Consider the function f(x)={1x2sin1x2,x≠00,x=0

. Then which of the following is not true
a.

None of these
b.

Limit does not exist at x=0

c.

Limit exists at x=0

d.

Function is discontinuous at x=0

The function is given as:

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

### Key points:
1. **Check for the limit at \( x = 0 \)**:
   \[
   \lim_{x \to 0} \frac{1}{x^2} \sin\left(\frac{1}{x^2}\right)
   \]
   We need to check if this limit exists or not. For large values of \( \frac{1}{x^2} \), the sine function oscillates between -1 and 1. However, since the term \( \frac{1}{x^2} \) grows very large as \( x \to 0 \), the oscillations become more rapid, and there is no single value the function approaches. Thus, **the limit does not exist at \( x = 0 \)**.

2. **Check for continuity at \( x = 0 \)**:
   Since the limit does not exist at \( x = 0 \), the function cannot be continuous at this point. For a function to be continuous at \( x = 0 \), the limit as \( x \to 0 \) must exist and be equal to \( f(0) \), which is 0. Since the limit doesn't exist, the function is **discontinuous** at \( x = 0 \).

### Now, let's analyze the options:

- **Option (a)**: None of these.  
  This means all the other options are incorrect. We'll check this last.
  
- **Option (b)**: Limit does not exist at \( x = 0 \).  
  This is **true** as explained above.
  
- **Option (c)**: Limit exists at \( x = 0 \).  
  This is **false** because the limit does not exist at \( x = 0 \).
  
- **Option (d)**: Function is discontinuous at \( x = 0 \).  
  This is **true** because the limit does not exist, so the function is discontinuous.

### Conclusion:
The option that is **not true** is **(c)**: "Limit exists at \( x = 0 \)". 

---

Would you like further details or explanations? Here are some related questions:

1. What is the definition of continuity for a function at a point?
2. How do oscillating functions behave near singularities?
3. What are some examples of discontinuous functions?
4. How is the sine function involved in periodic behavior?
5. How can limits fail to exist in certain cases?

**Tip**: When dealing with limits involving oscillatory functions, it's helpful to consider whether the oscillations dampen or grow as you approach the limit point.

Consider the function f(x) = { 1/x^2 sin(1/x^2), x ≠ 0; 0, x = 0 }. Then which of the following is not true: a. None of these, b. Limit does not exist at x=0, c. Limit exists at x=0, d. Function is discontinuous at x=0.

Explore the limit and continuity properties of the function f(x) = 1/x^2 sin(1/x^2) for x ≠ 0 and 0 for x = 0. This problem examines oscillatory behavior near x=0, determining whether the limit exists and if the function is continuous. Suitable for undergraduate calculus students.

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