Let `f(x)={((tan^2[x])/(x^2-[x]^2),),((1)/(sqrt(|x|cot|x|)),):} {:(,"for",xgt0,),(,"for",x=0,),(,"for",xlt0,):} "where" [x]` and `{x}` denote respectively the greatest integer less than equal to x and the fractional part of x, then

Options

A. (a)`lim_(xto0^(+))f(x)=1`
B. (b)`lim_(xto0^(-))f(x)=1`
C. (c)`cot^(-1)(lim_(xto0^(-))f(x))^(2)=1`
D. (d)None of the above

Answer

To solve the problem step by step, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} \frac{\tan^2(\{x\})}{x^2 - [x]^2} & \text{for } x > 0 \\ \frac{1}{\sqrt{|x| \cot |x|}} & \text{for } x = 0 \\ \frac{1}{\sqrt{|x| \cot |x|}} & \text{for } x < 0 \end{cases} \]

Where \([x]\) is the greatest integer less than or equal to \(x\) and \(\{x\} = x - [x]\) is the fractional part of \(x\).

### Step 1: Find the Right-Hand Limit (RHL) as \(x\) approaches 0 from the positive side.

We need to compute:

\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\tan^2(\{x\})}{x^2 - [x]^2} \]

For \(x > 0\), \([x] = 0\) and \(\{x\} = x\). Thus, we can rewrite the limit as:

\[ \text{RHL} = \lim_{x \to 0^+} \frac{\tan^2(x)}{x^2 - 0^2} = \lim_{x \to 0^+} \frac{\tan^2(x)}{x^2} \]

Using the limit property \(\lim_{x \to 0} \frac{\tan(x)}{x} = 1\), we have:

\[ \lim_{x \to 0^+} \frac{\tan^2(x)}{x^2} = \left(\lim_{x \to 0^+} \frac{\tan(x)}{x}\right)^2 = 1^2 = 1 \]

### Step 2: Find the Left-Hand Limit (LHL) as \(x\) approaches 0 from the negative side.

We compute:

\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{\sqrt{|x| \cot |x|}} \]

For \(x < 0\), \(|x| = -x\) and \(\cot(-x) = -\cot(x)\). Thus, we can rewrite the limit as:

\[ \text{LHL} = \lim_{x \to 0^-} \frac{1}{\sqrt{-x \cdot \cot(-x)}} = \lim_{x \to 0^-} \frac{1}{\sqrt{-x \cdot (-\cot(x))}} = \lim_{x \to 0^-} \frac{1}{\sqrt{x \cot(x)}} \]

Using the property \(\cot(x) = \frac{\cos(x)}{\sin(x)}\) and knowing that \(\lim_{x \to 0} \cot(x) = \infty\), we find:

\[ \text{LHL} = \lim_{x \to 0^-} \frac{1}{\sqrt{x \cdot \infty}} = 0 \]

### Step 3: Compare the Limits

Now we compare the right-hand limit and the left-hand limit:

- RHL = 1
- LHL = 0

Since \( \text{RHL} \neq \text{LHL} \), the limit of \(f(x)\) as \(x\) approaches 0 does not exist.

### Step 4: Conclusion

The function \(f(x)\) does not have a limit as \(x\) approaches 0. Therefore, we conclude that:

\[ \lim_{x \to 0} f(x) \text{ does not exist.} \]