The function f : R rightarrow R defined by f( | vedclass

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(B) To find the continuity of $f(x)$,we evaluate the limit $f(x) = \lim_{n \rightarrow \infty} \frac{\cos(2 \pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} -

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$R - \{-1, 1\}$

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(B) To find the continuity of $f(x)$,we evaluate the limit $f(x) = \lim_{n ightarrow \infty} \frac{\cos(2 \pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} - x^{2n}}$.Case $1$: $|x| Case $2$: $x = 1$. $f(1) = \lim_{n ightarrow \infty} \frac{\cos(2 \pi) - 1^{2n} \sin(0)}{1 + 1^{2n+1} - 1^{2n}} = \frac{1 - 0}{1 + 1 - 1} = 1$.Case $3$: $x = -1$. $f(-1) = \lim_{n ightarrow \infty} \frac{\cos(-2 \pi) - (-1)^{2n} \sin(-2)}{1 + (-1)^{2n+1} - (-1)^{2n}} = \frac{1 - 1 \cdot \sin(-2)}{1 - 1 - 1} = \frac{1 + \sin 2}{-1} = -(1 + \sin 2)$.Case $4$: $|x| > 1$. Divide numerator and denominator by $x^{2n}$: $f(x) = \lim_{n ightarrow \infty} \frac{\frac{\cos(2 \pi x)}{x^{2n}} - \sin(x-1)}{\frac{1}{x^{2n}} + x - 1} = \frac{0 - \sin(x-1)}{0 + x - 1} = \frac{-\sin(x-1)}{x-1}$.Checking continuity at $x=1$: $\lim_{x ightarrow 1^-} f(x) = \cos(2 \pi) = 1$. $\lim_{x ightarrow 1^+} f(x) = \lim_{x ightarrow 1^+} \frac{-\sin(x-1)}{x-1} = -1$. Since $1 
eq -1$,$f(x)$ is discontinuous at $x=1$.Checking continuity at $x=-1$: $\lim_{x ightarrow -1^-} f(x) = \frac{-\sin(-2)}{-2} = \frac{\sin 2}{-2} = -\frac{\sin 2}{2}$. $\lim_{x ightarrow -1^+} f(x) = \cos(-2 \pi) = 1$. Since $-\frac{\sin 2}{2} 
eq 1$,$f(x)$ is discontinuous at $x=-1$.Thus,$f(x)$ is continuous for all $x \in R - \{-1, 1\}$.

The function $f : R \rightarrow R$ defined by $f(x) = \lim_{n \rightarrow \infty} \frac{\cos(2 \pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} - x^{2n}}$ is continuous for all $x$ in.

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