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Step-by-Step Solution
Step 1: Understand the Given Expression
We need to evaluate
$ \displaystyle \lim_{x \to 0} \frac{\tan\bigl(\pi \sin^2 x \bigr) \;+\;\bigl(|x| - \sin\bigl(x [\,x\,]\bigr)\bigr)^2}{x^2}. $
Here, $[x]$ is the greatest integer less than or equal to $x$.
Step 2: Check the Right-Hand Limit (RHL) as $x \to 0^+$
For $x \to 0^+$, we have $[x] = 0$. Thus, $x [x] = x \cdot 0 = 0$ and $\sin\bigl(x [x]\bigr) = \sin(0) = 0$. Also, $|x| = x$ since $x$ is positive.
Substitute $[x] = 0$ into the expression inside the limit:
$$
\tan\bigl(\pi \sin^2 x\bigr) + (|x| - \sin(x [x]))^2
= \tan\bigl(\pi \sin^2 x\bigr) + (x - 0)^2
= \tan\bigl(\pi \sin^2 x\bigr) + x^2.
$$
Hence, for the right-hand limit,
$$
\lim_{x \to 0^+} \frac{\tan\bigl(\pi \sin^2 x\bigr) + x^2}{x^2}.
$$
As $x \to 0$, $\sin x \approx x$, so $\sin^2 x \approx x^2$. Therefore, $\tan\bigl(\pi \sin^2 x\bigr) \approx \tan\bigl(\pi x^2\bigr)$. For very small $x$, $\tan\bigl(\pi x^2\bigr) \approx \pi x^2$. Thus,
$$
\frac{\tan\bigl(\pi \sin^2 x\bigr)}{x^2} \approx \pi.
$$
So the right-hand limit becomes
$$
\lim_{x \to 0^+} \left( \frac{\tan\bigl(\pi \sin^2 x\bigr)}{x^2} + \frac{x^2}{x^2} \right)
= \pi + 1.
$$
Step 3: Check the Left-Hand Limit (LHL) as $x \to 0^-$
For $x \to 0^-$, $[x] = -1$ because $x$ is a small negative number. Hence, $x [x] = x \cdot (-1) = -x$. Also, $|x| = -x$ (since $x$ is negative).
Then $\sin\bigl(x [x]\bigr) = \sin(-x) = -\sin x$. Therefore, $|x| - \sin\bigl(x [x]\bigr) = -x - \bigl(-\sin x\bigr) = -x + \sin x.$
Substitute $[x] = -1$ into the expression:
$$
\tan\bigl(\pi \sin^2 x\bigr) + \bigl(|x| - \sin(x [x])\bigr)^2
= \tan\bigl(\pi \sin^2 x\bigr) + (-x + \sin x)^2.
$$
So the left-hand limit is
$$
\lim_{x \to 0^-} \frac{\tan\bigl(\pi \sin^2 x\bigr) + (-x + \sin x)^2}{x^2}.
$$
As before, $\tan\bigl(\pi \sin^2 x\bigr) \approx \pi x^2$ for small $x$. Now consider $(-x + \sin x)^2$. For small $x$, $\sin x \approx x - \frac{x^3}{6} + \dots$, so
$$
-x + \sin x \approx -x + \left(x - \frac{x^3}{6}\right) = -\frac{x^3}{6} + \dots \,.
$$
Squaring this gives $\bigl(-\frac{x^3}{6}\bigr)^2 = \frac{x^6}{36}$, which is of a higher order term compared to $x^2$. Thus, when divided by $x^2$, it vanishes as $x \to 0$.
Therefore, the dominant term in the numerator is $\tan\bigl(\pi \sin^2 x\bigr) \approx \pi x^2$. Hence,
$$
\lim_{x \to 0^-} \frac{\tan\bigl(\pi \sin^2 x\bigr)}{x^2} = \pi,
$$
and the squared term contributes nothing in the limit. Thus the left-hand limit is
$$
\pi + 0 = \pi.
$$
Step 4: Compare the Two One-Sided Limits
We found:
Right-Hand Limit (as $x \to 0^+$): $\pi + 1$
Left-Hand Limit (as $x \to 0^-$): $\pi$
Step 5: Conclude the Overall Limit
Since the right-hand limit and the left-hand limit are not equal ($\pi + 1 \neq \pi$), the limit
$
\displaystyle \lim_{x \to 0} \frac{\tan\bigl(\pi \sin^2 x\bigr) + (|x| - \sin(x [x]))^2}{x^2}
$
does not exist.
