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To determine if the graph of the given curve \( f(x) \) has a tangent at the origin, we need to analyze the differentiability of \( f(x) \) at \( x = 0 \). 

### Given Function:
\[
f(x) = 
\begin{cases} 
x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0, \\
0 & \text{if } x = 0.
\end{cases}
\]

### Condition for Tangent at the Origin:
For \( f(x) \) to have a tangent at \( x = 0 \), the derivative \( f'(0) \) must exist, which is defined as:
\[
f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - f(0)}{h}.
\]
Given that \( f(0) = 0 \), this simplifies to:
\[
f'(0) = \lim_{h \to 0} \frac{f(h)}{h}.
\]

### Analyzing the Limit:
For \( h \neq 0 \),
\[
f(h) = h^2 \sin\left(\frac{1}{h}\right).
\]
Thus, the limit becomes:
\[
f'(0) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right).
\]

### Behavior of the Limit:
- Since \( |\sin\left(\frac{1}{h}\right)| \leq 1 \), we have \( |h \sin\left(\frac{1}{h}\right)| \leq |h| \).
- As \( h \) approaches 0, \( |h| \) also approaches 0. Therefore, the limit:
\[
\lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0.
\]
This shows that \( f'(0) = 0 \).

### Conclusion:
The limit exists, meaning the derivative exists and equals 0. Hence, **the graph of the curve does have a tangent at the origin**.

The correct answer is **A. Yes, because the limit exists.**

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Would you like more details on this solution, or do you have any other questions?

Here are some related questions you might consider:
1. What is the geometrical interpretation of the tangent at \( x = 0 \) for this function?
2. How would the behavior of \( \sin\left(\frac{1}{x}\right) \) alone affect the differentiability of similar functions?
3. Can you describe a function that is continuous but not differentiable at a point?
4. What happens if the function were modified to \( x^3 \sin\left(\frac{1}{x}\right) \) for \( x \neq 0 \)?
5. What role does the limit definition of the derivative play in determining the tangent line?

**Tip:** When analyzing limits involving trigonometric functions, remember that bounding the trigonometric part between -1 and 1 is often a helpful first step.

Does the graph of the following curve have a tangent at the origin? Give reasons for the answer.
f(x) = { x^2 sin(1/x), x ≠ 0 
 0, x = 0 }

Analyze the differentiability of the piecewise function f(x) = { x^2 sin(1/x), x ≠ 0; 0, x = 0 } at the origin. The problem explores the limit definition of the derivative and applies the Squeeze Theorem, suitable for students in Grades 11-12.

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