Question

If $f\left(x\right)=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{\left(1+x^2\right)}+...+\frac{x^2}{\left(1+x^2\right)}+....,$

then at x = 0, f (x)
(a) has no limit
(b) is discontinuous
(c) is continuous but not differentiable
(d) is differentiable

Answer 1

(b) is discontinuous

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{\left(1+x^2\right)}+...+\frac{x^2}{\left(1+x^2\right)}+...., \\ \mathrm{When}x=0\mathrm{then}{x}^2=0 \\ \mathrm{and}\frac{x^2}{1+x^2}=0 \\ \therefore f\left(0\right)=0+0+0+0....... \\ \Rightarrow f\left(0\right)=0 \\ \mathrm{When},x\ne 0 \\ \mathrm{Then},x^2>0 \\ \mathrm{and}1+x^2>x^2 \\ \Rightarrow 0<\frac{x^2}{1+x^2}<1 \\ \therefore \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(x^2+\frac{x^2}{1+x^2}+\frac{x^2}{\left(1+x^2\right)}+...+\frac{x^2}{\left(1+x^2\right)}+....,\right) \\ =\underset{x\to 0}{\lim }\left[x^2\left(1+\frac{1}{1+x^2}+\frac{1}{\left(1+x^2\right)}+...+\frac{1}{\left(1+x^2\right)}+....,\right)\right] \\ =\underset{x\to 0}{\lim }\left[x^2\left(\frac{1}{1-\frac{1}{1+x^2}}\right)\right]\left[\mathrm{Sum}\mathrm{of}\mathrm{infinite}\mathrm{series}\mathrm{where},r=\frac{1}{1+x^2}\right] \\ =\underset{x\to 0}{\lim }\left[x^2\left(\frac{1+x^2}{x^2}\right)\right] \\ =\underset{x\to 0}{\lim }\left(1+x^2\right) \\ =1 \\ \therefore \underset{x\to 0}{\lim }f\left(x\right)\ne f\left(0\right) \\ \therefore f\left(x\right)\mathrm{is}\mathrm{discontinuous}\mathrm{at}x=0 \end{gathered}$$