x \) approaches 0, both \( -x \) and \( x \) approach 0. Therefore, by the Squeeze Theorem:

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

### Step 5: Conclude the limit

Thus, we conclude:

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

### Final Answer

The limit is:

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