Key Concepts and Formulas

• Trigonometric Identities:
  • $1 - \cos(2\theta) = 2\sin^2(\theta)$
• Taylor Series Expansions (around x=0):
  • $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
  • $\tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$
• Limit Properties: The ability to substitute the limit value directly if the expression is continuous at that point, or to use algebraic manipulation and series expansions to evaluate indeterminate forms.

Step-by-Step Solution

Step 1: Simplify the numerator using a trigonometric identity. The numerator is $(1 - \cos 2x)^2$. We can use the identity $1 - \cos(2\theta) = 2\sin^2(\theta)$. Let $\theta = x$. Then $1 - \cos(2x) = 2\sin^2(x)$. So, the numerator becomes $(2\sin^2(x))^2 = 4\sin^4(x)$. $\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}} = \mathop {\lim }\limits_{x \to 0} \,{{{{\left( {2{{\sin }^2}x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}} = \mathop {\lim }\limits_{x \to 0} \,{{{4{{\sin }^4}x}} \over {2x\,\tan x\, - x\tan 2x}}$

Step 2: Use Taylor series expansions for $\sin(x)$ and $\tan(x)$ around $x=0$. As $x \to 0$, we can approximate the functions using their Taylor series expansions. For the numerator, $\sin(x) \approx x$. Thus, $\sin^4(x) \approx x^4$. For the denominator, we need the expansions for $\tan(x)$ and $\tan(2x)$: $\tan(x) = x + \frac{x^3}{3} + O(x^5)$ $\tan(2x) = (2x) + \frac{(2x)^3}{3} + O((2x)^5) = 2x + \frac{8x^3}{3} + O(x^5)$

Substitute these into the limit expression: $\mathop {\lim }\limits_{x \to 0} \,{{{4{{\sin }^4}x}} \over {2x\,\tan x\, - x\tan 2x}} \approx \mathop {\lim }\limits_{x \to 0} \,{{{4x^4}} \over {2x\left( {x + \frac{x^3}{3}} \right) - x\left( {2x + \frac{8x^3}{3}} \right)}}$ Note: We are using the first few terms of the series. Higher-order terms will not affect the limit as they will be of higher powers of $x$ and will vanish faster than the lowest order terms.

Step 3: Expand and simplify the denominator. $\mathop {\lim }\limits_{x \to 0} \,{{{4x^4}} \over {2x\left( {x + \frac{x^3}{3}} \right) - x\left( {2x + \frac{8x^3}{3}} \right)}} = \mathop {\lim }\limits_{x \to 0} \,{{{4x^4}} \over {\left( {2x^2 + \frac{2x^4}{3}} \right) - \left( {2x^2 + \frac{8x^4}{3}} \right)}}$ Combine like terms in the denominator: $\mathop {\lim }\limits_{x \to 0} \,{{{4x^4}} \over {2x^2 + \frac{2x^4}{3} - 2x^2 - \frac{8x^4}{3}}} = \mathop {\lim }\limits_{x \to 0} \,{{{4x^4}} \over {\frac{2x^4}{3} - \frac{8x^4}{3}}}$ $= \mathop {\lim }\limits_{x \to 0} \,{{{4x^4}} \over {-\frac{6x^4}{3}}} = \mathop {\lim }\limits_{x \to 0} \,{{{4x^4}} \over {-2x^4}}$

Step 4: Cancel out the common term $x^4$ and evaluate the limit. Since $x \to 0$ but $x \neq 0$, we can cancel $x^4$ from the numerator and the denominator. $\mathop {\lim }\limits_{x \to 0} \,{{{4x^4}} \over {-2x^4}} = \mathop {\lim }\limits_{x \to 0} \,{{4} \over {-2}}$ The expression is now a constant, so the limit is the constant itself. ${{4} \over {-2}} = -2$

Common Mistakes & Tips

• Incorrect Taylor Series: Ensure you are using the correct terms in the Taylor series expansion. For $\tan(x)$, the first term is $x$, and the next is $\frac{x^3}{3}$. For $\tan(2x)$, remember to substitute $2x$ into the series.
• Algebraic Errors: Be careful with signs and fractions when simplifying the denominator. Expanding the terms and then combining them is crucial.
• Order of Terms: When using Taylor series for limits, always look for the lowest power of $x$ in both the numerator and the denominator after simplification. These terms will dominate the limit. In this case, after simplification, the $x^4$ terms were the lowest order in both.

Summary

The problem requires evaluating a limit of an indeterminate form as $x \to 0$. We first simplify the numerator using the double angle identity for cosine. Then, we employ the Taylor series expansions for $\sin(x)$ and $\tan(x)$ around $x=0$ to approximate the functions. By substituting these expansions into the limit expression and simplifying the denominator, we arrive at a form where the lowest power of $x$ in the numerator and denominator can be identified and cancelled. This leads to a constant value, which is the result of the limit.

The final answer is $\boxed{-2}$.