1. Key Concepts and Formulas

• Limits of sequences involving powers: For a real number $a$, the limit of $a^n$ as $n \rightarrow \infty$ behaves as follows:
  • If $|a| < 1$, then $\lim_{n \rightarrow \infty} a^n = 0$.
  • If $|a| > 1$, then $\lim_{n \rightarrow \infty} a^n = \infty$.
  • If $a = 1$, then $\lim_{n \rightarrow \infty} a^n = 1$.
  • If $a = -1$, then $\lim_{n \rightarrow \infty} a^n$ does not exist (oscillates between 1 and -1).
• Continuity of a function: A function $f(x)$ is continuous at a point $c$ if $\lim_{x \rightarrow c} f(x) = f(c)$. This requires the limit to exist, the function value to exist, and for them to be equal. For continuity on an interval, this must hold for all points in the interval.
• Piecewise function limits: When evaluating limits of piecewise functions, especially those defined by a limit process, we need to consider the limit from the left and the limit from the right at points where the definition of the function might change.

2. Step-by-Step Solution

We are given the function $f(x)=\lim\limits_{n \rightarrow \infty} \frac{\cos (2 \pi x)-x^{2 n} \sin (x-1)}{1+x^{2 n+1}-x^{2 n}}$. We need to find the values of $x$ for which $f(x)$ is continuous. We will analyze the limit by considering different cases for $|x|$.

Step 1: Analyze the behavior of $x^{2n}$ and $x^{2n+1}$ as $n \rightarrow \infty$. The behavior of $x^{2n}$ and $x^{2n+1}$ as $n \rightarrow \infty$ depends on the value of $x$.

• If $|x| < 1$, then $\lim_{n \rightarrow \infty} x^{2n} = 0$ and $\lim_{n \rightarrow \infty} x^{2n+1} = 0$.
• If $|x| > 1$, then $\lim_{n \rightarrow \infty} x^{2n} = \infty$ and $\lim_{n \rightarrow \infty} x^{2n+1} = \infty$.
• If $x = 1$, then $\lim_{n \rightarrow \infty} x^{2n} = 1$ and $\lim_{n \rightarrow \infty} x^{2n+1} = 1$.
• If $x = -1$, then $\lim_{n \rightarrow \infty} x^{2n} = 1$ (since $2n$ is even) and $\lim_{n \rightarrow \infty} x^{2n+1} = -1$ (since $2n+1$ is odd).

Step 2: Evaluate $f(x)$ for $|x| < 1$. For $|x| < 1$, we have $\lim_{n \rightarrow \infty} x^{2n} = 0$ and $\lim_{n \rightarrow \infty} x^{2n+1} = 0$. Substituting these into the expression for $f(x)$: $f(x) = \lim\limits_{n \rightarrow \infty} \frac{\cos (2 \pi x)-x^{2 n} \sin (x-1)}{1+x^{2 n+1}-x^{2 n}} = \frac{\cos (2 \pi x) - 0 \cdot \sin (x-1)}{1+0-0} = \cos (2 \pi x)$ The function $\cos(2 \pi x)$ is continuous for all real numbers. Therefore, $f(x)$ is continuous for all $x$ in $(-1, 1)$.

Step 3: Evaluate $f(x)$ for $|x| > 1$. For $|x| > 1$, we have $\lim_{n \rightarrow \infty} x^{2n} = \infty$ and $\lim_{n \rightarrow \infty} x^{2n+1} = \infty$. To evaluate the limit, we can divide the numerator and the denominator by the highest power of $x$ in the denominator, which is $x^{2n+1}$ or $x^{2n}$. Let's divide by $x^{2n}$: $f(x) = \lim\limits_{n \rightarrow \infty} \frac{\frac{\cos (2 \pi x)}{x^{2n}} - \sin (x-1)}{\frac{1}{x^{2n}}+x-1}$ As $n \rightarrow \infty$, $\frac{\cos (2 \pi x)}{x^{2n}} \rightarrow 0$ because $|\cos(2 \pi x)| \leq 1$. So, $f(x) = \frac{0 - \sin (x-1)}{0 + x - 1} = \frac{-\sin (x-1)}{x-1}$ This function is a ratio of continuous functions, and the denominator $x-1$ is non-zero for $|x| > 1$. Thus, $f(x) = \frac{-\sin (x-1)}{x-1}$ is continuous for all $x$ such that $|x| > 1$.

Step 4: Evaluate $f(x)$ at $x = 1$. We need to check for continuity at $x=1$. We must evaluate the limit as $x \rightarrow 1$ and compare it with $f(1)$. First, let's find $f(1)$ by substituting $x=1$ into the original limit expression: $f(1) = \lim\limits_{n \rightarrow \infty} \frac{\cos (2 \pi (1))-1^{2 n} \sin (1-1)}{1+1^{2 n+1}-1^{2 n}} = \lim\limits_{n \rightarrow \infty} \frac{\cos (2 \pi)-\sin (0)}{1+1-1} = \frac{1-0}{1} = 1$ Now, we need to evaluate the limit as $x \rightarrow 1$. We should consider the left-hand limit ($x \rightarrow 1^-$) and the right-hand limit ($x \rightarrow 1^+$). For $x \rightarrow 1^-$, we are in the case $|x| < 1$. So, $f(x) = \cos(2 \pi x)$. $\lim\limits_{x \rightarrow 1^-} f(x) = \lim\limits_{x \rightarrow 1^-} \cos (2 \pi x) = \cos (2 \pi) = 1$ For $x \rightarrow 1^+$, we are in the case $|x| > 1$. So, $f(x) = \frac{-\sin (x-1)}{x-1}$. $\lim\limits_{x \rightarrow 1^+} f(x) = \lim\limits_{x \rightarrow 1^+} \frac{-\sin (x-1)}{x-1}$ Let $y = x-1$. As $x \rightarrow 1^+$, $y \rightarrow 0^+$. $\lim\limits_{y \rightarrow 0^+} \frac{-\sin y}{y} = -1$ Since $\lim\limits_{x \rightarrow 1^-} f(x) = 1$ and $\lim\limits_{x \rightarrow 1^+} f(x) = -1$, the limit $\lim\limits_{x \rightarrow 1} f(x)$ does not exist. Therefore, $f(x)$ is discontinuous at $x=1$.

Step 5: Evaluate $f(x)$ at $x = -1$. We need to check for continuity at $x=-1$. We must evaluate the limit as $x \rightarrow -1$ and compare it with $f(-1)$. First, let's find $f(-1)$ by substituting $x=-1$ into the original limit expression: $f(-1) = \lim\limits_{n \rightarrow \infty} \frac{\cos (2 \pi (-1))-(-1)^{2 n} \sin (-1-1)}{1+(-1)^{2 n+1}-(-1)^{2 n}}$ Since $2n$ is even, $(-1)^{2n} = 1$. Since $2n+1$ is odd, $(-1)^{2n+1} = -1$. $f(-1) = \lim\limits_{n \rightarrow \infty} \frac{\cos (-2 \pi)-1 \cdot \sin (-2)}{1+(-1)-1} = \frac{1 - \sin (-2)}{1-1-1} = \frac{1 + \sin (2)}{-1} = -(1 + \sin 2)$ Now, we need to evaluate the limit as $x \rightarrow -1$. We should consider the left-hand limit ($x \rightarrow -1^-$) and the right-hand limit ($x \rightarrow -1^+$). For $x \rightarrow -1^+$, we are in the case $|x| < 1$. So, $f(x) = \cos(2 \pi x)$. $\lim\limits_{x \rightarrow -1^+} f(x) = \lim\limits_{x \rightarrow -1^+} \cos (2 \pi x) = \cos (-2 \pi) = 1$ For $x \rightarrow -1^-$, we are in the case $|x| > 1$. So, $f(x) = \frac{-\sin (x-1)}{x-1}$. $\lim\limits_{x \rightarrow -1^-} f(x) = \lim\limits_{x \rightarrow -1^-} \frac{-\sin (x-1)}{x-1} = \frac{-\sin (-1-1)}{-1-1} = \frac{-\sin (-2)}{-2} = \frac{\sin (2)}{-2} = -\frac{\sin 2}{2}$ Since $\lim\limits_{x \rightarrow -1^+} f(x) = 1$ and $\lim\limits_{x \rightarrow -1^-} f(x) = -\frac{\sin 2}{2}$, the limit $\lim\limits_{x \rightarrow -1} f(x)$ does not exist. Therefore, $f(x)$ is discontinuous at $x=-1$.

Step 6: Combine the results.

• $f(x)$ is continuous for $|x| < 1$ (i.e., in $(-1, 1)$).
• $f(x)$ is continuous for $|x| > 1$ (i.e., in $(-\infty, -1) \cup (1, \infty)$).
• $f(x)$ is discontinuous at $x=1$.
• $f(x)$ is discontinuous at $x=-1$.

Therefore, $f(x)$ is continuous for all $x \in \mathbb{R} - \{-1, 1\}$.

3. Common Mistakes & Tips

• Incorrectly handling limits of powers: Be very careful with the cases $|x|<1$, $|x|>1$, $x=1$, and $x=-1$. The behavior of $x^{2n}$ and $x^{2n+1}$ is crucial.
• Division by zero: When evaluating limits for $|x|>1$, ensure that the denominator does not become zero after simplification. In this case, for $|x|>1$, $x-1$ is never zero.
• Checking continuity at boundary points: Always check continuity at points where the definition of the function changes or where the limit behavior changes. This involves comparing the left-hand limit, the right-hand limit, and the function value at that point.

4. Summary

To determine the continuity of the given function, we analyzed its behavior by considering different intervals based on the magnitude of $x$. For $|x|<1$, $x^{2n}$ and $x^{2n+1}$ tend to 0, simplifying $f(x)$ to $\cos(2\pi x)$, which is continuous. For $|x|>1$, we divided by $x^{2n}$ to find $f(x) = \frac{-\sin(x-1)}{x-1}$, also continuous in this range. We then specifically checked the points $x=1$ and $x=-1$. At $x=1$, the left-hand limit (from $|x|<1$ case) and the right-hand limit (from $|x|>1$ case) were different (1 and -1 respectively), indicating discontinuity. At $x=-1$, the left-hand limit (from $|x|>1$ case) and the right-hand limit (from $|x|<1$ case) were also different ($-\frac{\sin 2}{2}$ and 1 respectively), confirming discontinuity. Thus, the function is continuous everywhere except at $x=1$ and $x=-1$.

5. Final Answer

The final answer is $\boxed{R-\{-1,1\}}$.