Let `lim_(xto0) ([x]^(2))/(x^(2))=m,` where `[.]` denotes greatest integer. Then, m equals to a. `-(1)/(sqrt(2))` b. `(1)/(sqrt(2))` c. `sqrt(2)` d. Limit doesn't exist

Options

A. `-(1)/(sqrt(2))`
B. `(1)/(sqrt(2))`
C. `sqrt(2)`
D. `-sqrt(2)`

Explanation

To solve the limit \( \lim_{x \to 0} \frac{[x]^2}{x^2} = m \), where \([x]\) denotes the greatest integer function, we will analyze the behavior of the function as \( x \) approaches 0 from both the right and the left.

### Step 1: Analyze the Right-Hand Limit (RHL)

When \( x \) approaches 0 from the right (i.e., \( x \to 0^+ \)):
- For \( 0 < x < 1 \), the greatest integer function \([x]\) is 0.
- Therefore, we have:
\[ [x]^2 = 0^2 = 0 \]
- Substituting this into the limit gives:
\[ \lim_{x \to 0^+} \frac{[x]^2}{x^2} = \lim_{x \to 0^+} \frac{0}{x^2} = 0 \]

### Step 2: Analyze the Left-Hand Limit (LHL)

Now, consider \( x \) approaching 0 from the left (i.e., \( x \to 0^- \)):
- For \( -1 < x < 0 \), the greatest integer function \([x]\) is -1.
- Therefore, we have:
\[ [x]^2 = (-1)^2 = 1 \]
- Substituting this into the limit gives:
\[ \lim_{x \to 0^-} \frac{[x]^2}{x^2} = \lim_{x \to 0^-} \frac{1}{x^2} \]
- As \( x \) approaches 0 from the left, \( x^2 \) approaches 0, which means:
\[ \frac{1}{x^2} \to +\infty \]

### Step 3: Conclusion

Since the right-hand limit (RHL) is 0 and the left-hand limit (LHL) approaches \( +\infty \), we conclude that:
\[ \lim_{x \to 0} \frac{[x]^2}{x^2} \text{ does not exist.} \]

Thus, the value of \( m \) is:
\[ \text{Option D: Limit doesn't exist.} \]