Question

The function \(f : R→R\) defined by \(f(x)=lim_{n→∞}\frac{cos2πx-x^{2n}sinx-1}{1+x^{2n+1}-x^{2n}}\) is continuous for all x in

• A. R-{-1}
• B. R-{-1,1}
• C. R-{1}
• D. R-{0}

Answer 1

The Correct Option is B

\(\begin{array}{l} f\left(x\right)=\displaystyle \lim_{n \to \infty}\frac{\cos\left(2\pi x\right)-x^{2n}\sin\left(x-1\right)}{1+x^{2n+1}-x^{2n}}\end{array}\)

\(\begin{array}{l}\text{For} \left|x\right| < 1, f\left(x\right) = cos2\pi x, \text{continuous function}\end{array}\)

|x| > 1,

\(\begin{array}{l} f\left(x\right)=\displaystyle \lim_{n \to \infty}\frac{x^{\frac{1}{{2n}}\cos2\pi x-\sin\left(x-1\right)}}{x^{\frac{1}{2n}+x-1}}\end{array}\)

\(\begin{array}{l} =\frac{-\sin\left(x-1\right)}{x-1},\text{continuous}\end{array}\)

For |x| = 1,

\(\begin{array}{l} f\left(x\right)=\left\{\begin{matrix}1 & \text{if}&x=1 \\-1\left(1+\sin2\right)&\text{if} &x=-1 \\\end{matrix}\right. \end{array}\)

Now, \(\begin{array}{l} \displaystyle \lim_{x \to 1^+} f\left(x\right)=-1,~~~\displaystyle \lim_{x \to 1^-}f\left(x\right)=1,\end{array}\)so discontinuous at x = 1

\(\begin{array}{l} \displaystyle \lim_{x \to -1^+}f\left(x\right)=1,~~\displaystyle \lim_{x \to -1^-}f\left(x\right)=-\frac{\sin2}{2},\end{array}\)

so discontinuous at x = –1

\(\begin{array}{l} \therefore f\left(x\right)\text{is continuous for all}~ x \in R – \{-1, 1\}\end{array}\)