We define a function h(x) on (-limit,+limit) by the following rules h(x)={ sin (1/x), x doesn’t equal 0. 0, x=0.  Compute the limit as x approaches 0 of k(x) = xh(x). Explain why is it invalid to simply compute the limit as k(x) = 0•h(0)=0•0=0?

We are given the function \( h(x) \), which is defined as:

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

We are tasked with computing the limit:

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

### Step-by-Step Solution:

#### 1. Behavior of \( h(x) \) as \( x \to 0 \)

For \( x \neq 0 \), \( h(x) = \sin\left(\frac{1}{x}\right) \). The function \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1 as \( x \to 0 \), because as \( x \) gets smaller, \( \frac{1}{x} \) becomes very large, and the sine function keeps oscillating. Thus, \( \sin\left(\frac{1}{x}\right) \) does not approach any specific value but continues to oscillate wildly.

#### 2. Computing the limit of \( k(x) = x h(x) \)

Now, we compute the limit of \( k(x) = x \sin\left(\frac{1}{x}\right) \) as \( x \to 0 \). Although \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1, we can bound the product \( x \sin\left(\frac{1}{x}\right) \) as follows:

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

By the Squeeze Theorem (or Sandwich Theorem), since both \( -|x| \) and \( |x| \) tend to 0 as \( x \to 0 \), the limit of \( x \sin\left(\frac{1}{x}\right) \) must also tend to 0:

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

Thus, the limit is:

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

### Why is it incorrect to compute the limit as \( k(0) = 0 \cdot h(0) = 0 \cdot 0 = 0 \)?

This approach is invalid because:

1. **The definition of a limit**: The value of \( k(x) \) at \( x = 0 \) is irrelevant for finding the limit as \( x \to 0 \). The limit considers the behavior of the function as \( x \) approaches 0, not the value at \( x = 0 \) itself.
   
2. **Discontinuity and oscillations**: The function \( \sin\left(\frac{1}{x}\right) \) is highly oscillatory as \( x \to 0 \), and while the product \( x \sin\left(\frac{1}{x}\right) \) approaches 0 due to the Squeeze Theorem, the function \( h(x) \) is not well-behaved at \( x = 0 \) in a simple sense. Hence, directly substituting \( x = 0 \) and calculating \( 0 \cdot 0 \) misses the important behavior of \( h(x) \) as \( x \to 0 \).

In conclusion, the limit calculation must take into account the oscillatory nature of \( h(x) \), and the product \( x \sin\left(\frac{1}{x}\right) \) must be handled carefully, rather than simply substituting values directly.

---

Would you like any further details on this explanation?

Here are 5 related questions for deeper exploration:

1. How does the Squeeze Theorem work in the context of limits?
2. What is an example of a function where direct substitution works for limits?
3. Can we find limits for functions with oscillations, like \( \sin\left(\frac{1}{x}\right) \), without the Squeeze Theorem?
4. What does it mean for a function to oscillate infinitely as it approaches a point?
5. What are some common misconceptions when computing limits involving multiplication by 0?

**Tip**: Always analyze the behavior of a function as you approach the point of interest, rather than focusing solely on its value at that point.

We define a function h(x) on (-limit,+limit) by the following rules h(x)={ sin (1/x), x doesn’t equal 0. 0, x=0. Compute the limit as x approaches 0 of k(x) = xh(x). Explain why is it invalid to simply compute the limit as k(x) = 0•h(0)=0•0=0?

Explore the limit of the function k(x) = x * sin(1/x) as x approaches 0, using the Squeeze Theorem to solve this oscillatory function. Understand why direct substitution is invalid in such cases. Suitable for advanced high school or early college students.

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