Solveeit Logo
Login

Question

Mathematics Question on Continuity and differentiability

The function f:R→Rf : R→R defined by f(x)=limn→∞cos2πx−x2nsinx−11+x2n+1−x2nf(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

R-{-1,1}

Explanation

Solution

f(x)=lim⁡n→∞cos⁡(2πx)−x2nsin⁡(x−1)1+x2n+1−x2n\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}

For∣x∣<1,f(x)=cos2πx,continuous function\begin{array}{l}\text{For} \left|x\right| < 1, f\left(x\right) = cos2\pi x, \text{continuous function}\end{array}

|x | > 1,

f(x)=lim⁡n→∞x12ncos⁡2πx−sin⁡(x−1)x12n+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}

=−sin⁡(x−1)x−1,continuous\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, lim⁡x→1+f(x)=−1,   lim⁡x→1−f(x)=1,\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

lim⁡x→−1+f(x)=1,  lim⁡x→−1−f(x)=−sin⁡22,\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

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