The integer $n$ for which $\lim_{x \to 0} \frac{(\sin x - x)\tan x}{x^n}$ is a finite non zero number, is

Looking at this limit problem, we need to determine the value of $n$ that makes the limit finite and non-zero.

To analyze the behavior near $x = 0$, we'll use Taylor series expansions:

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$

$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - ...$

Therefore:

$\sin x - x = -\frac{x^3}{6} + \frac{x^5}{120} - ...$

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For $\tan x$:

$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + ...$

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Now we compute the product:

$(\sin x - x)\tan x = \left(-\frac{x^3}{6} + \frac{x^5}{120} - ...\right)\left(x + \frac{x^3}{3} + ...\right)$

Expanding the leading terms:

$(\sin x - x)\tan x = -\frac{x^3}{6} \cdot x - \frac{x^3}{6} \cdot \frac{x^3}{3} + \frac{x^5}{120} \cdot x + ...$

$(\sin x - x)\tan x = -\frac{x^4}{6} - \frac{x^6}{18} + \frac{x^6}{120} + ...$

$(\sin x - x)\tan x = -\frac{x^4}{6} + \text{higher order terms}$

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The numerator behaves like $-\frac{x^4}{6}$ as $x \to 0$.

For the limit to be finite and non-zero, we need the denominator to have the same order as the numerator.

Therefore $n = 4$.

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We can verify:

$\lim_{x \to 0} \frac{(\sin x - x)\tan x}{x^4} = \lim_{x \to 0} \frac{-\frac{x^4}{6} + \text{higher order terms}}{x^4} = -\frac{1}{6}$

This is indeed finite and non-zero.

Therefore, the answer is $n = 4$.

Note: The stated correct answer is Option 1 (3), which appears to be inconsistent with the mathematical derivation shown above. The calculation demonstrates that $n = 4$ (Option 3) yields a finite non-zero limit.